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MATH/STAT
LIBRARY
 
DIOPHANTUS OF ALEXANDRIA
CAMBRIDGE UNIVERSITY PRESS
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Al rights reserved
DIOPHANTUS OF ALEXANDRIA:

A STUDY IN THE HISTORY

OF

GREEK ALGEBRA

BY

SIR THOMAS 1. HEATH, K.CB,

SC.D., SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE
SECOND EDITION

WITH A SUPPLEMENT CONTAINING AN ACCOUNT OF FERMAT’S
THEOREMS AND PROBLEMS CONNECTED WITH DIOPHANTINE
ANALYSIS AND SOME SOLUTIONS OF DIOPHANTINE
PROBLEMS BY EULER

Cambridge :

at the University Press
1010
Cambridge:
PRINTED BY JOHN CLAY, M.A,

AT THE UNIVERSITY PRESS
XA 3
Ds4 H4
19:0 a
MATH

In compliance with current copyright law,
U. C. Library Bindery produced this replacement volume on
paper that meets the ANSI Standard Z39.48-1984 to replace

the irreparably deteriorated original.

1993
 
PREFACE

HE first edition of this book, which was the first English
Diophantus, appeared in 1885, and has long been out of

print. Inquiries made for it at different times suggested to me
that it was a pity that a treatise so unique and in many respects
so attractive as the Arithmetica should once more have become
practically inaccessible to the English reader. At the same time
I could not but recognise that, after twenty-five years in which so
much has been done for the history of mathematics, the book
needed to be brought up to date, Some matters which in 1885
were still subject of controversy, such as the date of Diophantus,
may be regarded as settled, and some points which then had to
be laboured can now be dismissed more briefly. Practically the
- whole of the Introduction, except the chapters on the editions of
Diophantus, his methods of solution, and the porisms and other
assumptions found in his work, has been entirely rewritten and
much shortened, while the chapters on the methods and on the
porisms etc., have been made fuller than before. The new text of
Tannery (Teubner 1893, 1895) has enabled a number of obscure
passages, particularly in Books V and VI, to be cleared up and,
as a basis for a reproduction of the whole work, is much superior to
the text of Bachet. I have taken the opportunity to make my
version of the actual treatise somewhat fuller and somewhat closer
to the language of the original. In other respects also I thought
I could improve upon a youthful work which was my first essay in
the history of Greek mathematics. When writing it I was solely
concerned to make Diophantus himself known to mathematicians,

239006
vi PREFACE

and I did not pay sufficient attention to Fermat's notes on the
various problems. It is well known that it is in these notes that
many of the great propositions discovered by Fermat in the
theory of numbers are enshrined; but, although the notes are
literally translated in Wertheim’s edition, they do not seem to
have appeared in English; moreover they need to be supple-
mented by passages from the correspondence of Fermat and from
the Doctrinae analyticae Inventum Novum of Jacques de Billy.
The histories of mathematics furnish only a very inadequate
description of Fermat's work, and it seemed desirable to attempt
to give as full an account of his theorems and problems in
or connected with Diophantine analysis as it is possible to
compile from the scattered material available in Tannery and
Henry's edition of the Oeuvres de Fermat (1891—1896). So much
of this material as could not be conveniently given in the notes
to particular problems of Diophantus I have put together in
the Supplement, which is thus intended to supply a missing
chapter in the history of mathematics. Lastly, in order to make
the book more complete, I thought it right to add some of the
more remarkable solutions of difficult Diophantine problems given
by Euler, for whom such problems had a great fascination ; the last
section of the Supplement is therefore devoted to these solutions.

TT Lo1.

October, 1910.
CHAP. I.
lL

ln
IV

s. VL

Boor I.

es IL
wo HI
ow IV

V.

5 V1

CONTENTS

INTRODUCTION

Diophantus and his Works .

The MSS. of and writers on Diophantus

Notation and definitions of Diophantus
Diophantus’ methods of solution

The Porisms and other assumptions in Diophantus

The place of Diophantus

THE ARITHMETICA

On Polygonal numbers

Conspectus of the Arithmetica

SECTIONS I—V. NOTES, THEOREMS AND PROBLEMS BY FERMAT.

I.
11.
111.

IV.
V.

SUPPLEMENT

On numbers separable into integral squares
Equation 22—Ay?=1
Theorems and Problems on rational right-angled

triangles . : :
Other problems by Fermat
Fermat’s Triple-equations

SECTION VI. SOME SOLUTIONS BY EULER .

INDEX: I.

Si I,

GREEK .

ENGLISH

PAGES
1I—13
14—31
333
54—98
99—1I10
111—127

129 —143
143—156
156—168
168—199
200—22§
226—246
247—259
260—266

267—277
277—292

293—318
318—320
321—328
329—380
381—382
382—387
 
INTRODUCTION

CHAPILER 1
DIOPHANTUS AND HIS WORKS

THE divergences between writers on Diophantus used to begin,
as Cossali said?, with the last syllable of his name. There is now,
however, no longer any doubt that the name was Diophantos, not
Diophantes?.

The question of his date is more difficult. Abu’lfaraj, the
Arabian historian, in his History of the Dynasties, places Diophantus
under the Emperor Julian (A.D. 361-3), but without giving any
authority ; and it may be that the statement is due simply to a
confusion of our Diophantus with a rhetorician of that name,
mentioned in another article of Suidas, who lived in the time of
Julian, On the other hand, Rafael Bombelli in his Algebra,

1 Cossali, Origine, trasporto in Italia, primi progressi in essa dell’ Algebra (Parma,
1797-9), I. p. 61: “Su la desinenza del nome comincia la diversita tra gli scrittori.”

2 Greek authority is overwhelmingly in favour of Diophantos. The following is the
evidence, which is collected in the second volume of Tannery's edition of Diophantus
(henceforward to be quoted as ‘Dioph.,” “Dioph. 11. p. 36” indicating page 36 of
Vol. 11., while “ Dioph. 11. 20” will mean proposition 20 of Book 11.): Suidas s.v.
“Prarie (Dioph. 11. p. 36), Theon of Alexandria, on Ptolemy’s Synfaxis Book I. c.g
(Dioph. 11. p. 35), Anthology, Epigram on Diophantus (Ep. X1v. 126; Dioph. II. p. 60),
Anonymi prolegomena in Introductionem arithmeticam Nicomachi (Dioph. II. p. 73),
Georgii Pachymerae paraphrasis (Dioph. II. p. 122), Scholia of Maximus Planudes
(Dioph. 11. pp. 148, 177, 178 etc.), Scholium on Iamblichus 7» Nicomackhi arithm. introd.,
ed. Pistelli, p. 127 (Dioph. 11. p. 72), a Scholium on Dioph. 11. 8 from the MS. “4”
(Dioph. 11. p. 260), which is otherwise amusing (H vx? cov, Abpavre, etn perd Tod
Zatava ¢veka Tis duakolias Tv Te dNAwy gov OewpnudTwy kal 09) kal ToD wapbvros Bewprh-
matos, ‘“ Your soul to perdition, Diophantus, for the difficulty of your problems in general
and of this one in particular”) ; John of Jerusalem (1othc.) alone (Vita Joannis Damas-
ceni XI. : Dioph. 11. p. 36), if the reading of the MS. Parisinus 13359 is right, wrote, in
the plural, os ITvfaybpar 7 Awbgpavrar, where however Awbpavrac is clearly a mistake for
Abpavror.

3 ABdyios, copiarhs 'Avrioxevs, T@v éml 'Toukwavod Tob BagiNéws xpbrwy, kal pméxp
Oeodoglov Tol mpeaPBurépov: Pagyaviov warpbs, pabnrhs Awpdrrov.

H. D. x
2 INTRODUCTION

published in 1572, says dogmatically that Diophantus lived under
Antoninus Pius (138-161 A.D.), but there is no confirmation of this
date either.

The positive evidence on the subject can be given very shortly.
An upper limit is indicated by the fact that Diophantus, in his
book on Polygonal Numbers, quotes from Hypsicles a definition
of such a number! Hypsicles was also the writer of the sup-
plement to Euclid’s Book XIII on the Regular Solids known as
Book XIV. of the Elements; hence Diophantus must have written
later than, say, 150 B.C. A lower limit is furnished by the fact that
Diophantus is quoted by Theon of Alexandria?; hence Diophantus
wrote before, say, 350 A.D. There is a wide interval between
150 B.C. and 350 A.D., but fortunately the limits can be brought
closer. We have a letter of Psellus (11th c.) in which Diophantus
and Anatolius are mentioned as writers on the Egyptian method
of reckoning. *“Diophantus,” says Psellus? “dealt with it more
accurately, but the very learned Anatolius collected the most
essential parts of the doctrine as stated by Diophantus in a
different way (reading érépws) and in the most succinct form,
dedicating (mpooepwrvnoe) his work to Diophantus.” It would
appear, therefore, that Diophantus and Anatolius were contem-
poraries, and it is most likely that the former would be to the
latter in the relation of master to pupil. Now Anatolius wrote
about 278-9 A.D., and was Bishop of Laodicea about 280 A.D. We
may therefore safely say that Diophantus flourished about 250 A.D.
or not much later. This agrees well with the fact that he is not
quoted by Nicomachus (about 100 A.D.), Theon of Smyrna (about
130 A.D.) or lamblichus (end of 3rd c.).

1 Dioph. I. p. 470-2.

2 Theo Alexandrinus in primum librum Ptolemaei Mathematicae Compositionis (on c.

IX.) : see Dioph. II. p. 35, kad’ d kal Awbpavtés gmat’ THs yap povddos dueradérov olons
kal égTons wdvToTe, TO mOANamAagialbuevoy eldos ém’ avr avTd TO €ldos érTaL K.T.E.

3 Dioph. II. p. 38-9: mepl 8¢ 77s alyvmriakfs pefbdov TavTns Abpavros uév diéhafev
dkpiBéoTeporv, 6 8¢ Noytwraros 'Avarbhios Ta cuvekTikdTaTa wépn Ths kar’ ékelvov émioThuns
dmrohekduevos éTépw (? éTépws or éTalpy) AogdrTy guvorTikdTara mposepdvnoe. The MSS.
read érépw, which is apparently a mistake for érépws or possibly for éralpwp. Tannery con-
jectures 7¢ éralpy, but this is very doubtful ; if the article had been there, Awopdrre 73
éralpey would have been better. On the basis of éralppy Tannery builds the further
hypothesis that the Dionysius to whom the A7ithmetica is dedicated is none other than
Dionysius who was at the head of the Catechist school at Alexandria 232-247 and was
Bishop there 248-265 A.D. Tannery conjectures then that Diophantus was a Christian
and a pupil of Dionysius (Tannery, ‘‘ Sur la religion des derniers mathématiciens de
Pantiquité,” Extrait des Annales de Philosophie Chrétienne, 1896, p. 13 sqq.). It is
however difficult to establish this (Hultsch, art. ¢‘ Diophantos aus Alexandreia” in Pauly-
Wissowa’s Real- Encyclopidie der classischen Altertumswissenchaften).
DIOPHANTUS AND HIS WORKS 3

The only personal particulars about Diophantus which are
known are those contained in the epigram-problem relating to him
in the Anthology’ The solution gives 84 as the age at which he
died. His boyhood lasted 14 years, his beard grew at 21, he
married at 33; a son was born to him five years later and died, at
the age of 42, when his father was 80 years old. Diophantus’ own
death followed four years later? It is clear that the epigram was
written, not long after his death, by an intimate personal friend
with knowledge of and taste for the science which Diophantus
made his life-work?.

The works on which the fame of Diophantus rests are:

(1) The Arithmetica (originally in thirteen Books).
(2) A tract On Polygonal Numbers.
+ Six Books of the former and part of the latter survive,

Allusions in the Arithmetica imply the existence of

(3) A collection of propositions under the title of Porisms;
in three propositions (3, 5 and 16) of Book v. Diophantus quotes
as known certain propositions in the Theory of Numbers, prefixing
to the statement of them the words “We have it in the Porisms
that... 7 (éxouev év Tois Ioplopaoiy Ti k.T.E.).

A scholium on a passage of Iamblichus where he quotes a
dictum of certain Pythagoreans about the unit being the dividing
line (uefopiov) between number and aliquot parts, says “thus
Diophantus in the Moriastica®‘...... for he describes as ‘parts’ the
progression without limit in the direction of less than the unit.”
Tannery thinks the Mopiagricd may be ancient scholia (now
lost) on Diophantus I. Def. 3 sqq.’; but in that case why should
Diophantus be supposed to be speaking? And, as Hultsch

1 Anthology, Ep. x1v. 126; Dioph. II. pp. 6o-1.

2 The epigram actually says that his boyhood lasted } of his life; his beard grew
after {5 more; after 3 more he married, and his son was born five years later; the son
lived to half his father’s age, and the father died four years after his son. Cantor (Gesch.
d. Math. 13, p. 465) quotes a suggestion of Heinrich Weber that a better solution is
obtained if we assume that the son died at the time when his father’s age was double his,
not at an age equal to half the age at which his father died. In that case

(xt fxt+ix+5+4(x—4)+4=x, or 3x=196 and x=0654.
This would substitute 103 for 14, 16} for 21, 25% for 33, 30% for 42, 614 for 8o,
and 653% for 84 above. I do not see any advantage in this solution. On the contrary,
I think the fractional results are an objection to it, and it is to be observed that the
scholiast has the solution 84, derived from the equation
tr tdhr prt shat g=a

$ Hultsch, art. Diophantos in Pauly-Wissowa’s Real- Encyclopidie.

* Iamblichus Zn Nicomachi arithm. introd. p. 127 (ed. Pistelli) ; Dioph. 11. p. 72.

® Dioph. 11. p. 72 note.

Y wee 2
4 INTRODUCTION

remarks, such scholia would more naturally have been quoted
as oxéMa and not by the separate title Mopiagmika’. It may
have been a separate work by Diophantus giving rules for reckon-
ing with fractions; but I do not feel clear that the reference
may not simply be to the definitions at the beginning of the
Avrithmetica.

With reference to the title of the Arithmetica, we may observe
that the meaning of the word apcfuntica here is slightly different
from that assigned to it by more ancient writers. The ancients
drew a marked distinction between aptBuyTicy and NoyiaTikd,
though both were concerned with numbers. Thus Plato states
that aptfunTikyy is concerned with the abstract properties of
numbers (as odd and even, etc.), whereas AoyioTikr deals with the
same odd and even, but in relation to one another? Geminus also
distinguishes the two terms? According to him apfunTwksy deals
with numbers 7 themselves, distinguishing linear, plane and solid
numbers, in fact all the forms of number, starting from the unit,
and dealing with the generation of plane numbers, similar and
dissimilar, and then with numbers of three dimensions, etc.
NoyioTikn on the other hand deals, not with the abstract properties
of numbers in themselves, but with numbers of concrete things
(alaOntdv, sensible objects), whence it calls them by the names of
the things measured, eg. it calls some by the names up\itys and
¢drarityst. But in Diophantus the calculations take an abstract
form (except in V. 30, where the question is to find the number
of measures of wine at two given prices respectively), so that the
distinction between \oyioTikn and dpiBumTikn is lost.

We find the Arithmetica quoted under slightly different titles.
Thus the anonymous author of prolegomena to Nicomachus’
Introductio Arithmetica speaks of Diophantus’ “ thirteen Books of
Arithmetic®” A scholium on Iamblichus refers to “the last
theorem of the first Book of Diophantus’ Elements of Arithmetic

1 Hultsch, loc. cit.

2 Gorgias, 451 B, C: Ta pév &\\a kabdmwep 1 dpilbunTicy 1) NoyioTiky Exe mepl TO avrd
ydp éaTi, TO Te dpTiov kal To meprTéy: Otagéper ¢ TocobTor, STL kal wpds alTd Kal mwpds
EMA mds Exer mAHBous émiaromel TO mepTTOY Kal TO EpTiov H Nota Tk.

8 Proclus, Comment. on Euclid I., p. 39, 14-40, 7.

4 Cf. Plato, Laws 819 B, C, on the advantage of combining amusement with instruction
in arithmetical calculation, e.g. by distributing apples or garlands (widwr 7¢ Twwr
duavopal kal oTepdvwr) and the use of different bowls of silver, gold, or brass etc. (pudhas
apa xploov Kal xalkol kai dpylpov kal TowobTwy Twdr ENNwy Kepavvivres, ol ¢ Shas wws
duadidvTes, 6mep elmwov, els madiaw évapubrTovtes Tas TOY dvaykalwy dplbudv xpioets).

5 Dioph. II. p. 73, 26.
DIOPHANTUS AND HIS WORKS 5

(apibunTikiis oTovyetwaews).” A scholium on one of the epigrams
in Metrodorus’ collection similarly speaks of the “ Elements of
Diophantus®.”

None of the MSS. which we possess contain more than the
first six Books of the Arithimetica, the only variation being that
some few divide the six Books into seven? while one or two give
the fragment on Polygonal Numbers with the number vii. The
idea that Regiomontanus saw, or said he saw, a MS. containing
the thirteen Books complete is due to a misapprehension. There
is no doubt that the missing Books were lost at a very early date.
Tannery* suggests that Hypatia’s commentary extended only to
the first six Books, and that she left untouched the remaining
seven, which accordingly were first forgotten and then lost; he
compares the case of Apollonius’ Conics, the first four Books of
which were preserved by Eutocius, who wrote a commentary on
them, while the rest, which he did not include in his commentary,
were lost so far as the Greek text is concerned. While, however,
three of the last four Books of the Conics have fortunately reached
us through the Arabic, there is no sign that even the Arabians
ever possessed the missing Books of Diophantus. Thus the
second part of an algebraic treatise called the Fakhri by Abt
Bekr Muh. b. al-Hasan al-Karkhi (d. about 1029) is a collection of
problems in determinate and indeterminate analysis which not
only show that their author had deeply studied Diophantus, but in
many cases are taken direct from the A7:t/imetica, with the change,
occasionally, of some of the constants. In the fourth section of
this work, which begins and ends with problems corresponding to
problems in Diophantus Books IIL and IIL respectively, are 25
problems not found in Diophantus; but the differences from
Diophantus in essential features (eg. several of the problems lead
to equations giving irrational results, which are always avoided
by Diophantus), as well as other internal evidence, exclude the
hypothesis that we have here a lost Book of Diophantus®. Nor is
there any sign that more of the work than we possess was known

1 Dioph. 1I. p. 72, 17 ; Iamblichus (ed. Pistelli), p. 132, 12.

2 Dioph. 11. p. 62, 23.

3 e.g. Vaticanus gr. 200, Scorialensis 2-I1-15, and the Broscius MS. in the University
Library of Cracow ; the two last divide the first Book into two, the second beginning
immediately after the explanation of the sign for minus (Dioph. 1. p. 14, 1).

4 Dioph. 11. p. xvii, xviii.

5 See F. Woepcke, Extrait du Fakhri, traité d’Algébre par Abou Bekr Mohammed

ben Alhacan Alkarkhi (manuscrit 932, supplément arabe de la bibliothegue Impériale), Paris,
1853. .
6 INTRODUCTION

to Abw’l Wafa al-Biizjani (940-998 A.D.), who wrote a “commentary
(tafsir) on the algebra of Diophantus” as well as a “Book of
proofs of the propositions used by Diophantus in his work...”
These facts again point to the conclusion that the lost Books were
lost before the 10th c.

Tannery’s suggestion that Hypatia’s commentary was limited
to the six Books, and the parallel of Eutocius’ commentary on
Apollonius’ Conics, imply that it is the Jasz seven Books, and the
most difficult, which are lost. This view is in strong contrast to
that which had previously found most acceptance among com-
petent authorities. The latter view was most clearly put, and
most ably supported, by Nesselmann?, though Colebrooke? had
already put forward a conjecture to the same effect ; and historians
of mathematics such as Hankel, Moritz Cantor, and Giinther have
accepted Nesselmann’s conclusions, which, stated in his own
words, are as follows: (1) that much less of Diophantus is wanting
than would naturally be supposed on the basis of the numerical
proportion of 6 to 13; (2) that the missing portion is not to be
looked for at the end but in the middle of the work, and indeed
mostly between the first and second Books. Nesselmann’s general
argument is that, if we carefully read the last four Books, from the
third to the sixth, we find that Diophantus moves in a rigidly
defined and limited circle of methods and artifices, and that any
attempts which he makes to free himself are futile; “as often as
he gives the impression that he wishes to spring over the magic
circle drawn round him, he is invariably thrown back by an
invisible hand on the old domain already known ; we see, similarly,
in half-darkness, behind the clever artifices which he seeks to use
in order to free himself, the chains which fetter his genius, we hear
their rattling, whenever, in dealing with difficulties only too freely
imposed upon himself, he knows of no other means of extricating
himself except to cut through the knot instead of untying it.”
Moreover, the sixth Book forms a natural conclusion to the whole,
in that it consists of exemplifications of methods explained and
used in the preceding Books. The subject is the finding of right-
angled triangles in rational numbers such that the sides and area
satisfy given conditions, the geometrical property of the right-angled
triangle being introduced as a fresh condition additional to the
purely arithmetical conditions which have to be satisfied in the

Y Algebra der Griechen, pp. 264-273.
2 Algebra of the Hindus, Note M, p. Ixi.
DIOPHANTUS AND HIS WORKS 7

problems of the earlier Books. But, assuming that Diophantus’
resources are at an end in the sixth Book, Nesselmann has to
suggest possible topics which would have formed approximately
adequate material for the equivalent of seven Books of the
Arithmetica. The first step is to consider what is actually wanting
which we should expect to find, either as foreshadowed by the
author himself or as necessary for the elucidation or completion of
the whole subject. Now the first Book contains problems leading
to determinate equations of the first degree; the remainder of the —
work is a collection of problems which, with few exceptions, lead
to indeterminate equations of the second degree, beginning with
simpler cases and advancing step by step to more complicated
questions. There would have been room therefore for problems
involving (1) determinate equations of the second degree and (2)
indeterminate equations of the firsz. There is indeed nothing to
show that (2) formed part of the writer's plan; but on the other
hand the writer's own words in Def. 11 at the beginning of the
work promise a discussion of the solution of the complete or
adfected quadratic, and it is clear that he employed his method of
solution in the later Books, where in some cases he simply states
the solution without working it out, while in others, where the
roots are “irrational,” he gives approximations which indicate
that he was in possession of a scientific method. Pure quadratics”
Diophantus regarded as simple equations, taking no account of the
negative root. Indeed it would seem that he adopted as his
ground for the classification of quadratics, not the index of the
highest power of the unknown quantity contained in it, but the
number of terms left in it when reduced to its simplest form. His
words are!: “If the same powers of the unknown occur on both
sides, but with different coefficients (un opomAn6y 8¢), we must
take like from like until we have one single expression equal to
another. If there are on both sides, or on either side, any terms
with negative coefficients (év éA\eivreai Twa €ldn), the defects must
be added on both sides until the terms on both sides have:
none but positive coefficients (évvrapyovra), when we must again
take like from like until there remains one term on each side.
This should be the object aimed at in framing the hypotheses of
propositions, that is to say, to reduce the equations, if possible,
until one term is left equated to one term. But afterwards I will

1 Dioph. 1. Def. 11, p. 14.
8 INTRODUCTION

show you also how, when two terms are left equal to one term,
such an equation is solved.” That is to say, reduce the quadratic,
i possible, to one of the forms ax®= bx, ax*=c, or br=c: 1 will
show later how to solve the equation when three terms are left of
which any two are equal to the third, Ze. the complete quadratic
ar’+ bx + c= 0, excluding the case ax? + bx +c=o0. The exclusion
of the latter case is natural, since it is of the essence of the work
to find rational and positive solutions. Nesselmann might have
added that Diophantus’ requirement that the equation, as finally
stated, shall contain only positive terms, of which two are equated
to the third, suggests that his solution would deal separately with
the three possible cases (just as Euclid makes separate cases of the
equations in his propositions VI. 28, 29), so that the exposition
might occupy some little space. The suitable place for it would
be between the first and second Books. There is no evidence
tending to confirm Nesselmann’s further argument that the six
Books may originally have been divided into even more than
seven Books. He argues from the fact that there are often better
natural divisions in the middle of the Books (eg. at II. 19) than
between them as they now stand; thus there is no sign of a
marked division between Books I. and II. and between Books II.
and 111, the first five problems of Book II. and the first four of
Book III. recalling similar problems in the preceding Books
respectively. But the latter circumstances are better explained,
as Tannery explains them, by the supposition that the first
problems of Books II. and III are interpolated from some ancient
commentary. Next Nesselmann points out that there are a
number of imperfections in the text, Book V. especially having
been “treated by Mother Time in a very stepmotherly fashion”;
thus it seems probable that at Vv. 19 three problems have dropped
out altogether. Still he is far from accounting for seven whole
Books; he has therefore to press into the service the lost
“Porisms” and the tract on Polygonal Numbers.

If the phrase which, as we have said, occurs three times in
Book v., “We have it in the Porisms that...” indicates that the
“Porisms” were a definite collection of propositions concerning
the properties of certain numbers, their divisibility into a certain
number of squares, and so on, it is possible that it was from the
same collection that Diophantus took the numerous other pro-
positions which he assumes, either explicitly enunciating them, or
implicitly taking them for granted. May we not then, says
DIOPHANTUS AND HIS WORKS 9

b3

Nesselmann, reasonably suppose the “Porisms” to have formed
an introduction to the indeterminate and semi-determinate analy-
sis of the second degree which forms the main subject of the
Avwrithmetica, and te have been an integral part of the thirteen
Books, intervening, probably, between Books I. and II.? Schulz, on
the other hand, considered this improbable, and in recent years
Hultsch! has definitely rejected the theory that Diophantus filled
one or more Books of his Arithmetica exclusively with Porisms.
Schulz’s argument is, indeed, not conclusive. It is based on the
consideration that “ Diophantus expressly says that his work deals
with arithmetical problems®”; but what Diophantus actually says is
“ Knowing you, O Dionysius, to be anxious to learn the solution
(or, perhaps, ‘discovery,’ elpecw) of problems in numbers, I have
endeavoured, beginning from the foundations on which the study is
built up, to expound (UmooTijcar =to lay down) the nature and
force subsisting in numbers,” the last of which words would easily
cover propositions in the theory of numbers, while propositions,”
not “ problems,” is the word used at the end of the Preface, where
he says, “let us now proceed to the propositions (mporacers) ......
which have been treated in thirteen Books.”

On reconsideration of the whole matter, I now agree in the
view of Hultsch that the Porisms were not a separate portion of
the Arithmetica or included in the Aritimetica at all. If they had
been, I think the expression “ we have it in the Porisms” would
have been inappropriate. In the first place, the Greek mathe-
maticians do not usually give references in such a form as this
to propositions which they cite when they come from the same
work as that in which they are cited ; as a rule the propositions
are quoted without any references at all. The references in this
case would, on the assumption that the Porisms were a portion of
the thirteen Books, more naturally have been to particular pro-
positions of particular Books (cf. Eucl. XII 2, “ For it was proved

1 Hultsch, Joc. cit.

2 The whole passage of Schulz is as follows (pref. xxi): * Es ist daher nicht unwahr-
scheinlich, dass diese Porismen eine eigene Schrift unseres Diophantus waren, welche
vorziiglich die Zusammensetzung der Zahlen aus gewissen Bestandtheilen zu ihrem
Gegenstande hatten. Kénnte man diese Schrift als einen Bestandtheil des grossen in
dreizehn Biichern abgefassten arithmetischen Werkes ansehen, so wire es sehr erklirbar,
dass gerade dieser Theil, der den blossen Liebhaber weniger anzog, verloren ging. Da
indess Diophantus ausdriicklich sagt, sein Werk behandele arzthmetische Probleme, so hat
wenigstens die letztere Annahme nur einen geringen Grad von Wahrscheinlichkeit.”
10 INTRODUCTION

in the first theorem of the roth Book that...”). But a still vaguer
reference would have been enough, even if Diophantus had chosen
to give any at all ; if the propositions quoted had preceded those
in which they are used, some expression like TodTo gdp rpo-
yéypamrar, “for this has already been proved,” or 8édewkrar yap
rodro, “for this has been shown,” would have sufficed, or, if the
propositions occurred later, some expression like ws é€ijs SevyOnoerar
or SevxOraerar vp nudv HoTepoy, “as will be proved in due course”
or “later.” The expression “we have it in the Porisms” (in the
plural) would have been still more inappropriate if the “Porisms”
had been, as Tannery supposes! not collected together as one or
more Books of the Arithmetica, but scattered about in the work as
corollaries to particular propositions. And, as Hultsch says, it is
hard, on Tannery’s supposition, to explain why the three particular
theorems quoted from “the Porisms” were lost, while a fair
number of other additions survived, partly under the title mopiopa
(cf. L. 34, 1. 38), partly as “lemmas to what follows,” AMjuua els 10
éChs (cf. lemmas before 1V. 34, 35,36, V. 7, 8, VI. 12, 15). On the
other hand, there is nothing improbable in the supposition that
Diophantus was induced by the difficulty of his problems to give
place in a separate work to the “porisms” necessary to their
solution.

The hypothesis that the Porisms formed part of the Arithmet-
ica being thus given up, we can hardly hold any longer to
Nesselmann’s view of the contents of the lost Books and their
place in the treatise; and I am now much more inclined to the
opinion of Tannery that it is the last and the most difficult Books
which are lost. Tannery’s argument seems to me to be very
attractive and to deserve quotation in full, as finally put in the
preface to Vol. II. of his Diophantus®. He replies first to the
assumption that Diophantus could not have proceeded to problems
more difficult than those of Book v. “ But if the fifth or the sixth
Book of the Arithmetica had been lost, who, pray, among us would
have believed that such problems had ever been attempted by the
Greeks? It would be the greatest error, in any case in which a

1 Dioph. II. p. xix.

# Thus Tannery holds (oc. ¢z2.) that the solution of the complete quadratic was given
in the form of corollaries to I. 27, 30; and he refers the three ‘‘porisms” quoted in v. 3,

5, 16 respectively to a second (lost) solution of 111. 10, to III. 15, and to IV. 1, 2.
8 Dioph. II. p. xx.
DIOPHANTUS AND HIS WORKS %

thing cannot clearly be proved to have been unknown to all the
ancients, to maintain that it could not have been known to some
Greek mathematician. If we do not know to what lengths
Archimedes brought the theory of numbers (to say nothing of
other things), let us admit our ignorance. But, between the
famous problem of the cattle and the most difficult of Diophantus’
problems, is there not a sufficient gap to require seven Books to
fill it? And, without attributing to the ancients what modern
mathematicians have discovered, may not a number of the things
attributed to the Indians and Arabs have been drawn from
Greek sources? May not the same be said of a problem solved by
Leonardo of Pisa, which is very similar to those of Diophantus but
is not now to be found in the A#ithmetica? In fact, it may fairly
be said that, when Chasles made his reasonably probable restitution
of the Porisms of Euclid, he, notwithstanding the fact that he had
Pappus’ lemmas to help him, undertook a more difficult task than
he would have undertaken if he had attempted to fill up seven
Diophantine Books with numerical problems which the Greeks
may reasonably be supposed to have solved.”

On the assumption that the lost portion came at the end of the
existing six Books, Schulz supposed that it contained new methods
of solution in addition to those used in Books I. to VI, and in
particular extended the method of solution by means of the double
equation (SumAf) lootns or SumhoigdéTns). By means of the double
equation Diophantus shows how to find a value of the unknown
which will make two expressions (linear or quadratic) containing it
simultaneously squares. Schulz then thinks that he went on, in
the lost Books, to make #kree such expressions simultaneously
squares, z.e, advanced to a #riple equation. But this explanation
does not in any case take us very far. !

Bombelli thought that Diophantus went on to solve deter-
minate equations of the third and fourth degree!; this view,
however, though natural at that date, when the solution of cubic
and biquadratic equations filled so large a space in contemporary
investigations and in Bombelli’s own. studies, has nothing to
support it.

Hultsch? seems to find the key to the question in the fragment
of the treatise on Polygonal Numbers and the developments to

1 Cossali, I. pp. 78, 70- 2 Hultsch, Joc. cit.
12 : INTRODUCTION

‘which it might have been expected to lead. In this he differs
from Tannery, who says that, as Serenus’ treatise on the sections
of cones and cylinders was added to the mutilated Conics of
Apollonius consisting of four Books only, in order to make up a
convenient volume, so the tract on Polygonal Numbers was added
to the remains of the A#rithmetica, though forming no part of the
larger work’. Thus Tannery would seem to deny the genuineness
of the whole tract on Polygonal Numbers, though in his text he
only signalises the portion beginning with the enunciation of the
problem “ Given a number, to find in how many ways it can be
a polygonal number” as a “vain attempt by a commentator” to
solve this problem. Hultsch, on the other hand, thinks we may
conclude that Diophantus really solved the problem. He points
out moreover that the beginning of the tract is like the beginning
of Book I. of the Arithmetica in containing definitions and pre-
liminary propositions. Then came the difficult problem quoted,
the discussion of which breaks off in our text after a few pages;
and to this it would be easy to tack on a great variety of other
problems. Again, says Hultsch, the supplementary propositions
added by Bachet may serve to give an approximate idea of the
difficulty of the problems which were probably treated in Books VII.
and the following. And between these and the bold combination
of a triangular and a square number in the Cattle-Problem
stretches, as Tannery says, a wide domain which was certainly
not unknown to Diophantus, but was his hunting-ground for the
most various problems. Whether Diophantus dealt with plane
numbers, and with other figured numbers, such as prisms and
tetrahedra, is uncertain.

The name of Diophantus was used, as were the names of Euclid,
Archimedes and Heron in their turn, for the purpose of palming
off the compilations of much later authors. Tannery prints in
his edition three fragments under the head of “Diophantus
Pseudepigraphus.” The first? which is not “from the Arithmetic
of Diophantus ” as its heading states, is worth notice as containing
some particulars of one of “two methods of finding the square
root of any square number ”’; we are told to begin by writing the
number “according to the arrangement of the Indian method,” ze.
according to the Indian numerical notation which reached us
through the Arabs. The fragment is taken from a Paris MS.

1 Dioph. 11. p. xviii. 2 Dioph. 11. p. 3, 3=14-
DIOPHANTUS AND HIS WORKS 13

(Supplem. gr. 387), where it follows a work with the title "Apxn
ths peydlys kal Tvdukfis Yrndupoplas (Ze. yrndpopopias), written in
1252 and raided about half a century later by Maximus Planudes.
The second fragment! is the work edited by C. Henry in 1879 as
Opusculum de multiplicatione et divisione sexagesimalibus Diophanto
vel Pappo attvibuendum. The third? beginning with Acodpavrov
émumeSopetpia, is a compilation made in the Byzantine period out
of late reproductions of the yewuerpovueva and orepeoperpoipeva
of Heron. The second and third fragments, like the first, have
nothing to do with Diophantus.

I Dioph. 11. p. 3, 15-135, 17- 2 Dioph. 11. p. 15, 13-31, 22,
CHAPTER 11

THE MSS. OF AND WRITERS ON DIOPHANTUS

For full details of the various MSS. and of their mutual
relations, reference should be made to the prefaces to the first and
second volumes of Tannery’s edition’. Tannery’s account needs
only to be supplemented by a description given by Gollob? of
another MS. supposed by Tannery to be non-existent, but actually
rediscovered in the Library of the University of Cracow (Nr 544).
Only the shortest possible summary of the essential facts will be
given here.

After the loss of Egypt the work of Diophantus long remained
almost unknown among the Byzantines; perhaps one copy only
survived (of the Hypatian recension), which was seen by Michael
Psellus and possibly by the scholiast to Iamblichus, but of which
no trace can be found after the capture of Constantinople in 1204.
From this one copy (denoted by the letter @ in Tannery’s table of
the MSS.) another MS. (a) was copied in the 8th or gth century ;
this again is lost, but is the true archetype of our MSS. The
copyist apparently intended to omit all scholia, but, the distinction
between text and scholia being sometimes difficult to draw, he
included a good deal which should have been left out. For
example, Hypatia, and perhaps scholiasts after her, seem to have
added some alternative solutions and a number of new problems ;
some of these latter, such as IL. 1-7, 17, 18, were admitted into the
text as genuine.

The MSS. fall into two main classes, the ante-Planudes class,
as we may call it, and the Planudean. The most ancient and the
best of all is Matritensis 48 (Tannery's A), which was written in
the 13th century and belongs to the first class; it is evidently a
most faithful copy of the lost archetype (a). Maximus Planudes
wrote a systematic commentary on Books I. and II, and his scholia,

1 Dioph. I. pp. ili-v, II. pp. xxii—xxxiv.
2 Eduard Gollob, “Ein wiedergefundener Diophantuscodex ” in any Math.
w. Physik, XLV. (1899), hist.-litt. Abtheilung, pp. 137-140.
 

THE MSS. OF AND WRITERS ON DIOPHANTUS 15

which are edited by Tannery for the first time, are preserved in the
oldest representative which we possess of the Planudean class,
namely, Marcianus 308 (Tannery’s B5,), itself apparently copied
from an archetype of the 14th century now lost, with the exception
of ten leaves which survive in Ambrosianus Et 157 sup.
Tannery shows the relation of the MSS. in the following

diagram :

(2) Lost copy of the Hypatian recension.

(a) Lost copy, of eighth or ninth c.

(FIRST CLASS) (PLANUDEAN CLASS)

 

7 |
Matritensis 48 = 4, © 9. Lost MS. of the 14th c. of which ten leaves
13th c. are extant in Ambrosianus Et t57 sup.
o

—
.

 

Marcianus 308 = 5,

Vaticanus gr. 191=V, 10.
beginning of rth c.

second half of 15th c.

©

 

 

 

 

 

 

 

 

 

 

3.. Vaticanus or. 304,
beginning of 16th c. 11. Guelferbytanus 14. Ambrosianus

| Gudianus 1, 15th c. A gr1 sup.
(1545)

4. Parisinus 2379=C 12. Palatinus gr. 3971, 15. Vaticanus gr. 200
(after first two end of 16th c. i {1zaz)

Books), |
middle of 16th c. a

5. Parisinus 2378=2, 13. Reginensis 128, | 16. Scorialensis T-I-11
middle of 16th c. end of 16th c. | (1545)

6.. Neapolitanus 4. Parisinus 2370=C 17. Parisinus 2485=X4,
LC 17, (first two Books) middle of 16th c.
middle of 16th c.

| 18. Scorialensis

7. Urbinas gr. 74, 20. Taurinensis C ITI 16 R-III-18,

end of 16th c. middle of 16th c.
21. Parisinus Ars. 8406

8. Oxon. Baroccianus =X 19. Ambrosianus
166 (part of Book I. Q 121 sup. (part of
only) 22. Scorialensis Q-I-15, Book I.),

middle of 6th c. middle of 16th c.
23. Scorialensis R—-I1-3,
end of 16th c.
?

 

24. Oxon. Savilianus,
end of 16th c.

Auria’s recension made up out of MSS. 2, 3, 15 above and Xylander’s
translation : 25. Parisinus 2380=.0.
26. Ambrosianus E 5 sup.

27. MS. (Patavinus) of Broscius (Brozek) now at Cracow.
28. Lost MS. of Cardinal du Perron.
16 INTRODUCTION

The addition of a few notes as regards the most important and
interesting of the MSS., in the order of their numbers in Tannery's
arrangement, will now sufficiently complete the story.

1. The best and most ancient MS., that of Madrid (Tannery’s
A), was unfortunately spoiled at a late date by corrections made,
especially in the first two Books, from some MS. of the Planudean
class, in such a way that the original reading is sometimes entirely
erased or made quite illegible. In these cases recourse must be
had to the Vatican MS. 101.

2. The MS. Vaticanus graecus 191 was copied from 4 before
it had suffered the general alteration by means of a MS. of the
other class, though not before various other corrections had been
made in different hands not easily distinguished ; thus J some-
times has readings which Tannery found to have arisen from some
correction in 4. A appears to have been at Rome for a con-
siderable period at the time when I” was copied; for the librarian
who wrote the old table of contents! at the beginning of J inserted
in the margin in one place? the word dp&duevos, which had been
omitted, direct from the original (4).

3. Vat. gr. 304 was copied from V, not from A; Tannery
inferred this mainly from a collation of the scholia, and he notes
that the word apfauevos above mentioned is here brought into the
text by the erasure of some letters. This MS. 304, being very
clearly written, was used thenceforward to make copies from. The
next five MSS. do not appear to have had any older source.

4. The MS. Parisinus 2379 (Tannery’s C) was that used by
Bachet for his edition. It was written by one Ioannes Hydruntinus
after 1545, and has the peculiarity that the first two Books were
copied from the MS. Vat. gr. 200 (a MS. of the Planudean class),
evidently in order to include the commentary of Planudes, while
the MS. Vat. gr. 304 belonging to the pre-Planudes class was
followed in the remaining Books, no doubt because it was con-
sidered superior. Thus the class of which C is the chief repre-
sentative is a sort of mixed class.

5, 6. Parisinus 2378 = 2, and Neavaliiii mC 17, were
copied by Angelus Vergetius. In the latter Vergetius puts the

! The MS. 7 was made up of various MSS. before separated. The old table of
contents has Awgpdrrov dpibunricy: dpuovicd didgopa. The apuovika include the Zntro-
duction to Harmony by Cleonides, but without any author’s name. This fact sufficiently
explains the error of Ramus in saying, Schola mathematica, Bk 1. p. 35, ““Scripserat et
Diophantus harmonica.”

2 Dioph. 1. p. 2, 5-6.
THE MSS. OF AND WRITERS ON DIOPHANTUS 17

numbers A, B, T, A, E, Z, H at the top of the pages (as we put
headlines) corresponding to the different Books, implying that he
regarded the tract on Polygonal Numbers as Book VII.

The other MSS. of the first class call for no notice, and we pass
to the Planudean class.

9. Tannery, as he tells us, congratulated himself upon finding
in Ambrosianus Et 157 sup. ten pages of the archetype of the
class, and eagerly sought for new readings. So far, however, as he
was able to carry his collation, he found no difference from the
principal representative of the class (5,) next to be mentioned.

10. The MS. Marcianus 308 (= 5,) of the 15th century formerly
belonged to Cardinal Bessarion, and was seen by Regiomontanus
at Venice in 1464. It contains the recension by Planudes with his
commentary.

11. [It seems certain that the Wolfenbiittel MS. Guelferbytanus
Gudianus 1 (15th c.) was that which Xylander used for his
translation ; Tannery shows that, if this was not the MS. lent
to Xylander by Andreas Dudicius Sbardellatus, that MS. must
have been lost, and there is no evidence in support of the latter
hypothesis. It is not possible to say whether the Wolfenbiittel
MS. was copied from Marcianus 308 (B,) or from the com-
plete MS. of which Ambrosianus Et 157 sup. preserves the ten
leaves.

12. Palatinus gr. 391 (end of 16th c.) has notes in German in
the margin which show that it was intended to print from it; it
was written either by Xylander himself or for him. It is this MS.
of which Claudius Salmasius (Claude de Saumaise, 1588-1653)
told Bachet that it contained nothing more than the six Books,
with the tract on Polygonal Numbers.

13. Reginensis 128 was copied at the end of the 16th century
from the Wolfenbiittel MS.

14, 15. Ambrosianus A 91 sup. and Vaticanus gr. 200 both
come from B,; as they agree in omitting v. 28 of Diophantus, one
was copied from the other, probably the latter from the former.
They were both copied by the same copyist for Mendoza in 1545.
Vat. gr. 200 has headings which make eight Books ; according to
Tannery the first Book is numbered a’, the fourth §¥ ; before Vv. 20
(in Bachet’s numbering)—should this be IV. 20?—is the heading
Avopavrov €*, before the fifth Book Awogpavrov ¢%, before the sixth
Avopavrov §®, and before the tract on Polygonal Numbers
Awopavrov 7° ; this wrong division occurs in the next three MSS.

H. D. 2
18 INTRODUCTION

(16, 17, 18 in the diagram), all of which seem to be copied from
Vat. 200.

The MSS. numbered 20, 21, 22, 23 in the diagram are of the
hybrid class derived from Parisinus 2379 (C). Scorialensis Q-I1-15
and Scorialensis R-11-3, the latter copied from the former, have
the first Book divided into two (cf. p. 5 above), and so make
seven Books of the Arithmetica and an Felt Book of the
Polygonal Numbers.

27. The Cracow MS. has the same division into Books as the
MSS. last mentioned. According to Gollob, the collation of this
MS., so far as it was carried in 1899, showed that it agrees in the
main with 4 (the best MS.), B, (Marcianus 308) and C (Parisinus
2379); but, as it contains passages not found in the two latter, it
cannot have been copied from either of them.

25. Parisinus- 2380 appears to be the copy of Auria’s
Diophantus mentioned by Schulz as having been in the library of
Carl von Montchall and bearing the title “ Diophanti libri sex, cum
scholiis graecis Maximi Planudae, atque liber de numeris poly-
gonis, collati cum Vaticanis codicibus, et latine versi a Josepho
Aurial”

The first commentator on Diophantus of whom we hear is
Hypatia, the daughter of Theon of Alexandria ; she was murdered
by Christian fanatics in 415 A.D. According to Suidas she wrote
commentaries on Diophantus, on the Astronomical Canon (sc. of
Ptolemy) and on the Conics of Apollonius? Tannery suggests
that the remarks of Michael Psellus (11th c.) at the beginning of
his letter about Diophantus, Anatolius, and the Egyptian method
~ of arithmetical reckoning were taken bodily from some MS. of
Diophantus containing an ancient and systematic commentary ;
and he believes this commentary to have been that of Hypatia. I
have already mentioned the attractive hypothesis of Tannery that
Hypatia’s commentary extended only to our six Books, and that
this accounts for the loss of the rest. :

Georgius Pachymeres (1240 to about 1310) wrote in Greek a
paraphrase of at least a portion of Diophantus. Sections 25-44 of

1 Schulz, Diophantus, pref. xliii.

2 Suidas s.v. ‘Trarla: &ypayer vmbuvnua els Abpavrov, <els> Tov doTpovouikdy kavbra,
els 7a kwyika "AmoN\wvloy Vmbuvnua. So Tannery reads, following the best MSS. ; he
gives ample reasons for rejecting Kuster’s conjecture eis Awopdrrov 700 dorpovopukdy kavéva,
viz. (1) that the order of words would have been 7dr Atogpdrrov doTpovomkdy kavéva,
(2) that there is nothing connecting Diophantus with astronomy, while Suidas mentions,
s.v. O¢wr, a commentary els Tov IlToNeualov mpdxetpor kavéva.
THE MSS. OF AND WRITERS ON DIOPHANTUS 19

this survive and are published by Tannery in his edition of
Diophantus®. The chapters lost at the beginning may have con-
tained general observations and introductions to the first two
paragraphs of Book I.; section 25 begins with the third paragraph
(Def. 1), and the rest of the fragment takes us up to the problem
wil.sr 1, :

Soon afterwards Maximus Planudes (about 1260-1310) wrote
a systematic commentary on Books I., II. This is also included by
Tannery in his edition?

There are a number of other ancient scholia, very few of which
seemed to Tannery to be worth publication?

But in the meantime, and long before the date of Georgius
Pachymeres, the work of Diophantus had become known in Arabia,
where it was evidently the subject of careful study. We are told
in the Filrist, the main part of which was written in the year
937 A.D. (1) that Diophantus was a Greek of Alexandria who
wrote a book “On the art of algebra%” (2) that Abud’'l Wafa
al-Biizjani (940-998) wrote (@) a commentary (Zafsir) on the
algebra of Diophantus and (4) a book of “proofs to the pro-
positions used by Diophantus in his book and to that which
he himself (Abul Wafa) stated in his commentary?” (3) that
Qusta b. Luqa al-Ba‘labakki (died about 912) wrote a “com-
mentary on three and a half Books of Diophantus’ work on
arithmetical problems®” Qusta b. I.uqa, physician, philosopher,
astronomer, mathematician and translator, was the author of works
on Euclid and of an “introduction to geometry ” in the form of

question and answer, and translator of the so-called Books X1v., XV.
of Euclid; other Arabian authorities credit him with an actual
. “translation of the book of Diophantus on Algebra’” Lastly, we
‘are told by Ibn abi Usaibi‘a of “marginal glosses which Ishaq b.
Yinis (died about 1077), the physician of Cairo, after Ibn al-
Haitham, added to the book of Diophantus on algebraic problems.”
The title is somewhat obscure; probably Ibn al-Haitham (about
965-1039), who wrote several works on Euclid, wrote a commentary
on the Arithmetica and Ishaq b. Yunis added glosses to this
commentary? ;

! Dioph. 11. pp. 78-122. 2 Dioph. II. pp. 125-255.

3 The few that he gives are in Vol. II. pp. 256-260; as regards the collection in
general cf. Hultsch in Berliner philologische Wochenschrift, 1896, p. 615.

4 Filrist, ed. Suter, p. 22. 5 767d. p. 30 8 7bid. Pp. 43.

7 Suter, Die Mathematiker und Astronomen der Araber, 19oo, p. 41.
8 Suter, 0p. cit. pp. 107-8. Cf. Bibliotheca Mathematica 1Vy, 1903—4, p. 296.

2—2
20 INTRODUCTION

To Regiomontanus belongs the credit of being the first to call
attention to the work of Diophantus as being extant in Greek.
We find two notices by him during his sojourn in Italy, whither he
journeyed after the death of his teacher Georg von Peurbach,
which took place on the 8th April, 1461. In connexjon with
lectures on the astronomy of Alfraganus which he gave at Padua
he delivered an Oratio introductoria in omnes scientias mathe-
maticas’. In this he observed: “No one has yet translated from
the Greek into Latin the fine thirteen Books of Diophantus, in
which the very flower of the whole of Arithmetic lies hid, the ars
vei et census which to-day they call by the Arabic name of
Algebra?” Secondly, he writes to Bianchini, in answer to a letter
dated sth February, 1464, that he has found at Venice “Diofantus,”
a Greek arithmetician, who has not yet been translated into Latin;
that in his preface Diophantus defines the various powers up to
the sixth; but whether he followed out all the combinations of
these Regiomontanus does not know: “for not more than six
Books are found, though in the preface he promises thirteen. If
this book, which is really most wonderful and most difficult, could
be found entire, I should like to translate it into Latin, for the
knowledge of Greek which I have acquired while staying with my
most reverend master [Bessarion] would suffice for this....” He
goes on to ask Bianchini to try to discover a complete copy and,
in the meantime, to advise him whether he should begin to translate
the six Books®. The exact date of the Oratis is not certain.
Regiomontanus made some astronomical observations at Viterbo
in the summer and autumn of 1462. He is said to have spent a
year at Ferrara, and he seems to have gone thence to Venice.
Extant letters of his written at Venice bear dates from 27th July,
1463, to 6th July, 1464, and it may have been from Venice
that he made his visit to Padua. At all events the Oratio at
Padua must have been near in time to the discovery of the
MS. at Venice.

Notwithstanding that attention was thus called to the work, it

! Printed in the work Rudimenta astronomica Alfragani, Niirnberg, 1537.

% As the ars rei ef census, the solution of determinate quadratic equations, is not found
in our Diophantus, it would seem that at the time of the Oratio Regiomontanus had only
looked at the MS. cursorily, if at all.

3 The letter to Bianchini is given on p. 135 of Ch. Th. v. Murr’s Memorabilia,
Norimbergae, 1786, and partly in Doppelmayer’s Historische Nachricht von den Niirn-
bergischen Mathematicis und Kiinstlern (Niimberg, 1730), p. 5, note y.
THE MSS. OF AND WRITERS ON DIOPHANTUS 21

seems to have remained practically a closed book from the date of
Maximus Planudes to about 1570. Luca Paciuolo, towards the
end of the 15th c, Cardano and Tartaglia about the middle of the
16th, make no mention of it. Only Joachim Camerarius, in a
letter published in 1556% mentions’ that there is ‘ai MS.! of
Diophantus in the Vatican which he is anxious to see. Rafael
Bombelli was the first to find a MS. in the Vatican and to conceive
the idea of publishing the work. This was towards 1570, for in his
Algebra? published in 1572 Bombelli tells us that he had 7 z/e
years last past discovered a Greek book on Algebra written by “a
certain Diofantes, an Alexandrine Greek author, who lived in the
time of Antoninus Pius”; that, thinking highly of the contents of
the work, he and Antonio Maria Pazzi determined to translate it ;
that they actually translated five books out of the seven into
which the MS. was divided ; but that, before the rest was finished,
they were called away from it by other labours. Bombelli did not
carry out his plan of publishing Diophantus in a translation, but
he took all the problems of the first four Books and some of those
of the fifth, and embodied them in his Algebra, interspersing them
with his own problems. He took no pains to distinguish
Diophantus’ problems from his own; but in the case of the former
he adhered pretty closely to the original, so that Bachet admits his
obligations to him, remarking that in many cases he found

1 De Graecis Latinisque numerorunm notis et practerea Saracenis sew Indicis, etc. etc.,
studio Joachimi Camerarii, Papeberg, 1556.

2 Nesselmann tells us that he has not seen this work but takes his information about
it from Cossali. I was fortunate enough to find in the British Museum one of the copies
dated 1579 (really the same as the original edition of 1572 except that the title-page and
date are new, and a dedicatory letter on pp. 3-8 is reprinted; there were not two
separate editions). The title is 2’ Algebra, opera di Rafael Bombelli da Bologna diuisa in
tre Libri,..... In Bologna, Per Giovanni Rossi, MDLXX1X. The original of the passage
from the preface is :

““Questi anni passati, essendosi ritrouato una opera greca di questa disciplina nella
libraria di Nostro Signore in Vaticano, composta da un certo Diofante Alessandrino Autor
Greco, il quale fu a tempo di Antonin Pio, e havendomela fatta vedere Messer Antonio
Maria Pazzi Reggiano, publico lettore delle Matematiche in Roma, e giudicatolo con lui
Autore assai intelligente de’ numeri (ancorche non tratti de’ numeri irrationali, ma solo
in lui si vede vn perfetto ordine di operare) egli, ed io, per arrichire il mondo di cosl fatta
opera, ci dessimo a tradurlo, e cinque libri (delli sette che sono) tradutti ne habbiamo ; lo
restante non hauendo potuto finire per gli tranagli auenuti all’ uno, e all’ altro; e in detta
opera habbiamo ritrouato, ch’ egli assai volte cita gli Autori Indiani, col che mi ha fatto
conoscere, che questa disciplina appo gl’ indiani prima fu, che a gli Arabi.” The last
words stating that Diophantus often quotes from Indian authors are no doubt due to
Bombelli’s taking for part of Diophantus the tract of Maximus Planudes about the Indian
method of reckoning.
22 INTRODUCTION

Bombelli’s translation better than Xylander’s and consequently
very useful for the purpose of amending the latter. :

It may be interesting to mention a few points of notation in
this work of Bombelli. At the beginning of Book II. he explains
that he uses the word “tanto” to denote the unknown quantity,
not “cosa” like his predecessors; and his symbol for it is 1, the
square of the unknown (2?) is 2, the cube &; and so on. "For plus
and minus (pi and meno) he uses the initial letters p and
Thus corresponding to x + 6 we should find in Bombelli 11 2. 6,
and for x2 + 5x —4, 12 p. 51 m. 4. This notation shows, as will be
seen later, some advance upon that of Diophantus in one important
respect.

The next writer upon Diophantus was Wilhelm Holzmann who
published, under the Graecised form of his name, Xylander, by
which he is generally known, a work bearing the title: Diophanti
Alexandrini Rerum Awithmeticarum Libri sex, quorum primi duo
adiecta habent Scholia Maximi (ut coniectura est) Planudis. Item
Liber de Numeris Polygonis sew Multangulis. Opus incomparabile,
werae Avithmeticae Logisticae perfectionem continens, paucis adhuc
wisum. A Guil. Xylandro Auwugustano incredibili labore Latiné
redditum, et Commentariis explanatum, inque lucem editum, ad
Lllustriss. Principe Ludovicum Vuirtembergensem. Basileae per
Eusebium Episcopium, et Nicolai Fr. haevedes. MDLXXV. Xylander
was according to his own statement a “public teacher of Aristotelian
philosophy in the school at Heidelberg?” He was a man of almost
universal culture’, and was so thoroughly imbued with the classical
literature, that the extraordinary aptness of his quotations and his
wealth of expression give exceptional charm to his writing whenever
he is free from the shackles of mathematical formulae and techni-
calities. The ZEpistola Nuncupatoria is addressed to the Prince
Ludwig, and Xylander neatly introduces it by the line “ Offerimus
numeros, numeri sunt principe digni.” This preface is very quaint
and interesting. He tells us how he first saw the name of
Diophantus mentioned in Suidas, and then found that mention

: ! “Sed suas Diophanteis quaestionibus ita immiscuit, ut has ab illis distinguere non
sit in promptu, neque vero se fidum satis interpretem praebuit, cum passim verba
Diophanti immutet, hisque pleraque addat, pleraque pro arbitrio detrahat. In multis
nihilominus interpretationem Bombellii, Xilandriana praestare, et ad hanc emendandam
me adjuvisse ingenue fateor.”” Ad lectorem.

2 ““Publicus philosophiae Aristoteleae in schola Heidelbergensi doctor.”

3 Even Bachet, who, as we shall see, was no favourable critic, calls him Vir omnibus
disciplinis excultus.”
THE MSS. OF AND WRITERS ON DIOPHANTUS 23

had been made of his work by Regiomontanus as being extant
in an Italian library and having been seen by him. But, as the
book had not been edited, he tried to think no more of it but,
instead, to absorb himself in the study of such arithmetical books
as he could obtain, and in investigations of his own. Self-taught
except in so far as he could learn from published works such as
those of Christoff Rudolff (of the “ Coss”), Michael Stifel, Cardano,
Nufiez, he yet progressed so far as to be able to add to, modify
and improve what he found in those works. As a result he fell
into what Heraclitus called oimow, iepav voooy, that is, into the
conceit of “being somebody” in the field of Arithmetic and
“ Logistic”; others too, themselves learned men, thought him an
arithmetician of exceptional ability. But when he first became
acquainted with the problems of Diophantus (he continues) right
reason brought such a reaction that he might well doubt whether he
ought previously to have regarded himself as an object of pity or of
derision. He considers it therefore worth while to confess publicly his
own ignorance at the same time that he tries to interest others in
the work of Diophantus, which had so opened his eyes. Before this
critical time he was so familiar with methods of dealing with surds
that he had actually ventured to add something to the discoveries
of others relating to them ; the subject of surds was considered to
be of great importance in arithmetical questions, and its difficulty

1 I cannot refrain from quoting the whole of this passage : ‘“Sed cum ederet nemo :
cepi desiderium hoc paulatim in animo consopire, et eorum quos consequi poteram
Arithmeticorum librorum cognitione, et meditationibus nostris sepelive. Veritatis porro
apud me est autoritas, ut ei coniunctum etiam cum dedecore meo testimonium lubentissimée
perhibeam. Quod Cossica seu Algebrica (cum his enim reliqua comparata, id sunt quod
umbrae Homericé in Necya ad animam Tiresiae) ea ergo quod non assequebar modo,
quanquam mutis duntaxat usus preceptoribus caetera avrodldakros, sed et augere, uariare,
adeoque corrigere in loco didicissem, quae summi et fidelissimi in docendo uiri Christifer
Rodolphus Silesius, Micaelus Stifelius, Cardanus, Nonius, aliique litteris mandauerant :
incidi in olyow, lepav végov, ut scité appellauit Heraclitus sapientior multis aliis philoso-
phis, hoc est, in Arithmetica, et uera Logistica, putaui me esse aliquid: itaque de me
passim etiam a multis, iisque doctis uiris iudicatum fuit, me non de grege Arithmeticum
esse. Verum ubi primim in Diophantea incidi: ita me recta ratio circumegit, ut flendisne
mihi ipsi ante, an uerd ridendus fuissem, haud iniuria dubitauerim. Operae precium est
hoc loco et meam inscitiam inuulgare, et Diophantei operis, quod mihi nebulosam istam
caliginem ab oculis detersit, immo eos in coenum barbaricum defossos eleuauit et repur-
gauit, gustum aliquem exhibere. Surdorum ego numerorum tractationem ita tenebam,
ut etiam addere aliorum inuentis aliquid non poenitendum auderem, atque id quidem in
rebus arithmeticis magnum habetur, et difficultas istarum rerum multos a 1athematibus
deterret. Quanto autem hoc est praeclarius, in iis problematis, quae surdis etiam
numeris uix posse uidentur explicari, rem eo deducere, ut quasi solum arithmeticum
uertere iussi obsurdescant illi plane, et ne mentio quidem eorum in tractatione ingenio-
sissimarum quaestionum admittatur.”
24 INTRODUCTION

was even such as to deter many from the study of mathematics.
«But how much more splendid,” says Xylander, “in the case of
problems which seem to be hardly capable of solution even with
the help of surds, to bring the matter to the point that, while the
surds, when bidden (so to speak) to plough the arithmetic soil,
become true to their name and deaf to entreaty, they are not so
much as mentioned in these most ingenious solutions!” He then
describes the enormous difficulties which beset his work owing
to the corruptions in his text. In dealing, however, with the
mistakes and carelessness of copyists he was, as he says, no novice;
for proof of which he appeals to his editions of Plutarch, Stephanus
and Strabo. This passage, which is good reading, but too long
to reproduce here, I give in full in the notel. Next Xylander
tells us how he came to get possession of a manuscript of Dio-
phantus. In October of the year 1571 he made a journey to
Wittenberg ; while there he had conversations on mathematical
subjects with two professors, Sebastian Theodoric and Wolfgang
Schuler by name, who showed him a few pages of a Greek

1 Id uerd mihi accidit durum et uix superabile incommodum, quod mirificé deprauata
omnia inueni, cum neque problematum expositio interdum integra esset, ac passim numeri
(in quibus sita omnia esse in hoc argumento, quis ignorat?) tam problematum quam
solutionum siue explicationum corruptissimi. Non pudebit me ingenué fateri, qualem me
heic gesserim. Audacter, et summo cum feruore potius quam alacritate animi opus ipsum
initio sum aggressus, laborque mihi omnis uoluptati fuit, tantus est meus rerum arithmeti-
carum amor. quin et gratiam magnam me apud omnes liberalium scientiarum amatores ac
patronos initurum, et praeclare de rep. litteraria meriturum intelligebam, eamque rem
mihi laudi (quam a bonis profectam nemo prudens aspernatur) gloriaeque fortasse etiam
emolumento fore sperabam. Progressus aliquantulum, in salebras incidi: quae tantum
abest ut alacritatem meam retuderint, ut etiam animos mihi addiderint, neque enim mihi
novum aut insolens est aduersus librariorum incuriam certamen, et hac in re militaui, (ut
Horatii nostri uerbis utar) non sine gloria. quod me non arroganter dicere, Dio,
Plutarchus, Strabo, Stephanusque nostri testantur. Sed cum mox in ipsum pelagus
monstris scatens me cursus abripuit : non despondi equidem animum, neque manus dedi,
sed tamen saepius ad oram unde soluissem respexi, quam portum in quem esset enadendum
cogitando prospicerem, deprachendique non minus ueré quam eleganter ea cecinisse
Alcaeum, quae (si possum) Latiné in hac quasi uotiua mea tabula scribam.

Qui uela uentis uult dare, dum licet,

Cautus futuri praeuideat modum

Cursus. mare ingressus, marino

Nauiget arbitrio necesse est.
Sane quod de Echeneide pisce fertur, eum nauim cui se adplicet remorari, poené credibile
fecit mihi mea cymba tot mendorum remoris retardata. Expediui tamen me ita, ut facile
omnes mediocri de his rebus iudicio praediti, intellecturi sint incredibilem me laborem et
aerumnas difficilimas superasse : pudore etiam stimulatum oneris quod ultro mihi impos-
uissem, non perferendi. Paucula quaedam non plane explicata, studio et certis de causis
in alium locum reiecimus. Opus quidem ipsum ita absoluimus ut neque eius nos pudere
debeat, et Arithmeticae Logisticesque studiosi nobis se plurimum debere sint haud dubie
professuri.”
THE MSS. OF AND WRITERS ON DIOPHANTUS 25

manuscript of Diophantus and informed him that it belonged to
Andreas Dudicius whom Xylander describes as “Andreas Dudicius
Sbardellatus, hoc tempore Imperatoris Romanorum apud Polonos
orator.” On his departure from Wittenberg Xylander wrote out
and took with him the solution of a single problem of Diophantus,
to amuse himself with on his journey. This he showed at Leipzig
to Simon Simonius Lucensis, a professor at that place, who wrote to
Dudicius on his behalf. A few months afterwards Dudicius sent
the MS. to Xylander and encouraged him to persevere in his
undertaking to translate the Azithmetica into Latin. Accordingly
Xylander insists that the glory of the whole achievement belongs
in no less but rather in a greater degree to Dudicius than to
himself. Finally he commends the work to the favour of Prince
Ludwig, extolling the pursuit of arithmetical and algebraical
science and dwelling in enthusiastic anticipation on the influence
which the Prince’s patronage would have in helping and advancing
the study of Arithmetic, This Epistola Nuncupatoria bears the
date 14th August, 15742 Xylander died on the 10th of February
in the year following that of the publication, 1576.

Tannery has shown that the MS. used by Xylander was
Guelferbytanus Gudianus I. Bachet observes that he has not been
able to find out whether Xylander ever published the Greek text,
though parts of his commentary seem to imply that he had, or at
least intended to do so. It is now clear that he intended to bring
out the text, but did not carry out his intention. Tannery observes
that the MS. Palatinus gr. 391 seems to have been written either by
Xylander himself or for him, and there are German notes in the
margin showing that it was intended to print from it.

Xylander's achievement has been, as a rule, quite inadequately
appreciated. Very few writers on Diophantus seem to have studied
the book itself: a fact which may be partly accounted for by its
rarity. Even Nesselmann, whose book appeared in 1842, says that
he has never been able to find a copy. Nesselmann however seems
to have come nearest to a proper appreciation of the value of the
work : he says “Xylander’'s work remains, in spite of the various

1 “Hoc non modd tibi, Princeps Illustrissime, honorificum erit, atque gloriosum; sed
te labores nostros approbante, arithmeticae studium cim alibi, tum in tua Academia et
Gymnasiis, excitabitur, confirmabitur, prouehetur, et ad perfectam eius scientiam multi tuis
auspiciis, nostro labore perducti, magnam hac re tuis in remp. beneficiis accessionem
factam esse gratissima commemoratione praedicabunt.”

2 ¢“ Heidelberga. postrid. Eidus Sextiles cI0 10 LXX1V.”
26 INTRODUCTION

defects which are unavoidable in a first edition of so difficult an
author, especially when based on only one MS. and that full of
errors, a highly meritorious achievement, and does not deserve
the severe strictures which it has sometimes had passed upon it.
It is true that Xylander has in many places not understood his
author, and has misrepresented him in others; his translation is
often rough and un-Latin, this being due to a too conscientious
adherence to the actual wording of the original; but the result
was none the less brilliant on that account. The mathematical
public was - put in possession of Diophantus’ work, and the
appearance of the translation had an immediate and enormous
influence on the development and shaping of Algebra” As a
rule, the accounts of Xylander’s work seem to have been based
on what Bachet says about it and about his obligations to it.
When I came to read Bachet myself and saw how disparaging,
as a rule, his remarks upon Xylander were, I could not but suspect
that they were unfair. His repeated and almost violent repudiation
of obligation to Xylander suggested to me the very thing which he
disclaimed, that he was under too great obligation to his predecessor
to acknowledge it duly. I was therefore delighted at my good
fortune in finding in the Library of Trinity College, Cambridge,
a copy of Xylander, and so being able to judge for myself of
the relation of the later to the earlier work. The result was to
confirm entirely what I had suspected as to the unfair attitude
taken up by Bachet towards his predecessor. I found it every-
where; even where it is obvious that Xylander’s mistakes or
difficulties are due only to the hopeless state of his solitary MS.
Bachet seems to make no allowance for the fact. The truth is that
Bachet’s work could not have been as good as it was but for the
pioneer work of Xylander; and it is the great blot in Bachet’s
otherwise excellent edition that he did not see fit to acknowledge
the fact. - : or

I must now pass to Bachet’s work itself. It was the first
edition published which contained the Greek text, and appeared
in 1621 bearing the title: Dioplanti Alexandrini Avrithmeticovum
libri sex, et de numeris multangulis liber wunus. Nunc primiom
Graece et Latine editi, atque absolutissimis Commentariis illustrat.
Atictore Claudio Gaspare Bacheto Meziriaco Sebusiano, V.C. Lutetiae
Parisiorum, Sumptibus Hievonymi Drovart?, via Jacobaea, sub Scuto

! Nesselmann, p. 279-80.
% For ‘ sumptibus Hieronymi Drovart etc.” some copies have ¢ sumptibus Sebastiani
THE MSS. OF AND WRITERS ON DIOPHANTUS 27

Solari. MDCXX/I Bachet’s Greek text is based, as he tells us,
upon a MS. which he calls “codex Regius,” now in the Bibliotheque
Nationale at Paris (Parisinus 2379) ; this MS. is his sole authority,
except that Jacobus Sirmondus had part of a Vatican MS. (Vat.
gr. 304) transcribed for him. He professes to have produced a
good Greek text, having spent incalculable labour upon its emenda-
tion, to have inserted in brackets all additions which he made to it,
and to have given notice of all corrections, except those of an
obvious or trifling nature; a few passages he has left asterisked, in
cases where correction could not be safely ventured upon. He
is careful to tell us what previous works relating to the subject he
had been able to consult. First he mentions Xylander (he spells
the name as Xzlander throughout), who had translated the whole of
Diophantus, and commented upon him throughout, “except that
he scarcely touched a considerable part of the fifth book, the whole
of the sixth and the treatise on multangular numbers, and even
the rest of his. work was not very successful, as he himself admits
that he did not thoroughly understand a number of points.” Then
he speaks of Bombelli (as already mentioned) and of the Zetetica of
Vieta (in which the author treats in his own way a large number
of Diophantus’ problems: Bachet thinks that he so treated them
because he despaired of restoring the book completely). Neither
Bombelli nor Vieta (says Bachet) made any attempt to demonstrate
the difficult porisms and abstruse theorems in numbers which
Diophantus assumes. as known in many places, or sufficiently
explained the causes of his operations and artifices. All these
omissions on the part of his predecessors he thinks he has supplied
in his notes to the various problems and in the three books of
“Porisms” which he prefixed to the work’. As regards his Latin
translation, he says that he gives us Diophantus in Latin from the
version of Xylander most carefully corrected, in which he would
have us know that he has done two things in particular, first,

Cramoisy, via Jacobaea, sub Ciconiis.”” The copy (from the Library of Trinity College,
Cambridge) which I used in preparing my first edition has the former words; a copy in
the Library of the Athenaeum Club has the latter.

1 On the nature of some of Bachet’s proofs Nicholas Saunderson (formerly Lucasian
Professor) remarks in Zlements of Algebra, 1740, apropos of Dioph. 111. 15: ““ M. Bachet
indeed in the 16th and 17th props. of his second book of Porisms has given us demonstra-
tions, such as they are, of the theorems in the problem: but in the first place he
demonstrates but one single case of those theorems, and in the next place the demonstra-
tions he gives are only synthetical, and so abominably perplexed withal, that in each
demonstration he makes use of all the letters in the alphabet except I and O, singly to
represent the quantities he has there occasion for.”
28 INTRODUCTION

corrected what was wrong and filled the numerous lacunae,
secondly, explained more clearly what Xylander had given in
obscure or ambiguous language; “I confess however,” he says,
“that this made so much change necessary, that it is almost
fairer to attribute the translation to me than to Xilander. But if
anyone prefers to consider it as his, because I have held fast, tooth
and nail, to his words when they do not misrepresent Diophantus,
I have no objection.” Such sentences as these, which are no
rarity in Bachet’s book, are certainly not calculated to increase
our respect for the author. According to Montucla? “the historian
of the French Academy tells us” that Bachet worked at this edition
during the course of a quartan fever, and that he himself said that,
disheartened as he was by the difficulty of the work, he would never
have completed it, had it not been for the stubbornness which his
malady generated in him.

As the first edition of the Greek text of Diophantus, this work,
in spite of any imperfections we may find in it, does its author all
honour.

The same edition was reprinted and published with the addition
of Fermat's notes in 1670 : Diophanti Alexandrini Arithmeticorum
libri sex, et de numeris multangulis liber unus. Cum commentariis
C. G. Bacheti V.C. et obseruationibus D. P. de Fermat Senatoris
Tolosani. Accessit Doctrinae Analyticae tnuentum nouunt, collectum
ex varits ewusdem ID. de Fermat Epistolis. Tolosae, Excudebat
Bernardus Bosc, ¢ Regione Collegii Societatis Jesu. MDCLXX.
This edition was not published by Fermat himself, but by his
son after his death. S. Fermat tells us in the preface that this
publication of Fermat's notes to Diophantus® was part of an
attempt to collect together from his letters and elsewhere his
contributions to mathematics. The “Doctrinae Analyticae In-
uentum nouum” is a collection made by Jacobus de Billy*

! Deinde Latinum damus tibi Diophantum ex Xilandri versione accuratissime castigata,
in qua duo potissimum nos praestitisse scias velim, nam et deprauata correximus, hiantesque
passim lacunas repleuimus : et quae subobscure, vel ambigué fuerat interpretatus Xilander,
dilucidius exposuimus; fateor tamen, inde tantam inductam esse mutationem, vt prope-
modum aequius sit versionem istam nobis quam Xilandro tribuere. Si quis autem potius
ad eum pertinere contendat, quod eius verba, quatenus Diophanto fraudi non erant,
mordicus retinuimus, per me licet.”’ 21.1323,

% Now published in Zuvres de Fermat by P. Tannery and C. Henry, Vol. I. (1891),
pp- 289-342 (the Latin original), and Vol. 111. (1896), pp. 241-274 (French translation).

* Now published in Fuvres de Fermat, 111. 323-398 (French translation). De
Billy had already published in 1660 a book under the title Diophantus geometra sive
opus contextum ex arithmetica et geometria.
THE MSS. OF AND WRITERS ON DIOPHANTUS 29

from various letters which Fermat sent to him at different times.
The notes upon Diophantus’ problems, which his son hopes will
prove of value very much more than commensurate with their
bulk, were (he says) collected from the margin of his copy of
Diophantus. From their brevity they were obviously intended
for the benefit of experts!, or even perhaps solely for Fermat's
own, he being a man who preferred the pleasure which he had
in the work itself to any reputation which it might bring
him. Fermat never cared to publish his investigations, but was
always perfectly ready, as we see from his letters, to acquaint
his friends and contemporaries with his results. Of the notes
themselves this is not the place to speak in detail. This edition
of Diophantus is rendered valuable only by the additions in it
due to Fermat; for the rest it is a mere reprint of that of 1621.
So far as the Greek text is concerned, it is very much inferior
to the first edition. There is a far greater number of misprints,
omissions of words, confusions of numerals; and, most serious of
all, the brackets which Bachet inserted in the edition of 1621 to
mark the insertion of words in the text are in this later edition
altogether omitted. These imperfections have been already noticed
by Nesselmann? Thus the reprinted edition of 1670 is untrust-
worthy as regards the text.

In 1585 Simon Stevin published a French version of the first
four books of Diophantus®. It was based on Xylander and was
a free reproduction, not a translation, Stevin himself observing that
the MS. used by Xylander was so full of mistakes that the text of

L Lectori Beneuolo, p. iii: *‘ Doctis tantum quibus pauca sufficiunt, harum obserua-
tionum auctor scribebat, vel potius ipse sibi scribens, his studiis exerceri malebat quam
gloriari; adeo autem ille ab omni ostentatione alienus erat, vt nec lucubrationes suas
typis mandari curauerit, et suorum quandoque responsorum autographa nullo seruato
exemplari petentibus vltrd miserit; norunt scilicet plerique celeberrimorum huius saeculi
Geometrarum, quam libenter ille et quantd humanitate, sua iis inuenta patefecerit.”’

2 “Was dieser Abdruck an dusserer Eleganz gewonnen hat (denn die Bachet’sche
Ausgabe ist mit dusserst unangenehmen, namentlich Griechischen Lettern gedruckt), das
hat sie an innerm Werthe in Bezug auf den Text verloren. Sie ist nicht bloss voller
Druckfehler in einzelnen Worten und Zeichen (z. B. durchgehends = statt 3, goo)
sondern auch ganze Zeilen sind ausgelassen oder doppelt gedruckt (z. B. III. 12 eine
Zeile doppelt, 1v. 25 eine doppelt und gleich hinterher eine ausgelassen, 1vV. 52 eine
doppelt, v. 11 eine ausgelassen, desgleichen Vv. 14, 25, 33, VI. 8, 13 und so weiter), die
Zahlen verstiimmelt, was aber das Aergste ist, die Bachet’schen kritischen Zeichen sind
fast iiberall, die Klammer durchgingig weggefallen, so dass diese Ausgabe als Text des
Diophant véllig unbrauchbar geworden ist,” p. 283. :

3 Included in Z’Arithmetique de Simon Stevin de Bruges...A Leyde, De I'Imprimerie
de Christophle Plantin, C10. ID . LXXXV.
30 INTRODUCTION

Diophantus could not be given word for word® Albert Girard
added the fifth and sixth books to the four, and this complete
version appeared in 16252

In 1810 was published an excellent translation (with additions)
of the fragment upon Polygonal Numbers by Poselger : Diophantus
von A lexandyien iiber die Polygonal-Zallen. Uebersetst mit Zusdtzen
von F. Th. Poselger. Leipzig, 1810.

In 1822 Otto Schulz, professor in Berlin, published a very
meritorious German translation with notes: Diophantus wvon
Alexandria arithmetische Aufgaben nebst dessen Schrift iiber die
Polygon-Zaklen. Aus dem Griechischen iibersetst und mit An-
merkungen begleitet von Otto Schulz, Professor am Berliniscl-
Colnischen Gymnasium sum grauen Kloster. Berlin, 1822. [In der
Schlesingerschen Buch- und Musikhandlung. The work of Poselger
just mentioned was with the consent of its author incorporated in
Schulz’s edition along with his own translation and notes upon
the larger treatise, the Arithmetica. According to Nesselmann
Schulz was not a mathematician by profession; he produced,
however, a thoroughly useful edition, with notes chiefly upon
the matter of Diophantus and not on the text (with the exception
of a very few emendations): notes which, almost invariably correct,
help much to understand the author. Schulz’s translation is based
upon the edition of Bachet’s text published in 1670.

Another German translation was published by G. Wertheim
in 1890: Die Arithmetik und die Schrift iiber Polygonalzallen des
Diophantus von Alexandria.  Ubersetzt und mit Anmerkungen
begleitet von G. Wertheim (Teubner). Though it appeared before
the issue of Tannery’s definitive text, it is an excellent translation,
the translator being thoroughly equipped for his task; it is valuable
also as containing Fermat’s notes, also translated into German, with
a large number of other notes by the translator elucidating both
Diophantus and Fermat, and generalising a number of the problems
which, with very few exceptions, receive only particular solutions
from Diophantus himself. Wertheim has also included 46 epigram-
problems from the Greek anthology and the enunciation of the
famous Cattle-Problem attributed to Archimedes.

1 See Bibliotheca Mathematica V1ig, 1906-7, p. 59.

z L’ Arithmetique de Simon Stevin de Bruges, Reueué, corvigee & augmentee de plusieurs
travclez el annotation par Albert Girard Samieclois Mathematicien, A Leide, de
Imprimerie des Elzeviers c10.10.cxxv. Reproduced in the edition of Zes Euwvres
Mathematiques de Simon Stevin de Bruges. Par Albert Girard, Leyde, cin. ID. CXXXIV
THE MSS. OF AND WRITERS ON DIOPHANTUS 3

No description is necessary of the latest edition, by Tannery,
in which we at last have a definitive Greek text of Diophantus
with the ancient commentaries, etc., Diophanti Alexandrini opera
omnia cum Graects commentariis. LEdidit et Latine interpretatus
est Paulus Tannery (Teubner). The first volume (1893) contains
the text of Diophantus, the second (1895) the Pseudepigrapha,
Testimonia veterum, Pachymeres’ paraphrase, Planudes’ com-
mentary, various ancient scholia, etc., and 38 arithmetical epigrams
in the original Greek with scholia. Any further edition will neces-
sarily be based on Tannery, who has added all that is required in
the shape of introductions, etc.

Lastly we hear of other works on Diophantus which, if they
were ever written, are lost or remain unpublished. First, we find
it asserted by Vossius (as some have understood him) that the
Englishman John Pell wrote an unpublished Commentary upon
Diophantus. John Pell (1611-1685) was at one time professor
of mathematics at Amsterdam and gave lectures there on Dio-
phantus, but what Vossius says about his commentary may
well be only a recommendation to undertake a commentary,
rather than a historical assertion of its completion. Secondly,
Schulz states in his preface that he had lately found a note in
Schmeisser’'s Orthodidaktik der Mathematik that Hofrath Kausler
by command of the Russian Academy prepared an edition of
Diophantus’. This seems however to be a misapprehension on the
part of Schulz. Kausler is probably referring, not to a translation
of Diophantus, but to his memoir of 1798 published in Nova Acta
Acad. Petropol. X1. p. 125, which might easily be described as an
Ausarbeitung of Diophantus’ work.

I find a statement in the New American Cyclopaedia (New York,
D. Appleton and Company), Vol. v1., that “a complete translation
of his (Diophantus’) works into English was made by the late
Miss Abigail Lousada, but has not been published.”

! The whole passage of Schmeisser is: ‘Die mechanische, geistlose Behandlung der
Algebra ist ins besondere von Herrn Hofrath Kausler stark geriigt worden. In der
Vorrede zu seiner Ausgabe des Uflakerschen Exempelbuchs beginnt er so: ¢ Seit mehreren
Jahren arbeitete ich fiir die Russisch-Kaiserliche Akademie der Wissenschaften Diophants
unsterbliches Werk iiber die Arithmetik aus, und fand darin einen solchen Schatz von
den feinsten, scharfsinnigsten algebraischen Auflosungen, dass mir die mechanische,
geistlose Methode der neuen Algebra mit jedem Tage mehr ekelte u.s.w.”” (p. 33).
CHAPTER 11]
NOTATION AND DEFINITIONS OF DIOPHANTUS

As it is my intention, for the sake of brevity and per-
spicuity, to make use of the modern algebraical notation in giving
my account of Diophantus’ problems and general methods, it is
necessary to describe once for all the machinery which our author
uses for working out the solutions of his problems, or the notation
by which he expresses such relations as would be represented in
our time by algebraical equations, and, in particular, to illustrate
the extent to which he is able to manipulate unknown quantities.
Apart, however, from the necessity of such a description for the
proper and adequate comprehension of Diophantus, the general
question of the historical development of algebraical notation
possesses great intrinsic interest. Into the general history of this
subject I cannot enter in this essay, my object being the elucidation
of Diophantus ; I shall accordingly in general confine myself to an
account of his notation solely, except in so far as it is interesting
to compare it with the corresponding notation of his editors and
(in certain cases) that of other writers, as, for example, certain of
the early Arabian algebraists.

First, as to the representation of an unknown quantity. The
unknown quantity, which Diophantus defines as containing mA\#%6fos
povadwy adpiaTov, te. an undefined number of units (def. 2), is
denoted throughout by what was printed in the editions before
Tannery’s as the Greek letter s with an accent, thus ¢’, or in the
form ¢%. This symbol in verbal description he calls 0 apifuds, « the
number,” z.e.,, by implication, the number par excellence of the problem
in question. In the cases where the symbol is used to denote in-
flected forms, eg., the accusative singular or the dative plural, the
terminations which would have been added to the stem of the full
word dptfuos were printed above the symbol s in the manner of an
NOTATION AND DEFINITIONS OF DIOPHANTUS 33

exponent, thus s* (for apifuov, as v for Tov), s°°, the symbol being
in addition doubled in the plural cases, thus ¢¢%, ¢5°%, ¢¢%%, ¢5%, for
aptBpoi k.1.&. When the symbol is used in practice, the coefficient
is expressed by putting the required Greek numeral immediately
after it, thus ¢s® wa corresponds to 11%, "a to x and so on.

Tannery discusses the question whether in the archetype (a) of
the MSS. this duplication of the sign for the plural and this
addition of the terminations of the various cases really occurred.
He observes that any one accustomed to reading Greek MSS. will
admit that the marks of cases are common in the later MSS. but
are very frequently omitted in the more ancient. Further, the
practice of duplicating a sign to express the plural is more ancient
than that of adding the case-terminations. Tannery concludes that
the case-terminations (like the final syllables of abbreviations used
for othér words) were very-generally, if not always, wanting in the
archetype (@). If this seems inconsistent with the regularity with
which they appear in our MSS,, it has to be remembered that 4
and 7, do not represent the archetype (@) but the readings of a, the
copyist of which probably took it upon himself to substitute the
full word for the sign or to add the case-terminations. Tannery's
main argument is the frequent occurrence of instances where the
wrong case-ending has been added, e¢g., the nominative for the
genitive ; the conclusion is also confirmed by instances in which
different cases of the word apifucs, e.g. apifuod, aptfudr, and even
aplBudy written in full are put by mistake for kat owing to the
resemblance between the common abbreviation for «at and the
sign for aptfuss, and of course in such cases the abbreviation would
not have had the endings. As regards the duplication of the sign
for the plural, Tannery admits that this was the practice of the
Byzantines ; but he considers that the evidence is against sup-
“posing that Diophantus duplicated the sign; he does not do so
with any other of his technical abbreviations, those for wovds,
Svvauss, etc. Accordingly in his text of Diophantus Tannery has
omitted the case-endings and written the single sign for apifuos
whether in the singular or in the plural; in his second volume,
however, containing the scholia, etc., he has retained the duplicated
sign.

On the assumption that the sign was the Greek final sigma, it
was natural that Nesselmann should explain it by the supposition

! Dioph. II. pp. xxxiv—xxXix.
34 INTRODUCTION

that Diophantus, in search of a convenient symbol for his unknown
quantity, would select the only letter of the Greek alphabet which
was not already appropriated as a numeral’. But he made the
acute observation? that, as the symbol occurred in many places (of
course in Bachet’s text) for dptfuos used in the ordinary un-
technical sense, and was therefore, as it appeared, not exclusively
used to designate the unknown quantity, the technical apifuds, it
must after all be more of the nature of an abbreviation than an
algebraical symbol like our x. It is true that this uncertainty in
the use of the sign in the MSS. is put an end to by Tannery, who
uses it for the technical dptfuds alone and writes the untechnical
aptbuss in full ; but, even if Diophantus’ practice was as strict as
this, I do not think this argues any difference in the nature of the
abbreviation. There is also a doubt whether the final sigma, s,
was developed as distinct from the formr ¢ so early as the date of
the MSS. of Diophantus, or rather so early as the first copy of his
work, if the author himself really gave the explanation of the sign
as found in our text of his second definition. These considerations
suggested to me that the sign was not the final sigma at all, but
must be explained in some other way. I had to look for con-
firmation of this to the precise shape of the sign as found in extant
MSS. The only MS. which I had the opportunity of inspecting
personally was the MS. of the first ten problems of Diophantus in
the Bodleian; but here I found strong confirmation of my view in
the fact that the sign appeared as’¢3, quite different in shape from,
and much larger than, the final sigma at the end of words in the
same MS. (There is in the Oxford MS. the same irregularity as
was pointed out by Nesselmann in the use of the sign sometimes
for the technical, and sometimes for the untechnical, dpifuds?.)
But I found evidence that the sign appeared elsewhere in some-
what different forms. Thus Rodet in the Journal Asiatique of
January, 1878, quoted certain passages from Diophantus for the
purpose of comparison with the algebra of Muhammad b. Misa
al-Khuwarazmi. Rodet says he ccpied these passages exactly
from Bachet's MS. ; but, while he generally gives the sign as the
final sigma, he has in one case yy for apifuol. In this last case

! Nesselmann, pp. 29o-I. 2 4bid. pp. 300-1.

3 An extreme case is érata 70 Tob Sevrépov’ > dpiOuob évbs, where the sign (contrary to
what would be expected) means the untechnical dpfuds, and the technical is written in
full. Also in the definition 6 8¢ undéy TovTwy TOV (BlwpdTwy KTNTdUeEvOs...dplfuds KakeiTat

the word dpifués is itself denoted by the symbol, showing that the word and the symbol
are absolutely convertible.
NOTATION AND DEFINITIONS OF DIOPHANTUS 33

Bachet himself reads 55°. But the same form yy which Rodet
gives is actually found in three places in Bachet’s own edition.
(1) In his note to 1v. 3 he gives a reading from his own MS. which
he has corrected in his own text and in which the signs Ya and
YY» occur, evidently meaning apifuos @ and apifpot 7, though the
sign should have been that for apifposrov (= 1/2). (2) In the text of
Iv. 13 there is a sentence (marked by Bachet as interpolated) which
contains the expression yg, where the context again shows that
yy is for apibuoi. (3) At the beginning of Vv. 9 there is a difficulty
in the text, and Bachet notes that his MS. has wire 0 Sumrhaciov
avtod Y where a Vatican MS. reads apifudv (Xylander notes that
his MS. had in this place wjre 6 Svmhaciowv avrod ap uo a...)
It is thus clear that the MS. (Paris. 2370) which Bachet used
sometimes has the sign for @ptfuss in a form which is at least
sufficiently like | to be taken for it. Tannery states that the form
of the sign found in the Madrid MS. (4) is Y, while 5, has it in a
form (S) nearly approaching Bachet’s reproduction of it.

It appeared also that the use of the sign, or something like
it, was not confined to MSS. of Diophantus; on reference to .
Gardthausen, Griechische Palaeographie, 1 found under the head
“hieroglyphisch-conventionell ” an abbreviation §, ¢¢ for dpiBuds,
-ot, which is given as occurring in the Bodleian MS. of Euclid
(D’Orville 301) of the gth century. Similarly Lehmann® notes as
a sign for apifuos found in that MS. a curved line similar to that
which was used as an abbreviation for «ai. He adds that the
ending is placed above it and the sign is doubled for the plural.
Lehmann’s facsimile is like the form given by Gardthausen, but has
the angle a little more rounded. The form yy above mentioned
is also given by Lehmann, with the remark that it seems to be
only a modification of the other. Again, from the critical notes to
Heiberg’s texts of the Arenarius of Archimedes it is clear that the
sign for apiBuds occurred several times in the MSS. in a form
approximating to that of the final sigma, and that there was the
usual confusion caused by the similarity of the signs for aptfuos
and kai. In Hultsch’s edition of Heron, similarly, the critical
notes to the Geodaesia show that one MS. had an abbreviation for

1 Lehmann, Die tachygraphischen Abkiirzungen der griechischen Handschriften, 1880,
p- 107 : “Von Sigeln, welchen ich auch anderwirts begegnet bin, sind zu nennen dptfuds,
das in der Oxforder Euclidhandschrift mit einer der Note kai dhnlichen Schlangenlinie
bezeichnet wird.”

2 Cf. Heiberg, Quaestiones Archimedeae, pp. 172, 174, 187, 188, 191, 192 5; Archimedis
opera omnia, 11., pp. 268 sqq.

3—2
36 INTRODUCTION

dpifuds in various forms with the case-endings superposed ; some-
times they resembled the letter § sometimes p, sometimes O and
once £1. Lastly, the sign for dpifuds resembling the final sigma
evidently appeared in a MS. of Theon of Smyrna

All these facts strongly support the assumption that the sign
was a mere tachygraphic abbreviation and not an algebraical
symbol like our x, though discharging much the same function.
The next question is, what is its origin? The facts (1) that the
sign has the breathing prefixed in the Bodleian MS., which writes
’¢5 for apufuds, and (2) that in one place Xylander’s MS. read ap
tor the full word, suggested to me the question whether it could
be a contraction of the first two letters of «¢pifuos ; and, on con-
sideration, this seemed to me quite possible when I found a
contraction for ap given by Gardthausen, namely ¢p. It is easy to
see that a simplification of this in different ways would readily
produce signs like the different forms shown above. This then
was the hypothesis which I put forward twenty-five years ago, and
which I still hold to be the easiest and best explanation. Two
alternatives are possible. (1) Diophantus may not have made the
contraction himself. In that case I suppose the sign to be a cur-
sive contraction made by scribes ; and I conceive it to have come
about through the intermediate form $. The loss of the downward
stroke, or of the loop, would produce a close approximation to
the forms which we know. (2) Diophantus may have used a sign
approximately, if not exactly, like that which we find in the MSS.
For it is from a papyrus of 154 A.D., in writing of the class which
Gardthausen calls the “Majuskelcursive,” that the contraction ¢j? for
the two letters is taken. The great advantage of my hypothesis is
that it makes the sign for apfuos exactly parallel to those for the
powers of the unknown, eg, 4" for Svwaus and K* for «980s, and

to that for the unit wovas which is denoted by M, with the sole
difference that the letters coalesce into one instead of being
written separately.

Tannery’s views on the subject are, I think, not very con-
sistent, and certainly they do not commend themselves to me. He
seems to suggest that the sign is the ancient letter Koppa, perhaps
slightly modified ; he first says that the sign in Diophantus is
peculiar to him and that, although the word apfuds is very often

! Heron, ed. Hultsch, pp. 146, 148, 149, 150.
2 Theon of Smyrna, ed. Hiller, p. 56, critical notes.
NOTATION AND DEFINITIONS OF DIOPHANTUS 37

represented in mathematical MSS. by an abbreviation, it has much
oftener the form §' or something similar, closely resembling the
ancient Koppa. In the next sentence he seems to say that “on
the contrary the Diophantine abbreviation is an inverted di-
gamma”; yet lower down he says that the copyist of a (copied
from the archetype a) got the form Y by simplifying the more
complicated Koppa. And, just before the last remark, he has
stated that in the archetype a the form must have been S or very
like it, as is shown by the confusion with the sign for kai. (If this
is so, it can hardly have been peculiar to Diophantus, seeing that
the same confusion occurs fairly often in the MSS. of other
authors, as above shown.) I think the last consideration (the con-
fusion with «al) is very much against the Koppa-hypothesis ; and,
in any case, it seems to me very unlikely that a sign would be
used by Diophantus for the unknown which was already appro-
priated to the number go. And I confess I am unable to see in
the sign any resemblance to an inverted digamma.

Hultsch! regards it as not impossible that Diophantus may
have adopted one of the signs used by the Egyptians for their
unknown quantity Zax, which, if turned round from left to right,
would give Y; but here again I see no particular resemblance.
Prof. D’Arcy Thompson? has a suggestion that the sign might be
the first letter of cwpds, a heap. But, apart from the fact that the
final sigma (5) is not that first letter, there is no trace whatever
in Diophantus of such a use of the word cwpos; and, when
Pachymeres® speaks of a number being cwpela povadwr, he means
no more than the mA7fos povddwr which he is explaining : his
words have no connexion with the Egyptian au.

Notwithstanding that the sign is not the final sigma, I shall
not hesitate to use s for it in the sequel, for convenience of
printing. Tannery prints it rather differently as s.

We pass to the notation which Diophantus used to express the
different powers of the unknown quantity, corresponding to 2%, x?
and so on. He calls the square of the unknown quantity Swaps,
and denotes it by the abbreviation 4%. The word Suvauss,
literally “ power,” is constantly used in Greek mathematics for

1 Art. Diophantus in Pauly-Wissowa's Aecal-Encyclopidie der classischen Altertums-
wissenschaften.

2 Transactions of the Royal Society of Edinburgh, Vol. XXXVIII1. (1896), pp. 607-9.

3 Dioph. 11. p. 78, 4. Cf. Iamblichus, ed. Pistelli, p. 7,7; 34, 3; 81, 14, where cwpela,
is similarly used to elucidate mwA7fos.
38 INTRODUCTION

square’. With Diophantus, however, it is not any square, but only
the square of the unknown; where he speaks of any particular
square number, it is Terpdywvos apiBuss. The higher powers of the
unknown quantity which Diophantus makes use of he calls xvf3os,
SuvapoSivauss, SuvaudruBos, kuBdékuBos, corresponding respectively
to 2% a4, 2% 2% Beyond the sixth power he does not go, having
no occasion for higher powers in the solutions of his problems. For
these powers he uses the abbreviations K¥, 4¥d, 4dK%, K'K re-
spectively. There is a difference between Diophantus’use of the word
Svvaues and of the complete words for the third and higher powers,
namely that the latter are not always restricted like 8vauis to powers
of the unknown, but may denote powers of ordinary known num-
bers as well. This is no doubt owing to the fact that, while there
are two words 8vvaus and TeTpdywros which both signify “square,”
there is only one word for a third power, namely «JBos. It is
«important, however, to observe that the abbreviations ZO A%4,
AK”, KK, are, like 8twams and AY, only used to denote powers
of the unknown. The coefficients of the different powers of the
unknown, like that of the unknown itself, are expressed by the
addition of the Greek letters denoting numerals, e.g., AK” ks cor-
responds to 264°. Thus in Diophantus’ system of notation the signs
A" and the rest represent not merely the exponent of a power like
the 2 in 2%, but the whole expression 2? There is no obvious
connexion between the symbol 4 and the symbol s of which it is
the square, as there is between 2? and x, and in this lies the great
inconvenience of the notation. But upon this notation no advance
was made by Xylander, or even by Bachet and Fermat. They wrote
N (which was short for NVumerus) for the s of Diophantus, Q (Quad-
ratus) for 4%, C (Cubus) for K*, so that we find, for example,
10 + 5/V = 24, corresponding to x% + 5x =24. Other symbols were
however used even before the publication of Xylander’s Diophantus,
eg. in Bombelli’s Algebra. Bombelli denotes the unknown and its
powers by the symbols 1 2 2 and so on. But it is certain that
up to this time (1572) the common symbols had been R (Radix
or Res), Z (Zensus, te. square), C (Cubus). Apparently the first
important step towards 2% 1% etc, was taken by Vieta (1540—

! In Plato we have &twaus used for a square number (Zimaews, 31) and also
(Zeaetetus, 147 D) for a square root of a number which is not a complete square, 7.e. for
a surd ; but the commonest use is in geometry, in the form duvduer, “in square,” e.g. “AB
is durduer double of AC” means ‘“ 452 = 2 BC."
NOTATION AND DEFINITIONS OF DIOPHANTUS 39

1603), who wrote Ag, Ac, Aqq, etc. (abbreviated for A quadratus
and so on) for the powers of A. This system, besides showing the
con .exion between the different powers, has the infinite advantage
that by means of it we can use in one and the same solution any
number of unknown quantities. This is absolutely impossible with
the notation used by Diophantus and the earlier algebraists.
Diophantus in fact never uses more than one unknown quantity in
the solution of a problem, namely the aptfuds or s.

Diophantus has no symbol for the operation of multiplication ;
it is rendered unnecessary by the fact that his coefficients are all
definite numbers or fractions, and the results are simply put down
without any preliminary step which would call for the use of a
symbol. On the ground that Diophantus uses only numerical
expressions for coefficients instead of general symbols, it might
occur to a superficial observer that there must be a great want
of generality in his methods, and that his problems, being solved
with reference to particular numbers only, would possess the
attraction of a clever puzzle rather than any more general interest.
The answer to this is that, in the first place, it was absolutely
impossible that Diophantus should have used any other than
numerical coefficients, for the reason that the available symbols of
notation were already employed, the letters of the Greek alphabet
always doing duty as numerals, with the exception of the final s.
In the second place, it is not the case that the use of none but
numerical coefficients makes his solutions any the less general.
This will be clearly seen when I come to give an account of his
problems and methods.

Next as to Diophantus’ expressions for the operations of
addition and subtraction. For the former no symbol at all is
used : it is expressed by mere juxtaposition, thus Ka 4d" uyse-
corresponds to x?+ 13224 52. In this expression, however, there
is no absolute term, and the addition of a simple numeral, as
for instance B, directly after € the coefficient of 5, would cause
confusion. This fact makes it necessary to have some expression
to distinguish the absolute term from the variable terms. For this
purpose Diophantus uses the word wovades, or units, and denotes

them after his usual manner by the abbreviation #. The number
of units is expressed as a coefficient. Thus corresponding to
the expression #%+ 132% + 5x +2 we should find in Diophantus

Kz A wet MB As Bachet uses the sign + for addition, he -
40 INTRODUCTION

has no occasion for a distinct symbol to mark an absolute term.
He accordingly writes 1C+ 130 +5V +2. It is worth observing,
however, that the Italians do use a symbol in this case, namely NV
(Numero), the first power of the unknown being with them R
(Radice). Cossali! makes an interesting comparison between the
terms used by Diophantus for the successive powers of the unknown
and those employed by the Italians after their instructors, the
Arabians. He observes that Fra Luca (Paciuolo), Tartaglia, and
Cardano begin their scale of powers from the power o, not from the
power 1, as does Diophantus, and he compares the scales thus:

Scala Diofantea. Scala Araba.
Jess nent an vesmonkns us 1. Numero...il Noto.
x 1. Numero...’ Ignoto. 2. Cosa, Radice, Lato.
x2 2. Podesta. 3. Censo.
x2 3... Cubo. 4. Cubo,
x% 4. Podesta-Podesta. 5. Censo di Censo.
x5 5. Podesta-Cubo. 6. Relato 1°
a% 6. Cubo-Cubo. 7. Censo di Cubo, o Cubo di Censo.
Alin sh es In 8. Relato 2°.
ATL i Bia crete 9. Censo di Censo di Censo.
Ee Le to. Cubo di Cubo.

and so on. So far, however, as this is meant to be a comparison
between Diophantus and the early Arabian algebraists themselves
(as the title “ Scala Araba” would seem to imply), there appears to
be no reason why Cossali should not have placed some term to
express Diophantus’ povades in the same line with Numero in the
Arabian scale, and moved the numbers 1, 2, 3, etc. one place
upwards in the first scale, or downwards in the second. As
Diophantus does not go beyond the sixth power, the last three
places in the first scale are left blank. An examination of these
two scales will show also that the evolution of the successive
powers differs in the two systems. The Diophantine terms for
them are based on the addition of exponents, the Arabic on

1 Upon Wallis’ comparison of the Diophantine with the Arabian scale Cossali
remarks : ‘ma egli non ha riflettuto a due altre differenze tra le scale medesime. La
prima si ¢, che laddove Diofanto denomina con singolarita Numero il numero ignoto,
denominando Monade il numero dato di comparazione : gli antichi italiani degli arabi
seguaci denominano questo il Numero; e Radice, o Lato, o Cosa il numero sconosciuto.
La seconda ¢, che Diofanto comincia la scala dal numero ignoto; e Fra Luca, Tartaglia,

Cardano la incominciano dal numero noto. Ecco le due scale di rincontro, onde meglio
risaltino all’ occhio le differenze loro”, 1. p. 193.
NOTATION AND: DEFINITIONS OF DIOPHANTUS 41

their multiplication’. Thus the “ cube-cube” means in Diophantus
2°, while the Italian and Arabian system uses the expression “ cube
of cube” and applies it to 2%. The first system may (says Cossali)
be described as the method of representing each power by the
product of the two lesser powers which are the nearest to it, ze
method of multiplication ; the second the method of elevation, i.e. the
method which forms by the process of squaring and cubing all
powers which can be so formed, as the 4th, 6th, 8th, oth, etc.
The intermediate powers which cannot be so formed are called
in Italian Relati.. Thus the fifth. power is Relate 1°, #7 is
Relato 2°, 2 is Censo di Relato 1°, 22 is Relato 3°, and so on.
Another name for the Relati in use among European algebraists in
the 16th and 17th centuries was sursolida, with the variants super-
solida and surdesolida.

It is interesting to compare with these systems the Egyptian
method described by Psellus?. The next power after the fourth
(Suvapodivauis), te. 2° the Egyptians called “the first undescribed ”
(@M\oyos here apparently meaning that of which no account can
be given), because it is neither a square nor a cube; alternatively
they called it “the fifth number,” corresponding to the fifth power
of x. The sixth power they apparently called “cube-cube”; but
the seventh was “the second undescribed” (d\oyos SedTepos), as
being the product of the square and the “first undescribed,” or,
alternatively, the “seventh number.” The eighth power was the
“quadruple-square” (reTpam\i} Svuvaus), the ninth the “extended
cube” (kvBos éEeniktos). Thus the “first undescribed” and the
“ second undescribed” correspond to “ Relato 1°” and “ Relato 2°”
respectively, but the “quadruple-square” exhibits the additive
principle.

For subtraction Diophantus uses a symbol. His full term for
negation or wanting is Aetyrs, corresponding to dmapfis which
denotes the opposite. The symbol used to denote it in the MSS,
and corresponding to our — for maznus, is (Def. 9 kati 11s Aeirews
anuetov WV é\\mes kdTw vetov, A) “an inverted ¥ with the top

1 This statement of Cossali’s needs qualification however. There is at least one Arabian
algebraist, al-Karkhi (died probably about 1029), the author of the Fakkrz, who uses the
Diophantine system of powers of the unknown depending on the addition of exponents.
Al-Karkhi, namely, expresses all powers of the unknown above the third by means of
mal, his term for the square, and 4a'é, his term for the cube of the unknown, as follows.
~The fourth power is with him 247 mal, the fifth mal ka'é, the sixth ka'6 ka'b, the seventh
mal mal ka'b, the eighth mal ka'b ka'b, the ninth 4a'b ka'b ka'b, and so on. Among the

Italians too there was an exception, Leonardo of Pisa, who proceeded on the additive
principle (Bibliotheca Mathematica, Vi, 1905-6, p. 310). 2 Dioph. 11. p. 37-38.
42 INTRODUCTION

shortened, A.” As Diophantus uses no distinct sign for +, it
is clearly necessary, in order to avoid confusion, that all the
negative terms in an expression should be placed together
after all the positive terms. And so in fact he does place them.
Thus corresponding to 2 — 522+ 8x — 1, Diophantus would write

Kray ad alra With respect to this curious sign, given in
the MSS. as T and described as an inverted truncated W, I believe
that I was the first to suggest that it could not be what it is
represented as being. Even when, as in Bachet’s edition, the
sign was printed as yn I could not believe that Diophantus used
so fantastic a sign for minus as an inverted truncated ¥. In
the first place, an inverted ¥ seems too far-fetched; to one who
was looking for a symbol to express minus many others more
natural and less fantastic than sp must have suggested themselves.
Secondly, given that Diophantus used an inverted WV, why should he
truncate it? Surely that must have been unnecessary; we could
hardly have expected it unless, without it, confusion was likely
to arise; but \p could not well have been confused with anything.
This very truncation itself appears to throw doubt on the description
of the symbol as we find it in the MS. I concluded that the con-
ception of this symbol as an inverted truncated ¥ was a mistake,
and that the description of it as such is not Diophantus’ description,
but an explanation by a scribe of a symbol which he did not
understand’. I believe that the true explanation is the following.
Diophantus here took the same course as in the case of the other
symbols which we have discussed (those for apifuos, Svvauis, etc.).
As in those cases he took for his abbreviation the first letter of the
word with such an addition as would make confusion with numbers
impossible (namely the second letter of the word, which in each of
the cases happens to come later in the alphabet than the corre-
sponding first letter), so, in seeking an abbreviation for eins
and cognate inflected forms developed from Xm, he began by
taking the initial letter of the word. The uncial® form is A.
Clearly A by itself would not serve his purpose, since it denotes
a number. Therefore an addition is necessary. The second letter
is E, but AE is equally a number. The second letter of the stem

! I am not even sure that the description can be made to mean all that it is intended
to mean. eX\urés scarcely seems to be sufficiently precise. Might it not be applied to
A with any part cut off, and not only the top ?

* I adhere to the uncial form above for clearness’ sake. If Diophantus used the
‘“ Majuskelcursive ” form, the explanation will equally apply, the difference of form being
for our purpose negligible,
NOTATION AND DEFINITIONS OF DIOPHANTUS 43

Aeris 1, but Al is open to objection when so written. Hence
Diophantus placed the | zuside the A, thus, A. Of the possibility
of this I entertain no doubt, because there are undoubted cases
of combination, even in uncial writing, of two letters into one sign.
I would refer in particular to X, which is an uncial abbreviation for
TAAANTON. Now this sign, A, is an inverted and truncated W
(written in the uncial form, Vv); and we can, on this assumption,
easily account for the explanation of the sign for minus which is
given in the text.

The above suggestion, made by me twenty-five years ago,
seems to be distinctly supported by what Tannery says of the form
in which the sign appears in the MSS.! Thus he remarks (1) that
the sign in the MSS. is often made to lean to the right so that it
resembles the letter Lambda, (2) that Planudes certainly wrote Xx as
if he meant to write the first letter of Aelyrer, and (3) that the
letter A appears twice in 4 where it seems to mean Aouméds. Yet
in his edition of Diophantus Tannery did not adopt my explanation
or even mention it, but explained the sign as being in reality
adapted from the old letter Sampi (7), the objection to which
suggestion is the same as that to which the identification of s with
Koppa is open, namely that ? represented the number goo, as ¢
represented 9o. Tannery however afterwards? saw reason to
abandon his suggestion that-the symbol was originally an archaic
form of the Greek Sampi rather than “un monogramme se
rattachant a la racine de Neiyris.” The occasion for this change
of view was furnished by the appearance of the same sign in the
critical notes to Schone’s edition of the Metrica of Heron? which
led Tannery to re-examine the evidence of the MSS. of Diophantus
as to the sign and as to the exact word or words which it re-
presented in different places, as well as to search for any similar
expressions denoting subtraction which might occur in the works
of other Greek mathematicians. In the MSS. of Diophantus,
when the sign is resolved by writing a full word instead of it,
it is generally resolved into \elyrei, the dative of \efyrs; in such
cases the only grammatical possibility is to construct it with the
genitive case of the quantity subtracted, the meaning then being
“with the wanting, or deduction, of ...”. But the best MS. (A4)

1 Dioph. 11..p. xli.

2 Bibliotheca Mathematica Vy, 1904—5, pp. 5-8.

 Heronis Alexandrini opera, Vol. 111., 1903, pp, 156, 8, 10. The MS. reading is
povddwy 08 I 8’, the meaning of which is 74 — 4.
44 INTRODUCTION

has in some places the nominative Aeiyrcs, while in others it has the
symbol instead of parts of the verb \elmew, namely Mmwv or
Aeiyras and once even AMmwoe; hence we may conclude that in the
cases where 4 and B, have Aeiyre: followed by the accusative (which
is impossible grammatically) the sign was wrongly resolved, and
the full word should have been a participle or other part of the
verb Aelmew governing the accusative. The question therefore
arises whether Diophantus himself used the dative Aeiyrer at all
or whether it was introduced into the MSS. later. Certain it is
that the use is foreign to Classical Greek; but, even if it began
with Diophantus, it did not finally hold the field before the time of
Planudes. No evidence for it can be found in Greek mathe-
maticians before Diophantus. Ptolemy has in two places Aeinrav
and Aeimovoav respectively, followed by the accusative, and in
one case To amo THs [A Newpbév Imo Tod amo THs ZI (where the
meaning is Z[?- [A?). Consequently we cannot suppose that the
sign where it occurs in the Metrica of Heron represents the dative
Aetrer; it must rather stand for a participle, active or passive.
Tannery suggests that the full expression in that passage was
povdadwv 08 \ewpbévros TecoapakaiSexdrov, the participle being
passive and the construction being the genitive absolute; but I
think a perhaps better alternative would be wovddwv 08 Aenpracdv
Tecoaparaidékaror, where the active participle would govern
the accusative case of the term subtracted. From all this we
may infer that the sign had no exclusive reference to the sub-
stantive eis, still less to the dative case of that substantive, but
was a conventional abbreviation associated with the root of the
verb Aelmew. In these circumstances I think I may now fairly
claim Tannery as, substantially, a convert to my view of the
nature of the sign.

For division it often happens that no symbol is necessary,
ze. in the cases where the divisor divides the dividend without
a remainder. In other cases the quotient has to be expressed
as a fraction, whether the divisor is a specific number or contains
the variable. The case of division comes then under that of
fractions.

Fractions are represented in different ways according as they are
submultiples (fractions with unity as numerator) or not. In the
case of submultiples the Greeks did not write the numerator, but
only the denominator, distinguishing the submultiple from the
cardinal number itself by affixing a certain sign. In more recent
NOTATION AND DEFINITIONS OF DIOPHANTUS 453

MSS. a double accent was used for this purpose: thus 4” =3.
Diophantus follows this plan in the hypothesis and analysis of his
problems, though in the solutions he seems to have written the
numerator a and assimilated the notation to that used for other
fractions. The sign, however, added to the cardinal number to
express the submultiple takes somewhat different forms in A:
sometimes it is a simple accent, sometimes more elaborate, as ~
above the letter and to the right, or actually forming a continuation
of the numeral sign, eg. @ =}. Tannery adopts as the genuine
mark in Diophantus the affix X in place of the accent: thus 4X =1.
For 4 he writes L’ as being most suitable for the time of Diophantus,
though 4 has <—, sometimes without the dot.

Of the other class of fractions (numerator not unity) % stands by
itself, having a peculiar sign of its own ; curiously enough it occurs
only four times in Diophantus. A has a sign for it which was
confused with that for dptfuds in one place; Tannery judges from
the Greek mathematical papyrus of Achmim® that its original form
was ¢; he himself writes in his text the common form «. In the
rare cases where the first hand in the oldest MS. (4) has fractions
as such with numerator and denominator written in full, the
denominator is written above the numerator. Tannery therefore
adopts, in his text, this way of writing fractions, separating the

Ls
121

numerator and denominator by a horizontal line: thus pra =—.

K
It is however better to omit the horizontal line (cf. p in Kenyon
Papyri 11. No. cclxv. 40; also the fractions in Schone’s edition
of Heron's Metrica). Once we find in the same MS. (4) in the first
hand the form (e®=15. In this latter method of writing fractions
the denominator is written as we write exponents ; and this is the
method adopted by Planudes and by Bachet in his edition.
Another alternative is to write the numerator first, and then the
denominator after it in the same line, marking the denominator with
the submultiple sign in some form; thus 48 would mean £; this is
the most convenient method for purposes of printing. Or the de-
nominator may be written as an abbreviation for the ordzna/ number,
and the case-termination may be added higher up; eg. v xy" = 50
twenty-thirds. But the denominators are nearly always omitted

1 Published by Baillet in Mémoires publiés par les Membres de la Mission archéologique
Srangaise au Caire, T. 1x, Fascicule 1, pp. 1-88. Paris, 1892.
46 INTRODUCTION

altogether in the first hand of 4 ; in the first two Books B, and the
second hand of 4 give the denominator in the place in which we write
an exponent, following the method of Planudes; in the last four
Books both MSS. almost invariably omit the denominator. In
some cases the omission is not unnatural, ze. where the denominator
has once been given, and it is almost superfluous to repeat it
in other fractions immediately following which have the same
denominator ; in other cases it was probably omitted because the
superposed denominator was taken by the copyist to be an inter-
linear scholium. A few examples of fractions from Diophantus
may be added:

 

 

Ie] 17 bu 2456 a.ca

tl = 25 (Vv. 10); Busi (1v. 28); erm = Hr ry 0)
Byrd an 1

vy. SxXKQ = oe (IV. 16); = pia” awl’ = SEL {1v. 30);

[276] 1
Trl’ = 389% Vv. 2),
142

Diophantus however often expresses fractions by putting év
poplew or poplov between the numerator and denominator, ze. he

y asm
says one number divided by another. Cf. Mpv. {AmS popiov

r
ks. Bpud = 1507984/262144 (1V. 28), where of course M = pvpiddes
(tens of thousands); 8. EX €v poplw pKf Jaxe = 25600/1221025
(v. 22). As we said, the most orthodox way of writing a sub-
multiple was to omit the numerator (unity) and use the denominator
with a distinguishing sign attached, eg. sX or ¢'=%. But in his
solutions Diophantus often uses the form applicable to fractions
Pi I

other than submultiples; eg. he writes -a for oT (Iv. 28).

Numbers partly integral and partly fractional, where the
fraction is a submultiple or the sum of submultiples, are written
much as we write them, the fraction simply following the integer ;
eg. ayX=1}; in the Lemma to v. 8 we have BL’ ¢'=24} or 23,
where % is decomposed into submultiples as in Heron. Cf. also
(Hl. 11) rol. 1s X= 3704 +.

Before leaving the subject of numerical notation, it may be
convenient to refer to the method of writing large numbers.

>
Myriads (tens of thousands) are expressed by AZ, myriads to the
NOTATION AND DEFINITIONS OF DIOPHANTUS 44

second power by MM" or, in words, Sevrépa pvpids. The de-
nominator 187474560 in V. 8 would thus be written popiov Sevrépas

updos a kal pupiddwv mpodTov mut kal M SHE, and the fraction
131200224/1629586560 would be written Sevrépa pvpias a mpwTar

(nvpiades) yp Mm Hakd poplov Sevrépwv wupiidwv 1s TpGTWY

(uvpradwv) B Avy M SHEL

But there is another kind of fraction, besides the purely
numerical one, which is continually occurring in the Arithmetica,
such fractions namely as involve the unknown quantity in some
form or other in their denominators. The simplest case is that in
which the denominator is merely a power of the unknown, s.
Concerning fractions of this kind Diophantus says (Def. 3): “As
fractions named after numbers have similar names to those of the
numbers themselves (thus a third is named from three, a fourth
from four), so the fractions homonymous’ with the numbers just
defined are called after them; thus from dpfucs we name
the fraction apifpootév [te. 1/x from x], 70 Suvamootév from
Svvauts, 70 kvBooTov from «VBos, To JSuvvauoduvvapooTov from’
Suvapoduvaus, To SuvapmorvBosTov from SuvvamoxvBos, and To
kvBokvBootov from xvBoxkvBos. And every such fraction shall
have, above the sign for the homonymous number, a line to
indicate the species.” Thus we find, for example, 1v. 3, ¢X % cor-
responding to 8/x and, Iv. 15, sX Ae for 35/2. Cf. 4" av for 250/22

Where the denominator is a compound expression involving the
unknown and its powers, Diophantus uses the expedient which he
often adopts with numerical fractions when the numerators and
denominators are large numbers, namely the insertion of év uopiw
or pmoplov between the expressions for the numerator and de-
nominator. Thus in VI. 12 we have

4% EM Binley opie 4” 4 a M A AE
= (604% + 2520)/(x* + 900 — 6012),
and in VL 14
AT BAM opie 4" 4 aM ne A 4" 8
; = (152° — 36)/(2* + 36 — 122%).
For loos, equal, connecting the two sides of an equation, the

sign in the archetype seems to have been «7; but copyists intro-

1 Hultsch, Joc. cit.
48 INTRODUCTION

duced a sign which was sometimes confused with the sign y for
aptBuos ; this was no doubt the same abbreviation Y as that shown
(with terminations of cases added above) in the list given at
the end of Codex Parisinus 2360 (Archimedes) of contractions
found in the “very ancient” MS. from which it was copied and
which was at one time the property of Georgius Vallal,

Diophantus evidently put down his equations in the ordinary
course of writing, ze. they were written straight on, as are the
steps in the propositions of Euclid, and not put in separate lines for
each step in the process of simplification. In the scholia of
Maximus Planudes however we find conspectuses of the problems
with steps in separate lines which, except for the slightly more
cumbrous notation, make the work scarcely more difficult to follow
than it is in our notation. Though in the MSS. we have the
abbreviation ¢* to denote equality, Bachet makes no use of any
symbol for the purpose in his Latin translation. He uses
throughout the full Latin word. It is interesting however to observe
that in the notes to his earlier translation (1575) Xylander had
already used a symbol to denote equality, namely ||, two short
vertical parallel lines. Thus we find, for example (p. 76),

10+12]|1Q+6N +09,
which we should express by 22+ 12 =22+ 6x + Q.

Now that we have described in detail Diophantus’ method of
expressing algebraical quantities and relations, it is clear that it is
essentially different in its character from the modern notation.
While in modern times signs and symbols have been developed

! Heiberg, Quaestiones Archimedeae, p. 115.

2 One instance will suffice. On the left Planudes has abbreviations for the words
showing the nature of the steps or the operations they involve, e.g. &0. = éfeats (setting-
out), TeTp. =TeTpaywriouds (squaring), givl. =oirbeots (adding), dp. =dgalpests (subtrac-
tion), uep. = pepiopés (division), vr. =Umrapées (resulting fact).

Dioph. 1. 28.
Planudes. Modern equivalent.
K on [Given numbers] 20, 208
£0. sa uo HOTA sa Put for the numbers x + 10, 10 — x.
rerp.  AYasskuop AY uop A ssk Squaring, we have x*+ 20x + 100,
x%+ 100 — 20x.

awd. AYBues 1° pogy Adding, 2242+ 200=1208.
ag. 4%B i poy Subtracting, 2x2=8.
pep. 478 i ued Dividing, x2=4.

sa $9 nop x=0.

vr. uel non Result: [the numbers are] 12, 8.
NOTATION AND DEFINITIONS OF DIOPHANTUS 49

which have no intrinsic relationship to the things which they
represent, but depend for their use upon convention, the case
is quite different in Diophantus, where algebraic notation takes
the form of mere abbreviation of words which are considered as
pronounced or implied.

In order to show in what place, in respect of systems of
algebraic notation, Diophantus stands, Nesselmann observes that
we can, as regards the form of exposition of algebraic operations
and equations, distinguish three historical stages of development,
well marked and easily discernible. (1) The first stage Nessel-
mann represents by the name Rhetorical Algebra or “reckoning by
complete words.” The characteristic of this stage is the absolute
want of all symbols, the whole of the calculation being carried on
by means of complete words, and forming in fact continuous prose.
As representatives of this first stage Nesselmann mentions Iambli-
chus (of whose algebraical work he quotes a specimen in his fifth
chapter) “and all Arabian and Persian algebraists who are at
present known.” In their works we find no vestige of algebraic
symbols; the same may be said of the oldest Italian algebraists
and their followers, and among them Regiomontanus. (2) The
second stage Nesselmann proposes to call the Syncopated Algebra.
This stage is essentially #/keforical, and therein like the first in
its treatment of questions; but we now find for often-recurring
operations and quantities certain abbreviational symbols. To
this stage belong Diophantus and, after him, all the later
Europeans until about the middle of the seventeenth century
(with the exception of Vieta, who was the first to establish,
under the name of Logistica speciosa, as distinct from Logistica
numerosa, a regular system of reckoning with letters denoting
magnitudes and not numbers only). (3) To the third stage
Nesselmann gives the name Symbolic Algebra, which uses a com-
plete system of notation by signs having no visible connexion
with the words or things which they represent, a complete language
of symbols, which supplants entirely the #/ketorical system, it being
possible to work out a solution without using a single word of the
ordinary written language, with the exception (for clearness’ sake)
of a connecting word or two here and there, and so on’. Neither

1 It may be convenient to note here the beginnings of some of our ordinary algebraical
symbols. © The signs + and — first appeared in print in Johann Widman’s arithmetic
(1489), where however they are scarcely used as regular symbols of operation ; next they
are found in the Rechenbuch of Henricus Grammateus (Schreiber), written in 1518 but
perhaps not published till 1521, and then regularly in Stifel’s 4rithmetica integra (1544)

H. D, 4
50 INTRODUCTION

is it the Europeans from the middle of the seventeenth century
onwards who were the first to use symbolic forms of Algebra.
In this they were anticipated by the Indians.

Nesselmann illustrates these three stages by three examples,
quoting word for word the solution of a quadratic equation
by Muhammad b. Misa as an example of the first stage, and
the solution of a problem from Diophantus as representing the
second.

First Stage. Example fromm Muhammad b. Misa (ed. Rosen,
p- 5). “A square and ten of its roots are equal to nine and thirty
dirhems, that is, if you add ten 7oofs to one square, the sum is equal
to nine and thirty. The solution is as follows. Take half the number
of roots, that is in this case five; then multiply this by itself, and
the result is five and twenty. Add this to the nine and thirty,
which gives sixty-four; take the square root, or eight, and subtract
from it half the number of roofs, namely five, and there remain
three: this is the roof of the square which was required, and the
square itself is nine.’

Here we observe that not even are symbols used for numbers,
so that this example is even more “7/etorical” than the work of
Iamblichus who does use the Greek symbols for his numbers.

 

as well as in his edition of Rudolff’s Coss (1553). Vieta (1540-1603) has, in addition,
= for ~. Robert Recorde (1510-1558) had already in his Algebra (Z%e Whetstone of
Witte, 1557) used =(but with much longer lines) to denote equality (* bicause noe .2.
thynges, can be moare equalle””). Harriot (1560-1621) denoted multiplication by a dot,
and also by mere juxtaposition of letters; Stifel (1487-1567) had however already
expressed the product of two magnitudes by the juxtaposition of the two letters represent-
ing them. Oughtred (1574-1660) used the sign x for multiplication. Harriot also
introduced the signs > and < for greater and less respectively. - for division is found
in Rahn’s Algebra (1659). Descartes introduced in his Geometry (1637) our method of
writing powers, as @% a* etc. (except a?, for which he wrote aa); but this notation was
practically anticipated by Pierre Hérigone (Cours mathématique, 1634), who wrote a2, a3,
a4, etc, and the idea is even to be found in the Rechenbuch of Grammateus above
mentioned, where the successive powers of the unknown are denoted by pri, se, ter, etc.
The use of x for the unknown quantity began with Descartes, who first used z, then y, and
then x for this purpose, showing that he intentionally chose his unknowns from the last
letters of the alphabet. / for the square root is traceable to Rudolff, with whom it had
only two strokes, the first (down) stroke being short, and the other relatively long.
! Thus Muhammad b. Miisa states in words the following solution.

x%+ 10x = 30,
x24 10x +25=064;
therefore’ x+5=8,

x=3.
NOTATION AND DEFINITIONS OF DIOPHANTUS gsi

Second Stage. As an example of Diophantus I give a trans-
lation word for word of 11. 8. So as to make the symbols correspond

exactly I use S (Square) for A¥ (8bvapss), N (Number) for 5, U

(Units) for M (novades).

“To divide the proposed square into two squares. Let it be
proposed then to divide 16 into two squares. And let the first be
supposed to be 1.5; therefore the second will be 16 U- 1S. Thus
16 U — 1.5 must be equal to a square. I form the square from any
number of /V’s minus as many U’s as there are in the side of
16 U’s. Suppose this to be 2V—4U. Thus the square itself will
be 485 16U-16A. These are equal to 16U/—1S. Add to each
the negative term (5 \etyrs, the deficiency) and take likes from
likes. Thus 3.5 are equal to 164, and the A is 16 fifths. One
[square] will be 236, and the other 4%, and the sum of the two
makes up 4%, or 16 U, and each of the two is a square.”

Of the third stage any exemplification is unnecessary.

To the form of Diophantus’ notation is due the fact that he
is unable to introduce into his solutions more than one unknown
quantity. This limitation has made his procedure often very dif-
ferent from our modern work. In the first place we can begin
with any number of unknown quantities denoted by different
symbols, and eliminate all of them but one by gradual steps in the
course of the work; Diophantus on the other hand has to perform
all his eliminations beforehand, as a preliminary to the actual
work, by expressing every quantity which occurs in the problem
in “terms of only one unknown. This is the case in the great
majority of questions of the first Book, which involve the solu-
tion of determinate simultaneous equations of the first degree
with two, three, or four variables; all these Diophantus expresses
in terms of one unknown, and then proceeds to find it from a
simple equation. Secondly, however, this limitation affects much of
Diophantus’ work injuriously ; for, when he handles problems which
are by nature indeterminate and would lead with our notation to an
indeterminate equation containing two or three unknowns, he is
compelled by limitation of notation to assume for one or other of
these some particular number arbitrarily chosen, the effect of the
assumption being to make the problem a determinate one. How-
ever, it is but fair to say that Diophantus, in assigning an arbitrary
value to a quantity, is careful to tell us so, saying, “for such and
such a quantity we put any number whatever, say such and such a

4—2
52 INTRODUCTION

number.” Thus it can hardly be said that there is (as a rule) any
loss of generality. We may say, then, that in general Diophantus is
obliged to express all his unknowns in terms, or as functions, of
. one variable. He compels our admiration by the clever devices
by which he contrives so to express them in terms of his single
unknown, s, as to satisfy by that very expression of them all
conditions of the problem except one, which then enables us to
complete the solution by determining the value of s. Another
consequence of Diophantus’ want of other symbols besides s to
express more variables than one is that, when (as often happens)
it is necessary in the course of a problem to work out a subsidiary
problem in order to obtain the coefficients etc. in the functions of
which express the numbers to be found, the unknown quantity
which it is the object of the new subsidiary problem to find is also
in its turn denoted by the same symbol s; hence we often have
in the same problem the same variable s used with two different
meanings. This is an obvious inconvenience and might lead to
confusion in the mind of a careless reader. Again we find two
cases, II. 28 and 29, where for the proper working-out of the
problem two unknowns are imperatively necessary. We should of
course use x and y; but Diophantus calls the first s as usual; the
second, for want of a term, he agrees to call “one unit” ie. 1.
Then, later, having completed the part of the solution necessary to
find 5, he substitutes its value, and uses s over again to denote
what he had originally called “1”—the second variable—and so
finds it. This is the most curious case of all, and the way in which
Diophantus, after having worked with this “1” along with other
numerals, is yet able to put his finger upon the particular place
where it has passed to, so as to substitute s for it, is very remark-
able. This could only be possible in particular cases such as those
which I have mentioned; but, even here, it seems scarcely possible
now to work out the problem by using x and 1 for the variables
as originally taken by Diophantus without falling into confusion.
Perhaps, however, in working out the problems before writing them
down as we have them Diophantus may have given the “1” which
stood for a variable some mark by which he could recognise it
and distinguish it from other numbers.

Diophantus will have in his solutions no numbers whatever
except “rational” numbers; and in pursuance of this restriction he
excludes not only surds and imaginary quantities, but also zegative
quantities. Of a negative quantity per se, 7.e. without some positive
NOTATION AND DEFINITIONS OF DIOPHANTUS 53

quantity to subtract it from, Diophantus had apparently no con-
ception. Such equations then as lead to surd, imaginary, or
negative roots he regards as useless for his purpose: the solution
is in these cases advvaros, impossible. So we find him (v. 2)
describing the equation 4 =4x + 20 as dromos, absurd, because it
would give x=—4. Diophantus makes it his object throughout
to obtain solutions in rational numbers, and we find him frequently
giving, as a preliminary, the conditions which must be satisfied in
order to secure a result rational in his sense of the word. In the
great majority of cases, when Diophantus arrives in the course of
a solution at an equation which would give an irrational result, he
retraces his steps and finds out how his equation has arisen, and
how he may, by altering the previous work, substitute for it
another which shall give a rational result. This gives rise, in
general, to a subsidiary problem the solution of which ensures
a rational result for the problem itself. Though, however, Dio-
phantus has no notation for a surd, and does not admit surd
results, it is scarcely true to say that he makes no use of quadratic
equations which lead to such results. Thus, for example, in V. 30
he solves such an equation so far as to be able to see to what
integers the solution would approximate most nearly.
CHAPTER IV

DIOPHANTUS’ METHODS OF SOLUTION

BEFORE I give an account in detail of the different methods
which Diophantus employs for the solution of his problems, so far
as they can be classified, it is worth while to quote some remarks
which Hankel has made in his account of Diophantus®. Hankel,
writing with his usual brilliancy, says in the place referred to, “The
reader will now be desirous to become acquainted with the classes
of indeterminate problems which Diophantus treats of, and with
his methods of solution. As regards the first point, we must observe
that included in the 130 (or so) indeterminate problems, of which
. Diophantus treats in his great work, there are over 50 different
classes of problems, strung together on no recognisable principle
of grouping, except that the solution of the earlier problems facili-
tates that of the later. The first Book is confined to determinate
algebraic equations; Books II. to V. contain for the most part
indeterminate problems, in which expressions involving in the first
or second degree two or more variables are to be made squares or
cubes. Lastly, Book VI. is concerned with right-angled triangles
regarded purely arithmetically, in which some linear or quadratic
function of the sides is to be made a square or a cube. That is all
that we can pronounce about this varied series of problems without
exhibiting singly each of the fifty classes. Almost more different
in kind than the problems are their solutions, and we are completely
unable to give an even tolerably exhaustive review of the different
turns which his procedure takes. Of more general comprehensive
methods there is in our author no trace discoverable: every
question requires a quite special method, which often will not
serve even for the most closely allied problems. It is on that

Y Zur Geschichte der Mathematik in Alterthum und Mittelalter, Leipzig, 1874,
pp. 104-5.
DIOPHANTUS METHODS OF SOLUTION 55

account difficult for a modern mathematician even after studying
100 Diophantine solutions to solve the 101st problem; and if we
have made the attempt, and after some vain endeavours read
Diophantus’ own solution, we shall be astonished to see how
suddenly he leaves the broad high-road, dashes into a side-path
and with a quick turn reaches the goal, often enough a goal with
reaching which we should not be content; we expected to have
to climb a toilsome path, but to be rewarded at the end by an
extensive view; instead of which our guide leads by narrow,
strange, but smooth ways to a small eminence; he has finished!
He lacks the calm and concentrated energy for a deep plunge
into a single important problem; and in this way the reader also
hurries with inward unrest from problem to problem, as in a
game of riddles, without being able to enjoy the individual one. \
Diophantus dazzles more than he delights. He is in a wonderful
measure shrewd, clever, quick-sighted, indefatigable, but does not
penetrate thoroughly or deeply into the root of the matter.” As
his problems seem framed in obedience to no obvious scientific
necessity, but often only for the sake of the solution, the solution
itself also lacks completeness and deeper signification. He is a
brilliant performer in the a77 of indeterminate analysis invented by
him, but the science has nevertheless been indebted, at least directly,
to this brilliant genius for few methods, because he was deficient
in the speculative thought which sees in the True more than the
Correct. That is the general impression which I have derived from
a thorough and repeated study of Diophantus’ arithmetic.”

It might be inferred from these remarks of Hankel that
Diophantus’ object was less to teach methods than to obtain a
multitude of mere results. On the other hand Nesselmann
observes! that Diophantus, while using (as he must) specific
numbers for numbers which are “given” or have to be arbitrarily
assumed, always makes it clear how by varying our initial as-
sumptions we can obtain any number of particular solutions of
the problem, showing “that his whole attention is directed to
the explanation of the method, to which end numerical examples
only serve as means”; this is proved by his frequently stopping
short, when the method has been made sufficiently clear, and
the remainder of the work is mere straightforward calculation.
The truth seems to be that there is as much in the shape of general

L Algebra der Griechen, pp. 308-9.
56 INTRODUCTION

methods to be found in Diophantus as his notation and the nature
of the subject admitted of. On this point I can quote no better
authority than Euler, who says’: “ Diophantus himself, it is true,
gives only the most special solutions of all the questions which he
treats, and he is generally content with indicating numbers which
furnish one single solution. But it must not be supposed that his
method was restricted to these very special solutions. In his
time the use of letters to denote undetermined numbers was not
yet established, and consequently the more general solutions which
we are now enabled to give by means of such notation could not
be expected from him. Nevertheless, the actual methods which he
uses for solving any of his problems are as general as those which
are in use today; nay, we are obliged to admit that there is
hardly any method yet invented in this kind of analysis of which
there are not sufficiently distinct traces to be discovered in Dio-
phantus.”

In his 8th chapter, entitled “ Diophantus’ treatment of equations?”
Nesselmann gives an account of Diophantus’ solutions of (1) Deter-
minate, (2) Indeterminate equations, classified according to their
kind. In chapter 9, entitled “Diophantus’ methods of solution?”
he classifies these “ methods” as follows*: (1) “ The adroit assump-
tion of unknowns,” (2) “Method of reckoning backwards and
auxiliary questions,” (3) “Use of the symbol for the unknown in
different significations,” (4) “Method of Limits,” (5) “Solution by
mere reflection,” (6) “Solution in general expressions,” (7) “ Arbi-
trary determinations and assumptions,” (8) “Use of the right-
angled triangle.”

At the end of chapter 8 Nesselmann observes that it is not
his solutions of equations that we have to wonder at, but the art,
amounting to virtuosity, which enabled Diophantus to avoid such
equations as he could not technically solve. We look (says Nessel-
mann) with astonishment at his operations, when he reduces the
most difficult problems by some surprising turn to a quite simple

Y Novi Commentarii Academiae Petropolitanae, 1756-7, Vol. VI. (1761), p. 155= Com-
mentationes arithmeticae collectae (ed. Fuss), 1849, I. p. 193.

? ¢ Diophant’s Behandlung der Gleichungen.”

3 “Diophant’s Auflésungsmethoden.”

* (1) ‘“Die geschickte Annahme der Unbekannten,” (2) ‘‘ Methode der Zuriick-
rechnung und Nebenaufgabe,” (3) “Gebrauch des Symbols fiir die Unbekannte in
verschiedenen Bedeutungen,” (4) * Methode der Grenzen,” (5) ‘“ Auflésung durch blosse
Reflexion,” (6) ‘ Auflosung in allgemeinen Ausdriicken,” (7) * Willkiihrliche Bestim-
mungen und Annahmen,” (8) * Gebrauch des rechtwinkligen Dreiecks.”
DIOPHANTUS METHODS OF SOLUTION 57

equation. Then, when in the gth chapter Nesselmann passes to the
“methods,” he prefaces it by saying: “To give a complete picture
of Diophantus’ methods in all their variety would mean nothing else
than copying his book outright. The individual characteristics of
almost every problem give him occasion to try upon it a peculiar
procedure or found upon it an artifice which cannot be applied to
any other problem....Meanwhile, though it may be impossible to
exhibit all his methods in any short space, yet I will try to describe
some operations which occur more often or are particularly re-
markable for their elegance, and (where possible) to bring out
the underlying scientific principle by a general exposition and by
a suitable grouping of similar cases under common aspects or
characters.” Now the possibility of giving a satisfactory account of
the methods of Diophantus must depend largely upon the meaning
we attach to the word “method.” Nesselmann’s arrangement seems
to me to be faulty inasmuch as (1) he has treated Diophantus’
solutions of equations, which certainly proceeded on fixed rules,
and therefore by “method,” separately from what he calls “methods
of solution,” thereby making it appear as though he did not
look upon the “treatment of equations” as “methods”; (2) the
classification of the “Methods of solution” seems unsatisfactory.
Some of the latter can hardly be said to be methods of solution at
all; thus the third, “ Use of the symbol for the unknown in different
significations,” might be more justly described as a “hindrance to
the solution”; it is an znconvenience to which Diophantus was
subjected owing to the want of notation. Indeed, on the as-
sumption of the eight “methods,” as Nesselmann describes them,
it is really not surprising that no complete account of them
could be given without copying the whole book. To take the
first, “the adroit assumption of unknowns.” Supposing that a
number of essentially different problems are proposed, the differences
make a different choice of an unknown in each case absolutely
necessary. That being so, how could a rule be given for all cases?
The best that can be done is to give a number of typical instances.
Precisely the same remark applies to “methods” (2), (5), (6), (7).
The case of (4), “ Method of Limits,” is different; here we have
a “method” in the true sense of the term, ze. in the sense of an
instrument for solution. And accordingly in this case the method
can be exhibited, as I hope to show later on; (8) also deserves
to some extent the name of a “method.”
8 INTRODUCTION

In one particular case, Diophantus formally states a method or
rule ; this is his rule for solving what he calls a “double-equation,”
and will be found in IL 11, where such an equation appears for the
first time. Apart from this, we do not find in Diophantus’ work
statements of method put generally as book-work to be applied to
examples. Thus we do not find the separate rules and limitations
for the solution of different kinds of equations systematically
arranged, but we have to seek them out laboriously from the
whole of his work, gathering scattered indications here and there,
and to formulate them in the best way that we can.

I shall now attempt to give a short account of those methods
running through Diophantus which admit of general statement.
For the reasons which I have stated, my arrangement will be
different from that of Nesselmann ; I shall omit some of the heads
in his classification of “methods of solution”; and, in accordance
with his remark that these “methods” can only be adequately
described by a transcription of the entire work, I shall leave them
to be gathered from a perusal of my reproduction of Diophantus’
book.

I shall begin my account with

I. DIOPHANTUS TREATMENT OF EQUATIONS.

This subject falls naturally into two divisions: (A) Determinate
equations of different degrees, (B) Indeterminate equations.

(A) Determinate equations.

Diophantus was able without difficulty to solve determinate
equations of the first and second degrees; of a cubic equation we
find in his Arithmetica only one example, and that is a very
special case. The solution of simple equations we may pass over;
we have then to consider Diophantus’ methods of solution in the
case of (1) Pure equations, (2) Adfected, or mixed, guadratics.

(1) Pure determinate equations.

By pure equations I mean those equations which contain only
one power of the unknown, whatever the degree. The solution is
effected in the same way whatever the exponent of the term in the
DIOPHANTUS’ METHODS OF SOLUTION 59

unknown; and Diophantus treats pure equations of any degree
as if they were simple equations of the first degree.

He gives a general rule for this case without regard to the
degree!: “If a problem leads to an equation in which any terms
are equal to the same terms but have different coefficients, we must
take like from like on both sides, until we get one term equal to
one term. But, if there are on one side or on both sides any negative
terms, the deficient terms must be added on both sides until all the
terms on both sides are positive. Then we must take like from like
until one term is left on each side.” After these operations have
been performed, the equation is reduced to the form Ax™= 5 and
is considered solved. The cases which occur in Diophantus are
cases in which the value of x is found to be a rational number,
integral or fractional. Diophantus only recognises one value of x
which satisfies this equation; thus, if 72 is even, he gives only the
positive value, excluding a negative value as “impossible.” In the
same way, when an equation can be reduced in degree by dividing
throughout by any power of x, the possible values, x=o0, thus
arising are not taken into account. Thus an equation of the form
2*= ax, which is of common occurrence in the earlier part of the
book, is taken to be merely equivalent to the simple equation r=a.

It may be observed that the greater proportion of the problems
in Book I. are such that more than one unknown quantity is sought.
Now, when there are two unknowns and two conditions, both
unknowns can easily be expressed in terms of one symbol. But,
when there are three or four quantities to be found, this reduction
is much more difficult, and Diophantus shows great adroitness in
effecting it: the ultimate result being that it is only necessary
to solve a simple equation with one unknown quantity.

2) Mixed quadratic equations.
q q

After the remarks in Def. 11 upon the reduction of equations
until we have one term equal to another term, Diophantus
adds?: “But we will show you afterwards how, in the case also
when two terms are left equal to a single term, such an equation
can be solved.” That is to say, he promises to explain the
solution of a mixed quadratic equation. In the Arithmetica,
as we possess the book, this promise is not fulfilled. The first

1 Def. 11.
2 lorepov 0¢é gou delfoper kal mds vo eld@v lowy évl kaTaketpfévTwy TO TowolTOV NUeTal.
60 INTRODUCTION

indications we have on the subject are a number. of cases in which
the equation is given, and the solution written down, or stated to
be rational, without any work being shown. Thus

(IV. 22) 2*=4x — 4, therefore r= 2;
(Iv. 31) 3252° = 3x + 18, therefore x = Ff or f;
(VL 6) 842+ 7x =7, whence xr =};
(VL. 7) 84x*— 7x=7, hence xr =1{;
(VL 9) 630x*— 73x =06, therefore x =F ;
and (VL 8) 630x%+ 73x =06, and x is rational.

These examples, though proving that Diophantus had somehow
arrived at the result, are not in themselves sufficient to show that
he was necessarily acquainted with a regular method for the
solution of quadratics; these solutions might (though their variety
makes it somewhat unlikely) have been obtained by mere #rial.
That, however, Diophantus’ solutions of mixed quadratics were not
merely empirical is shown by instances in v. 30. In this problem
he shows that he could approximate to the root in cases where it is
not “rational.” As this is an important point, I give the substance
of the passage in question: “This is not generally possible unless
we contrive to make x >4(#*—60) and <}(2*— 60). Let then
2? — 60 be > 5x, but 2? —60< 8x. Since then 2?— 60 > 52, let 60 be
added to both sides, so that 2? > 5x + 60, or 2? = 5x + some number
> 60; therefore + must not be less than 11.” In like manner
Diophantus concludes that “4? = 8x + some number less than 60;
therefore x must be found to be not greater than 12.”

~ Now, solving for the positive roots of these two equations, we
have

>} (5+4/265) and x < 4+ 4/76,
or 2> 100304... and .z< 127177...

It is clear that + may be < 11 or >12, and therefore Dio-
phantus’ limits are not strictly accurate. As however it was
doubtless his object to find znzegral limits, the limits 11 and 12
are those which are obviously adapted for his purpose, and are
a fortior: safe.

In the above equations the other roots obtained by prefixing
the negative sign to the radical are negative and therefore would
be of no use to Diophantus, In other cases of the kind occurring
DIOPHANTUS METHODS OF SOLUTION 61

in Book V. the equations have both roots positive, and we have to
consider why Diophantus took no account of the smaller roots in
those cases.

We will take first the equations in Vv. 10 where the inequalities

to be satisfied are
TAX NTA hh hd (1)

TZEXIO FIO eee (2).
Now, if a, B be the roots of the equation
2% — px + g = 0 (p, ¢ both positive),
and if a > 3, then
(a) in order that 22 — px + ¢ may be >o0
we must have 2 > a or <3,
and (4) in order that 2* — px + ¢ may be <o
we must have x <a and > 8.
(1) The roots of the equation
172 = 722417 =0

 

 

 

are 36% V1007. that is, 67:73. Sd P20
r 17 17 17
and, in order that 172? — 72x 4 17 may be <0, we must have
ra T0000
17 17

~ (2) The roots of the equation

104% 722 4- 10=0

are 36 £ v3; i Voss. that is, 66577... and Sue
I9 19 19
and, in order that 192? — 722 + 19 may be > 0, we must have
a 661577... i 5422...

19

Diophantus says that x must not be greater than $f or less than
86. These are again doubtless intended to be a fortiori limits;
but $§ should have been ¢7, and the more correct way of stating the
case would be to say that, if # is not greater than $f and not less
than $f, the given conditions are a fortiori satisfied.

Now consider what alternative (if any) could be obtained, on
Diophantus’ principles, if we used the lesser positive roots of the
62 INTRODUCTION

equations. If, like Diophantus, we were to take a fortiori limits,
we should have to say
x < {5 but > £,
which is of course an impossibility. Therefore the smaller roots
are here useless from his point of view.
This is, however, not so in the case of another pair of in-
equalities, used later in Vv. 30 for finding an auxiliary x, namely
2 +60>22x,
2+ 60< 242.
The roots of the equation
xt —22x+4+60=0
are 11 + V61; that's, 1331... and 318...;
and the roots of the equation 2% — 24x +60=0
are 12 + V34; that is, 21°16... and 2'33....
In order therefore to satisfy the above inequalities we must have
22> 1831...0r < 318...,
and x= 21°16 ... but > 2°33,

Diophantus, taking a fortiori integral limits furnished by the
greater roots, says that x must not be less than 19 but must be
less than 21. But he could also have obtained from the smaller
roots an integral value of x satisfying the necessary conditions,
namely the value x= 3; and this would have had the advantage
of giving a smaller value for the auxiliary x than that actually
taken, namely 20. Accordingly the question has been raised?
whether we have not here, perhaps, a valid reason for believing
that Diophantus only knew of the existence of roots obtained by
using the positive sign with the radical, and was unaware of the
solution obtained by using the negative sign. But in truth we
can derive no certain knowledge on this point from Diophantus’
treatment of the particular equations in question. Thus, eg, if he
chose to use the first of the two equations

72% > 1722417,
72x < 102% + 19,
for the purpose of obtaining an upper limit only, and the second

1 This is remarked by Loria (Ze scienze esatte “Yell antica Grecia, V. p. 128).
But in fact, whether we take 20 or 3 as the value of the auxiliary unknown, we get
the same value for the original x of the problem. For the original x has to be found
from x? —60= (x —m)? where 2 is the auxiliary x; and we obtain x=r114 whether we
put 2% —60=(x—20)% or x%-60=(x- 3)2

2 Loria, gp- cit. p. 129.
DIOPHANTUS METHODS OF SOLUTION 63

for the purpose of obtaining a lower limit only, he could only use
the values obtained by using the positive sign. Similarly, if, with

the equations
60> 222,

2+ 60< 247,
he chose to use the first in order to find a lower limit only, and
the second in order to find an upper limit only, it was not open to
him to use the values corresponding to the negative sign.

For my part, I find it difficult or impossible to believe that
Diophantus was unaware of the existence of two real roots in
such cases. The numerical solution of quadratic equations by the
Greeks immediately followed, if it did not precede, their geometrical
solution. We find the geometrical equivalent of the solution of
a quadratic assumed as early as the fifth century B.C.,, namely by
Hippocrates of Chios in his Quadrature of lunes? the algebraic form
of the particular equation being x? +4/3.ar=a® The complete
geometrical solution was given by Euclid in VI. 27-29: and the
construction of VI. 28 corresponds in fact to the megative sign
before the radical in the case of the particular equation there
solved, while a quite obvious and slight variation of the con-
struction would give the solution corresponding to the positive sign.
In VI. 29 the solution corresponds to the positive sign before the
radical; in the case of the equation there dealt with the other sign
would not give a “real” solution® It is true that we do not find
the negative sign taken in Heron any more than in Diophantus,
though we find Heron* stating an approximate solution of the
equation

x (14 —2)= 06720144, |
without showing how he arrived at it; x is, he says, approximately
equal to 84. It is clear however that Heron already possessed
a scientific method of solution. Again, the author of the so-called
Geometry of Heron’ practically states the solution of the equation

H+ 3Pr=212

A (154 x 212 + 841) — 29
II ?

.in the form r=

 

! Enestrom in Bibliotheca Mathematica 1X5, 1908-9, pp. 71-2.

* Simplicius, Comment. in Aristot. Phys., ed. Diels, p. 64, 18; Rudio, Der Bericht
des Simplicius iiber die Quadraturen des Antiphon und des Hippokrates, 1907, p. 58, 8-11.

8 The Thirteen Books of Euclid’s Elements, Cambridge, 1908, Vol. 11. pp. 257-267.

* Heron, Metrica, ed. Schone, pp. 148-151. The text has 8 as the approximate solu-
tion, but the correction is easy, as the inference immediately drawn is that 14 — x= 33.

® Heron, ed. Hultsch, p. 133, 10-23. See M. Cantor, Geschichte der Math. Ig, P. 405.
64 INTRODUCTION

showing pretty clearly the rule followed after the equation had
been written in the form
121224 638x = 212 X 154.

We cannot credit Diophantus with less than a similar uniform
method; and, if he did not trouble to give two roots where both
were real, this seems quite explicable when it is remembered that he
did not write a text-book of algebra, and that his object through-
out is to obtain a single solution of his problems, not to multiply
solutions or to show how many can be found in each case.

In solving. such an equation as

—bxr4+c=0
it is our modern practice to divide out by a in order to make the
first term a square. It does not appear that Diophantus divided
out by a; rather he multiplied by a so as to bring the equation
into the form
ax — abr + ac=0;
then, solving, he found

axr=1%0 + v (}0*— ac),
and wrote the solution in the form
_30 + VEE — ac)
mi
wherein his method corresponds to that of the Pseudo-Heron above
referred to.

From the rule given in Def. 11 for removing by means of addition
any negative terms on either side of an equation and taking equals
from equals (operations called by the Arabians a/jabr and almuka-
bala) it is clear that, as a preliminary to solution, Diophantus so
arranged his equation that all the terms were positive. Thus,
from his point of view, there are three cases of mixed quadratic
equations.

Case 1. Form mxa*+ px =gq; the root is

—32 + VG 2 + mg)

wm )
according to Diophantus. An instance is afforded by v1. 6. Dio-
phantus namely arrives at the equation 622+ 3x = 7, which, if it \is
to be of any service to his solution, should give a rational value
of x; whereupon he says “the square of half the coefficient! of x

! For “coefficient” Diophantus simply uses wAffos, multitude or number; thus
“number of dpfuol” = coeff. of x. The absolute term is described as the ‘‘ units.”
DIOPHANTUS’ METHODS OF SOLUTION 65

together with the product of the absolute term and the coefficient
of #* must be a square number; but it is not,” ze. Lp* + mg, or in
this case (§)*+ 42, must be a square in order that the root may be
rational, which in this case it is not.

Case 2. Form ma®=px +q. Diophantus takes

L_ 3p Nar mg)
m

An example is IV. 39, where 222 > 6x + 18. Diophantus says:
“To solve this take the square of half the coefficient of x, ze. 9, and
the product of the absolute term and the coefficient of 2%, ie. 36.
Adding, we have 45, the square root! of which is not®* <7. Add
half the coefficient of x, [and divide by the coefficient of #*]; whence
x is not <5.” Here the form of the root is given completely; and
the whole operation of finding it is revealed. Cf. 1v. 31, where
Diophantus remarks that the equation 52°= 3x + 18 “is not rational.
But 3, the coefficient of #7, is a square plus 1, and it is necessary
that this coefficient multiplied by the 18 units and then added to
the square of half the coefficient of x, namely 3, that is to say 2%,
shall make a square.”

Case 3. Form mx*+ g =px. Diophantus’ root is

12 +VGP mg)
m

Cf in v. 10 the equation already mentioned, 1722+ 17 < 72x,
Diophantus says: “ Multiply half the coefficient of x into itself and
we have 1296; subtract the product of the coefficient of 2? and the
absolute term, or 289. The remainder is 1007, the square root of
which is not? > 31. Add half the coefficient of x, and the result is
not >67. Divide by the coefficient of 2, and x is not >67/17.”
Here again we have the complete solution given. Cf. VI. 22, where,
having arrived at the equation 172x = 3362%+ 24, Diophantus
remarks that “this is not always possible, unless half the coefficient
of x multiplied into itself, 77¢nus the product of the coefficient of x2
and the units, makes a square.”

For the purpose of comparison with Diophantus’ solutions of
quadratic equations we may refer to a few of his solutions of

1 The “square root” is with Diophantus m\evpd, or ‘‘side.”

% 7, though not accurate, is clearly the nearest integral limit which will serve the
purpose. ;

3 As before, the nearest integral limit.

H. D. 5
66 INTRODUCTION

(3) Simultaneous equations involving quadratics.
Under this heading come the pairs of equations

(1.27)

£49 hig

fy=5 }’
ng (1 28)
ly b. (1. 30.)

I use Greek letters to distinguish the numbers which the
problem requires us to find from the one unknown which Dio-
phantus uses and which I shall call x

In the first two of the above problems, he chooses his x thus.
Let, he says, :
E—n=2¢r (E>)

Then it follows, by addition and subtraction, that

E=a+ax, n=a-—=zx

Consequently, in I 27,

En=@+x)(a—2x)=a*—2*=B,
and x is found from this “pure” quadratic equation.

If we eliminate £ from the original equations, we have

9p—2an+B=0,

which we should solve by completing the square (2 — 7)? whence
(a —n)y=a-B,

which is Diophantus’ ultimate equation with a — » for x.

Thus Diophantus’ method corresponds here again to the ordi- .
nary method of solving a mixed quadratic, by which we make it
into a pure quadratic with a different x. :

In I..30 Diophantus puts £ + n=2x, and the solution proceeds
in the same way as in I 27.

In 1. 28 the resulting equation in x is

E+n=(a+rP+(a—-2P=2(a"+2%=04A.
(4) Cubic equation.

There is no ground for supposing that Diophantus was acquainted
with the algebraical solution of a cubic equation. Itis true that there
is one cubic equation to be found in the Arithmetica, but it is only
a very particular case. In VI. 17 the problem leads to the equation

24274 3=2+ 3x — 32% — 1,
DIOPHANTUS METHODS OF SOLUTION 67

and Diophantus says simply “whence x is found to be 4.” All
that can be said of this is that, if we write the equation in true
Diophantine fashion, so that all the terms are positive,
2+ x =41+4.
This equation being clearly equivalent to
(4 D=4{F+1),

Diophantus no doubt detected the presence of the common factor
on both sides of the equation. The result of dividing by it
is x=4, which is Diophantus’ solution. Of the other two roots
x= + 4/(— 1) no account is taken, for the reason stated above.

It is not possible to judge from this example how far Dio-
phantus was acquainted with the solution of equations of a degree
higher than the second.

I pass now to the second general division of equations.

(B) Indeterminate equations.

As 1 have already stated, Diophantus does not, in his
Arithmetica as we have it, treat of indeterminate equations of the
first degree. Those examples in Book I. which would lead to such
equations are, by the arbitrary assumption of a specific value for
one of the required numbers, converted into determinate equations.
Nor is it likely that indeterminate equations of the first degree
were treated of in the lost Books. For, as Nesselmann observes,
while with indeterminate quadratic equations the object is to obtain
a rational result, the whole point in solving indeterminate simple
equations is to obtain a result in znzegral numbers. But Diophantus
does not exclude fractional solutions, and he has therefore only to
see that his results are positive, which is of course easy. Inde-
terminate equations of the first degree would therefore, from
Diophantus’ point of view, have no particular significance. We
take therefore, as our first division, indeterminate equations of
the second degree.

(a) Indeterminate equations of the second degree.

The form in which these equations occur in Diophantus is
invariably this: one or two (but never more) functions of the
unknown quantity of the form A*+ Bx + C or simpler forms
are to be made rational square numbers by finding a suitable
value for x. Thus the most general case is that of solving one or
two equations of the form Ax? + Bx + C= y~

5—2
68 INTRODUCTION

(1) Single equation.

The single equation takes special forms when one or more of
the coefficients vanish or satisfy certain conditions. It will be well
to give in order the different forms as they can be identified in
Diophantus, premising that for “= »*” Diophantus simply uses the
formula {oov TeTpaydve, “is equal to a square,” or Towel TeTpdywvov,
“ makes a square.”

1. Equations which can always be solved rationally. This is
the case when 4 or C or both vanish.

Form Bx=j%. Diophantus puts for y? any arbitrary square
number, say »?. Then x= m?B.

Ex. IIL 5: 2x =? 5? is assumed to be 16, and x= 8.

Form Br+ C= Diophantus puts for »* any square #2 and
x=(m*— C)/B. He admits fractional values of xz, only taking
care that they are “rational,” ze rational and positive.

Ex. IIL 6: 6x41 = j*= 121, say, and x= 20.

Form Ax*+ Bx =3. Diophantus substitutes for y any multiple

2
of x, as 2; whence Ax + B= > x, the factor » disappearing and
Bn?
=r
Exx. 11. 21: 42°+ 3x == (37), say, and r=4§.
IL 33: 16224 7x =p = (5x), say, and x=].

the root x = 0 being neglected as usual. Thus x=

2. Equations which can only be rationally solved if certain
conditions are fulfilled.

The cases occurring in Diophantus are the following.

Form Ax*+ C=y% This can be rationally solved according to
Diophantus

(«) When 4 is positive and a square, say a>
Thus @%2+ C=32 In this case 32 is put = (ar + my;

 

therefore a + C= (ax t m),
C — m2
and = z :
2ma

(m and the doubtful sign being always assumed so as to give x
a positive value).

(B) When Cis positive and a square number, say ¢%
Thus Ax® + 2=y% Here Diophantus puts y = (mx t ¢);
DIOPHANTUS’ METHODS OF SOLUTION 69

therefore Av =(mx +c),
2mc
and Xs tf for,
A — wm?

(y) When one solution is known, any number of other
solutions can be found. This is enunciated in the Lemma to VI. 15
thus, though only for the case in which C is negative: “Given two
numbers, if, when one is multiplied by some-square and the other
is subtracted from the product, the result is a square, then another
square also can be found, greater than the aforesaid square which
has the same property.” It is curious that Diophantus does not
give a general enunciation of this proposition, inasmuch as not
only is it applicable to the cases + 42° + C=72 but also to the
general form Ax*+ Bx + C=y~

Diophantus’ method of finding other greater values of x satisfy-
ing the equation Ax* — C'=* when one such value is known is as
follows.

Suppose that x, is the value already known and that ¢ is the
corresponding value of y.

Put x=2,+% in the original expression, and equate it to
(g — ££), where 4 is some integer.

Since A@+Er—C=(g—kEN,
it follows (because by hypothesis 4x, — C = ¢?) that
28 (Ax, + kg) = £ (#2 — 4),

2 (Az + fg)
whence f= ER ’
2 (A, + kg)
and X=a,+ SRL :

In the second Lemma to VI. 12 Diophantus does prove that the
equation Ax* + C=? has an infinite number of solutions when
A + C is a square, ze in the particular case where the value r=1
satisfies the equation. But he does not always bear this in mind;
for in III. 10 the equation 52x24 12 =? is regarded as impossible of
solution although 52 + 12 =64, a square, and a raticnal solution is
therefore possible. Again in IIL 12 the equation 26612 — 10 =? is
regarded as impossible though x = 1 satisfies it.

The method used by Diophantus in the second Lemma to
V1. 12 is like that of the T.emma to VI. 15.

Suppose that 4 + C = ¢~

Put 1 + £ for x in the original expression 42% + C, and equate it
to (¢ — ££)’, where £ is some integer.
70 is INTRODUCTION

 

 

Thus AQ + 2+ C= (g — 4EY,
and it follows that 28(4 + &g)=84*— 4),
z2(A +t hy)
so that fn
2(A4 + £9)
and he ae a

It is of course necessary to choose £2 such that #2 > 4.

It is clear that, if x =o satisfies the equation, C is a square, and
therefore this case (vy) includes the previous case (8).
It is to be observed that in VI. 14 Diophantus says that a rational

solution of the equation
Ari— C=

is impossible wiless A is the sum of two squares.

[In fact, if x= p/q satisfies the equation, and Ax? — = £2,
we have Ap? = cg? + kg?

aN. RIN

a 2=(77 + (J 1

Lastly, we have to consider

Form Ax 4 Bx + C="

This equation can be reduced by means of a change of variable

to the preceding form wanting the second term. For, if we put

pm ey the transformation gives

— Ds
Ans tAE =7’

Diophantus, however, treats: this form of the equation quite
separately from the other and less fully. According to him the
rational solution is only possible in the following cases.

(¢) When 4 is positive and a square, or the equation is

r+ Byrd C=1~,
Diophantus then puts 3? = (ax — m)?, whence
m?— C

=. Exx. 11 20, 22 etc.
ER (Exe. 11 20, 22 elec)

(8) When Cis positive and a square, or the equation is
Ax + Bx + =~
Diophantus puts 2 = (¢c — mx)’, whence

2mc+ B
m:— A’

 

{(Exx. Iv. 3, getc)
DIOPHANTUS’ METHODS OF SOLUTION 71

(y) When 1B? - AC is positive and a square number. Dio-
phantus never expressly enunciates the possibility of this case;
but it occurs, as it were unawares, in IV. 31. In that problem

3x + 18 — x?
is to be made a square. To solve this Diophantus assumes
LIS —ai=uar

which leads to the quadratic 3x + 18 — 522 = 0; but “the equation
is not rational.” Accordingly the assumption 44? will not do;
“and we must find a square [to replace 4] such that 18 times
(this square + 1) + (3) may be a square.” Diophantus then solves
the auxiliary equation 18(m*+1)+ $=? finding m=18. He
then assumes 3x + 18 — 22 =(18)2x2 which gives 32522— 3x — 18 =0,
“and x = J&, that is £&.* :

1 With this solution should be compared the much simpler solution of this case given
by Euler (4/gebra, tr. Hewlett, 1840, Part 11. Arts. 50-53), depending on the separation
of the quadratic expression into factors. (Curiously enough Diophantus does not separate
quadratic expressions into their factors except in one case, VI. 19, where however his
purpose is quite different : he has made the sum of three sides of a right-angled triangle
4x2 + 6x + 2, which has to be a cube, and, in order to simplify, he divides throughout by
x +1, which leaves 4x +2 to be made a cube.)

Since $B? — AC is a square, the roots of the quadratic 4x? + Bx + C=o0 are real, and
the expression has two real linear factors. Take the particular case now in question,
where Diophantus actually arrives at 3x +18 — x? as the result of multiplying 6 —x and
3 +x, but makes no use of the factors.

 

We have 3x +18 —x2=(6 — x) (3 +x).
Assume then (6-2) (3+) Li 6- ap,
and we have 226 -x)=¢%*(3 +x),
9

where p, 4 may be any numbers subject to the condition that 2p%>¢%. If p2=9, ¢*=16,

we have Diophantus’ solution p=

In general, if Ax? + Bx + C= (f+gx) (h+ kx),
wo tip (Free) kta) =; (Fg,
whence 7 (4+ kx) = 2 (+g),
and x i ~fy > .

kq® - gp*

This case, says Euler, leads to a fourth case in which 4x?+ Bx + C=? can be solved,
though neither 4 nor C is a square, and though B%- 44 C is not a square either. The
fourth case is that in which 4x*+ Bx + C is the sum of two parts, one of which is a square
and the other is the product of two factors linear in x. For suppose

Ax*+ Bx+C=2%+XY,
where Z=dx+te, X=fx+g Y=lx+téh
72 INTRODUCTION

It is worth observing that from this example of Diophantus we
can deduce the reduction of this general case to the form
Ari + CC =3°
wanting the middle term.
Assume, with Diophantus, that Ax* + Bx + C = m*x*: therefore

by solution we have
_—3BtNIB*—AC+ Cm?
I A —m? :
and x is rational provided that }B%?— AC + Cm? is a square. This
condition can be fulfilled if }B*— AC is a square, by the pre-

 

 

2
We can then put 2rxv=(242x),
2
whence r=2f 2:0 x,
g g
2? 2
or friimotidira its (ata)
that is, x (pf + 209d — g*h) = kg? — 2pge — pg.
Ex. 1. Equation 2x4%- 1=3% :
Put 2x2 —1=x2+(x+1)(x—1)= {r+ en} .
2
Therefore x — 1=2l 23 l (v4 1),
g
and *(P+aopg-)=- (+7).

As x? is alone found in our equation, we can take either the positive or negative sign
and we may put
Lo fag
hr See .
rtg—yq
Ex. 2. Equation 2x2 +2=7%
Here we put 2x? +2=g4+2(x+1) (x1).

2
Equating this to { 3 (x+ 0}

2
we have 2e-n)=42 +5 (r+ 1),
or 2(P*-2°)= — (af + apg + 2°),
ult grt og
= pant
20: =D
It is to be observed that this method enables us to solve the equation

and

Ax —2=y?
whenever it can be solved rationally, 7.e. whenever 4 is the sum of two squares (d2%+ e2,
say). For then

Ax? — A=d?x% + (ex —¢) (ex +c).

In cases not covered by any of the above rules our only plan is to try to discover one
solution empirically. If one solution is thus found, we can find any number of others; if
we cannot discover such a solution by trial (even after reducing the equation to the
simplest form 4'x"+ C=y'3), recourse must be had to the method of continued fractions
elaborated by Lagrange (cf. Oeupres, 11. pp. 377-535 and pp. 655—726 ; additions to
Euler's Algebra). 3
DIOPHANTUS' METHODS OF SOLUTION 73

ceding case. If }B*— AC is not a square, we have to solve
(putting, for brevity, D for }B* — AC) the equation

D + Cm? = p72,
and the reduction is effected.

(2)  Double-equation.

By the name “double-equation” Diophantus denotes the pro-
blem of finding one value of the unknown quantity which will make
two different functions of it simultaneously rational square numbers.
The Greek term for the “double-equation” occurs variously as dumrhoi-
corns, Surf) loons or Surhf lowaws. We have then to solve the
equations

mx tort a=11

n+ Bx + b= ol
in rational numbers. The necessary preliminary condition is that
each of the two expressions can severally be made squares. This
is always possible when the first term (in 22) is wanting. This is
the simplest case, and we shall accordingly take it first.

1. Double-equation of the first degree.

Diophantus has one general method of solving the equations
ar +a =u’
Br+b= re ?
taking slightly different forms according to the nature of the
coefficients.

(a) First method of solution of
ax + a=?
frame
This method depends upon the identity
22+) -H(2-F =22

If the difference between the two expressions in x can be separated
into two factors p, ¢, the expressions themselves are equated
to {3 (2 * ¢)}* respectively. Diophantus himself (11. 11) states his
rule thus.

“Observing the difference [between the two expressions], seek
two numbers such that their product is equal to this difference;
then equate either the square of half the difference of the two
factors to the lesser of the expressions or the square of half the
sum to the. greater.”
74 INTRODUCTION

We will take the general case and investigate to what particular
classes of cases the method is applicable, from Diophantus’ point
of view, remembering that his cases are such that the final quadratic
equation in x always reduces to a simple equation.

Take the equations

ax + a=u
Bx+ b= i :
Subtracting, we have
(a—PB)x+(a—b)=u>— wr

We have then to separate (a—B)x + (a— 0) into two factors;
let these be 2, {(a— B) x + (a — b)}/2.

We write accordingly

cleaMrto—sp
wt w= 2 ;
ut w=p.
Thus w=arra=t [BEREEEZ0,
therefore {(a=B)x+a—b+p42=4p* (ax + a),

or (a—Byx*+2x{(a— B)(a—0b+p") —2p*a} + (a— b+") — 4ap*=0,
that is, (a—B)x*+22{(a—p)(a—0)—7*(a+B)}
+(@—0by—2p (a+b) +p =o.
Now, in order that this equation may reduce to a simple
equation, either
(1) The coefficient of 2* must vanish, so that

a=j,
or (2) The absolute term must vanish, that is,
—2@t+b)+(a—by=o,
or {2 — (a+ 0) = 4a0,
so that a6 must be a square number.

Therefore either 2 and 4 are both squares, in which case we
may substitute ¢* and 4? for them respectively, p being then equal
to c + 4, or the ratio of a to & is the ratio of a square to a square.

With respect to (1) we observe that on one condition it is not
necessary that a — 3 should vanish, ze. provided we can, before
solving the equations, make the coefficients of x the same in both
expressions by multiplying either equation or both equations by
some square number, an operation which does not affect the
problem, since a square multiplied by a square is still a square.
DIOPHANTUS’ METHODS OF SOLUTION 75

In other words, it is only necessary that the ratio of a to 8B should
be the ratio of a square to a square’.

Thus, if a/8=m*n* or an®= Bm, the equations can be solved
by multiplying them respectively by #* and m?; we can in fact
solve the equations

an’x + a = 1?
any + b= 2 :
like the equations

ax +a=u"
ar+6=w?’

in an infinite number of ways.
Again, the equations under (2)
ax +E =u
Brt-di=
can be solved in two different ways according as we write them in
this form or in the form
ad’x + fd =u"
Beir + cd? = Wl ?
obtained by multiplying them respectively by 42, ¢ in order that
the absolute terms may be equal.
I shall now give those of the possible cases which we find solved
in Diophantus’ own work. These are equations

(1) of the form
om:x + a=
anir+ 6 = wl’

I Diophantus actually states this condition in the solution of 1v. 32 where, on arriving
at the equations 3
8 -—x=u?
8 -3x=w? ’
he says: “And this is not rational because the coefficients of x have not to one another
the ratio which a square number has to a square number.”

Similarly in the second solution of III. 15 he states the same condition along with an
alternative condition, namely that @ has to 4 the ratio of a square to a square, which is
the second condition arrived at under (2) above. On obtaining the equations

4x + 3=12
64x + 53 =w? |
Diophantus observes ‘But, since the coefficients in one expression are respectively greater
than those in the other, neither have they (in either case) the ratio which a square number
has to a square number, the hypothesis which we took is useless.”

Cf. also 1v. 39 where he says that the equations

8x+4= i;
bx + 4 =w?

are possible of solution because there is a square number of units in each expression.
76 INTRODUCTION

a case which includes the more common one where the coefficients
of x in both are equal;
(2) of the form
or: =u")
Br+d?= wf 2
"solved in two different ways according as they are written in this
form or in the alternative form
adix + cdi =u”
Ber+ cdi = ee :

General solution of Form (1) or
any + a=’
atx +t b=)’
Multiply by #? wn? respectively, and we have to solve the

equations ;
amy + and =u"?
anny + on = w?)

The difference is an®—bm?; suppose this separated into two
factors p, ¢.

Let wt w =p,

UWF w=gq,
therefore wWr=1(p+g) wi=1(p—9)
and am*r*xr+an*=%(p + q),
or am’ n’x + bn =} (p— gi

Either equation gives the same value of x, and

_H( +a) =k (an + bow)

x
arn?

 

since pg = an® — bm*.

Any factors p, ¢ may be chosen provided that the resulting
value of x is positive.

Ex. from Diophantus :

65 — 6x= 2

Ot = Ba? | : {1v. 32)
therefore 260 — 24x =u”

65 — 24x =)"

[The difference = 105=15. 13, say;
therefore 1(135—13¥=65— 247; that is, 242 =064, and z=}.
DIOPHANTUS METHODS OF SOLUTION 77

General solution (first method) of Form (2), or
ar + c= wu"
Br +d= ee :
In order to solve by this method, we multiply by #2
respectively and write
ad?x + cd? = u* |
Bex + cd =u)’
« being supposed to be the greater.
"The difference =(ad?—~ Sc) Let the factors of this be pv, g.
Therefore w=1(pr+9)
w=} (pr— gq).
Thus x is found from the equation
ad?x + 2d? =} (px +g)
This equation gives
p22 + 2x (pg — 2ad?) + ¢* — 4c2d*=0,
or, since pg = (ad® — B¢?),
Pt — 2x (ad? + Be) + ¢* — 42d? =o.
In order that this may reduce to a simple equation, as Dio-
phantus requires, the absolute term must vanish,
or J =4E43
and g=2cd.
Thus our method in this case furnishes us with only oze solution
of the double-equation, ¢ being restricted to the value 2¢d, and the
solution is

 

pe (ad + B®) 82d (ad? + Be)
2 (ad? id BEY
Ex. from Diophantus. This method is only used in one par-
ticular case (IV. 39), where ¢*=d* as the equations originally stand,
the equations being
Sr++ y=
6x+4 = i :
The difference is 2x, and ¢ is necessarily taken to be 24/4, or 4;
the factors are therefore 1x, 4.
Therefore 8r+4=13EFxr+4)7
and r=1I12.

General solution (second method) of Form (2) or
ax + =? :
BxA d=)"
78 INTRODUCTION

The difference = (a — B) x + (¢ — d?).

Let the factors of this be p, {(a — B) x + ¢* — d%}/p.

Then, as before proved (p. 74), # must be equal to (¢ + &).
Therefore the factors are

Se ON ctd,
ctd

and we have finally

artc=

 

 

Therefore CEY 2% + 4x f b= — af = 0,

which equation gives two possible valyes for . Thus in this case
we can find by our method zee values of x, since one of the factors
p may be either (c +d) or (c — 4d).

Ex. from Diophantus. To solve the equations

Sal fl]

$1 (11. 15.)

The difference is here 5x + 5, and Diophantus chooses as the
factors 5, x + 1. This case therefore corresponds to the value
c+ d of p. The solution is given by

(3x + 3) = 10x + 9, whence x = 28.

The other value, ¢ — d, of p is in this case excluded, because it
would lead to a negative value of x.

The possibility of deriving any number of solutions of a double-
equation when one solution is known does not seem to have been
noticed by Diophantus, though he uses the principle in certain
cases of the single equation (see above, pp. 69, 70). Fermat was the
first, apparently, to discover that this might always be done, if one
value a of x were known, by substituting x + a for x in the equa-
tions. By this means it is possible to find a positive solution, even
if a is negative, by successive applications of the principle.

But, nevertheless, Diophantus had certain peculiar artifices by
which he could arrive at a second value. One of these artifices
(which is made necessary in one case by the unsuitableness of the
value of x found by the ordinary method) gives a different way of
solving a double-equation from that which has been explained, and
is used only in one special case (IV. 39).
DIOPHANTUS’ METHODS OF SOLUTION 79

(B) Second method of solving a double-equation of the first
degree.
Consider only the special case

Ax mo =1r,
(f+ f)x + n?=ur
Take these expressions, and #2, and write them in order of
magnitude, denoting them for convenience by 4, B, C.
d=/+Nx+i?, B=lr+w, C=
Therefore 8 7, and .
Suppose now that /ww + 122 = (y +n);

therefore hx = 3° + 2ny,
and 4-8 L(y + 2ny),

a 32 J 2 .
or A =p Eppa? (5 + 2np);

thus it is only necessary to make this expression a square.
Assume therefore that
(1 +7) ton (24 y+ n= (py — ny;
and any number of values for y, and therefore for x, can be found,
by varying 2.
Ex. from Diophantus (the only one), 1v. 30.

~ In this case there is the additional condition of a limit to the
value of x. The double-equation

Sx + 4=10’
6x + 4 = uw?
has to be solved in such a manner that x < 2.
Here ons 1, and B is taken? to be (y + 2)%
B-C
Therefore A-B=1(r+4y);
therefore A= La) LP +4744

=Ar aii,
which must be made a square.

! Of course Diophantus uses the same variable x where I have for clearness used y.
Then, to express what I have called » later, he says: “I form a square from 3 minus
some number of x's, and x becomes some number multiplied by 6 and then added to 12,
divided by the difference by which the square of tke number exceeds 3.”
8o INTRODUCTION

If we multiply by ¥, we must make
3% + 12y + 9 = a square,
where y must be < 2.
Diophantus assumes
37° + 127 +9 =(3 — my,
6m + 12
E—3
and the value of # is then taken such as to make y < 2.
It is in a note on this problem that Bachet shows that the
double-equation

whence p=

ax + a =u’
Bx+b= 7

can be rationally solved by a similar method provided that the

coefficients satisfy either of two conditions, although none of the

coefficients are squares and neither of the ratios a: 8 and a:0 is

equal to the ratio of a square toa square. Bachet’s conditions are:

(1) That, when the difference between the two expressions
is multiplied or divided by a suitably-chosen number, and the
expression thus obtained is subtracted from the smaller of the
original expressions, the result is a square number, or

(2) That, when the difference between the two expressions
is multiplied or divided by a suitably-chosen number, and the
smaller of the two original expressions is subtracted from the
expression obtained by the said multiplication or division, the
result is a number bearing to the multiplier or divisor the ratio
of a square to a square’.

1 Bachet of course does not solve equations in general expressions (his notation does
not admit of this), but illustrates his conditions by equations in which the coefficients are
specific numbers. I will give one of his illustrations of each condition, and then set the
conditions out more generally.

 

Case (1). Equations 3x+13=2°
xk v= i
difference 2x+ 6

The suitably-chosen number (to dzzide by in this case) is 2 ;
% (difference) =x + 3,
and (lesser expression) — 4 (difference) =x +7 — (x + 3)=4, that is, a square.
We have then to find two squares such that
their difference =2 (difference between lesser and 4).

Assume that the lesser=(y + 2)% 2 being the square root of 4.
Therefore (greater square) =3 (lesser) — 8

=32+ 125+ 4.
DIOPHANTUS METHODS OF SOLUTION 81

2. Double-equation of the second degree,

or the general form
: Av + Brt C=
AL Brd C=)” .

These equations are much less thoroughly treated in Diophan-
tus than those of the first degree. Only such special instances

 

To make 33° + 12y + 4 a square we put

rtp ta=la—pf,
where » must lie between certain limits which have next to be found. The equation gives

75

In order that y may be positive, p* must be > 3; and in order that the second of the
original expressions, assumed equal to (y+ 2)? may be greater than 7 (it is in fact x +7),
we must have (y+ 2) > 2% (an a fortiors limit, since 24 >./7), or y>%.

 

Therefore P+12>3(F2-3),

or 169+ 57>3%
Suppose that 32*=16p+ 533, which gives p=73.
Therefore 2<7% but £°>3.

Put p=3 in the above equation ; therefore

3+ 1+ 4=(2- 37)

and Y=4.
Therefore x=(y+2) - 7=10.
Case (2). Equations b6x+25=27)
ax + s=aff?

difference 4x+22
The suitable-number (again to divide by) in this case is 2;
1 (difference) =2x+ 11,
and 4 (difference) — (lesser expression) =8=12. 4,
where 2 is the divisor used and 4 is the ratio of a square to a square.
Hence two squares have to be found such that
(their difference) =2 (sum of lesser and 8).
If the lesser is 3? the greater is 332+ 16 =(4 — p9)?, say.
Bachet gives, as limits for p,
2+ 4h *>3
and puts p=3. This gives y=4, so that x=6%.
Let us now state Bachet’s conditions generally.
Suppose the equations to be
ax+a=1 |
Bx +b=1w? "
The difference is (a — B) x + (a — 6).

This has to be multiplied by £ which is the ‘“suitable” factor in this case, and, if

we subtract the product from Bx +4, we obtain
ah or
b spl 0), or TE
82 INTRODUCTION

occur as can be easily solved by the methods which we have
described for equations of the first degree.
The following types are found.
(1) en
pit Prt o=ur)’ :
The difference is (a — 8) x + (a — 4), and, following the usual
course, we may, e.g, resolve this into the factors

 

(1) The first of Bachet’s conditions is that

v2 = a square = p?[¢?, say.
(2) The second condition is that
gran LB
9=f = unt’
aB—ab

or = a ratio of a square to a square.

B
It is to be observed that the first of these conditions can be obtained by considering
the equation
(@=B)x*+2x{(a=-B)(@-0) = (@a+B)} + (a= 0)’ = 2p (a +0) +p'=0,
obtained on page 74 above.

Diophantus only considers the cases in which this equation reduces to a simple
equation; but the solution of it as a mixed quadratic gives a rational value of x provided
that

{@-PB) (@=0)-2*(a+B)}*- (a= B)*{(a= 0)? 2° (a+ 6) +2} is a square,
that is, if

2{{a+B)% = (a=)? +282 {(a+0)(a—B)*>—(a®- B%) (a —b)} is a square,
which reduces to aff? +{a—B)(eb-aB)=a square .............coc0eeee-e. (A).
This can be solved (cf. p. 68 above), if

 

oo
= is a square. (Bachet’s first condition.)
Again take Bachet’s second condition
BN a square = 7 s
T= quare = = say,
and substitute 872s? for aB — aé in the equation (A) above.
Therefore app? —(a—B)B 5= a square,
or ap’? -- (a — B) B=a square.

This is satisfied by p'=1; therefore (p. 69) any number of other solutions can
be found. :
The second condition can also be obtained directly by eliminating x from the equations

ax +a=u? :
Bx+b=w?"’
for the result is St oe =

which can be rationally solved if

 

ap —ah
! - =a square,
8
DIOPHANTUS. METHODS OF SOLUTION 83

@- == Les;

 

as usual, we put

— 2
pritarta=1 (28011 1a-0),

4 \a—20
pt ll renal)
or pr bRr thE ak? a+b).

In order that + may be rational a condition is necessary; thus
x is rational if

This is the case in the only instance of the type where a is not

equal to 4, namely (IIL. 13)
42° + Se :
41 — 2x — 4 = WP]

the difference is 16x + 4, and the resolution of this into the factors
4, 4x + 1 solves the problem.

In the other cases of the type a=2¢4; the difference is then
(a — B) x, which is resolved into the factors

 

 

 

ee 2 2p7;
2p
: 1 /a—f ¥
and t 292 7 wm = (los 0 ;
we pu pr+oar+a PN 2p px
1.7 2
or pPr+prt+a=-= (552 - apr),
4
whence al og i — BY a.
2 4p
Exx. from Diophantus:
2tr—1=7
i 7 {1v..23.)
and FAIA + T=22
22 Tm 2 2

(2) The second type found in Diophantus? is

rtarta=1w
Pr+ta=a’
where one equation has no term ina, and p=1,a=0.
! Tt is perhaps worth noting that the method of the ‘‘ double-equation ” has a distinct
advantage in this type of case. The alternative is to solve by the method of Euler (who

does not use the * double-equation ’), 7.e. to put the linear expression equal to p* and then,
substituting the value of x (in terms of p) in the quadratic expression, to solve the

6—2
84 INTRODUCTION

The difference 2? + (« — B) x is resolved into the factors
r(x +a—p);
and we put Bxr+a=4%(a— BR)
which gives x.

 

resulting equation in p. But the difficulties would generally be great. Take the case of
v1. 6 where
x24 1
I4x+1
If rqr+1=2% x=(p2-1)[14;
{=
4
or 2-2? + 197 =a square.
This does not admit of solution unless we could somehow discover empirically ose
value of p which would satisfy the requirement, and this would be very difficult.
Let us take an easier case for solution by this method,
224+ 1=22
x+i1=w?f’

have to be made squares.

1)Z

therefore EE 5

 

+1 has to be made a square,

which is solved by Euler (4/gebra, Part 11. Art. 222), and let us compare the working of
the two methods in this case.

I. Euler's method. Assuming x + 1=p*and substituting 22 1 for x in the quadratic

expression, we have
2-22 + 2=a square.

This can only be solved generally if we can discover one possible value of p by trial ;
this however is not difficult in the particular case, for p=1 is an obvious solution.

To find others we put 1+¢ instead of in the expression to be made a square; this
gives

I+ 40% + 49° + 9* =a square.
This can be solved in several ways.

1. Suppose 1+ 472+ 40% +0 = (1 + ¢%)2;
thus 49’ +4¢°=12¢% whence ¢= -2, == and x=-3.
2 4
2. Suppose E+ 40% + 40% + 9 = (1 — 92;
thus 49° + 4¢°= — 24%, and g=-3, p=-1 and x= 2;
2 4
3. Suppose 440% +408 + gr = (1+ 2p = 7%);
and we find, in either case, that g= — 2, so that p= — 1, x=o.
4. Suppose I+ 40% + 48 +4 = (1 + 29%)? ;

2
and we have 4¢°+¢*=4¢% whence 7=1, sal and x= @ =1=i2
3 3 9
This value of x satisfies the conditions, for
2 2
x+1=(1) y pga iOl (2) a
3 81 9
The above five suppositions therefore give only two serviceable solutions

x=-3, yal

4 9

To find another solution we take one of the values of ¢ already found, say ¢= =1 and
2

 
DIOPHANTUS METHODS OF SOLUTION 85

Exx. from Diophantus :
2-12 =
63x — 12 = w?)’

(V. 1.)

22 1 == yf
14% + 1 Be (VL 6.)
Tete anid Fash
2464 = aff)’ ha

 

substitute 7» — : for g. This gives p=14+¢=7r+ I and we substitute this value for in the
2
expression pf — 242+ 2.

25 I ;
We have then = = 3 re 724273 +71 to be made a square, or

25 — 247 — 87% + 3273 + 16/4=a square.

1. We take 5+/7= 472 for the root, so that the absolute term and the term in #4
may disappear. We can make the term in » disappear also by putting 1of= —24 or
f=- . We then have

— 872+ 321% =12(f2 + 40) £8/7°.

(a) The upper sign gives

-8+327=40+/%+ 8/7,

21
and r=(f1+48)/(32- 8) =22

61
th cpl fh 2 EE
us 2 er and x=p2-1 je

(6) The lower sign gives
—8+327= — 404/287,

and r=(f-golfga+ of) =-L;
thus p=- 2 , and hil as before.
20 400
4 2 2
We have therefore x+1= (2) , and #2+1= (2) .
20 400

2. Another solution is found by assuming the root to be 5+ f+ g7* and determining
/f and g so that the absolute term and the terms in 7, 7? may vanish; the result is

 

 

wll pa lp  yieBl | S06
r= 5° i= 125’ fi 16-g2 861’
7 2
whence p= iy oo 2047837 fy 1=(222)
1722 20635284 1722
2
and pr r= (255365)
2965284
II. Method of “ double-equation.”
: 2 r=u2,
x + 1=us

The difference =x? — x.

(1) If we take as factors x, x—1 and, as usual, equate the square of half their

; I
difference, or —, to x+ 1, we have
4

Io

HAE,

or xX

4 fu +
86 INTRODUCTION

The absolute terms in the last case are made equal by multiply-
ing the second equation by (128)? or 16384.

(3) One separate case must be mentioned which cannot be
solved, from Diophantus’ standpoint, by the foregoing method,
but which sometimes occurs and is solved by a special artifice.

The form of double-equation is

ax’ + ar =u") tia Ane {1),
Br or mar]. ....oo0iiines NR, (2).
Diophantus assumes "=m,
whence, by (1), x =al(m® — a),

A!

©

|

(2) If we take 1a 2x — 2, as factors, half the sum of which is *x— 1, so that the

absolute terms may disappear in the resulting equation, we have

2 Lx
2-2x=a3,
6 2

2h
or Fra ’

© on

ttl,

9
(3) To find another value by means of the first, namely x= — 3 we substitute od
4

and

for x in the original expressions. We then have to solve

2-3 /
Pe Tye

25 2
—= = uc,

16

T 9
+—- =r.
3 4
Multiply the latter by = so as to make the absolute terms the same, and we must have

Subtract from the first expression, and the difference is 3? ~3L yey (- 2) then
4
equating the square of half the difference of the factors to the smaller expression,

we have
Na
4 (= mE

so that 961 = 400y + 100.

Therefore
> 2 2
$= ot wmpude BE rin f2Y oer e (2,
400 4 400 20 \ 400

(4) If we start from the known value > and put el for x in the equations, we

obtain Euler’s fourth value of x, namely zo47857 3
2065284
Thus all the four values obtained by Euler are more easily obtained by the method of
the ¢“ double-equation.”
DIOPHANTUS METHODS OF SOLUTION 87

and, by substitution in (2), we derive that

 

‘a ba
—— must be a squar
2 £ — 2 > 7m? — a 2 Mquare,

9 i a a square.
We have therefore to solve the equation
abm?+ a (af — ad) =’, :
and this form can or cannot be solved by the methods already
given according to the nature of the coefficients’. Thus it can
be solved if (a8 —ab)/a is a square or if a/b is a square.

or

Exx. from Diophantus:

Or’ 4 48 =
Ott j= i : (vr. 129
6x —'sp=yp
i Hy = il : (VL. 14.)

(0) Indeterminate equations of a degree higher than the second.

(1) Single equations.

These are properly divided by Nesselmann into two classes ;
the first comprises those problems in which it is required to make
a function of x, of a degree higher than the second, a square ; the
second comprises those in which a rational value of x has to be
found which will make any function of x, not a square, but a higher
power of some number. The first class of problems requires the
solution in rational numbers of

Az? + Beri +, . + Ket l =i, -
the second the solution of

Ax + Ber" 1A ... + Kx 4 L=g,
for Diophantus does not go beyond making a certain function of
x a cube. In no instance, however, of the first class does the index

n exceed 6, while in the second class # does not (except in a
special case or two) exceed 3.

! Diophantus apparently did not observe that the above form of double-equation can
be reduced to one of the first degree by dividing by x* and substituting y for 1/x, when it
becomes

at+ay=u"
B+ by=w".

Adapting Bachet’s second condition, we see that the equations can be rationally solved
if (Ba — ab)[a is a square, which is of course the same as one of the conditions under which
the above equation a@ém?+ a (af — ab) can be solved.
88 INTRODUCTION

First Class. Equation
Az? + Bx? ... + Kx + L =~
The forms found in Diophantus are as follows:

1. Equation Axd + Br + Cx + d* = y~

Here, as the absolute term is a square, we might put for y
the expression mx +4 d, and determine 2 so that the coefficient
of x in the resulting equation vanishes. In that case

oped =(, and m= C/2d;
and we obtain, in Diophantus’ manner, a simple equation for x,
giving
C? — 4d*B
=

Or we might put for y the expression #2? + nx + d, and deter-
mine m2, 2 so that the coefficients of x, #2 in the resulting equation
both vanish, in which case we should again have a simple equa-
tion for x. Diophantus, in the only example of this form of
equation which occurs (VI. 18), makes the first supposition. The
equation is

2-327 + 3x +1 =797,

and Diophantus assumes y = 3x + 1, whence x = 21,

2. Equation Axt+ Bri+(Cx*+ Dx + E=5~

In order that this equation may be solved by Diophantus’
method, either 4 or £ must be a square. If 4 is a square and
equal to @% we may assume y=art ry 7, determining 7 so
that the term in 2? vanishes. If £ is a square (= ¢*), we may write
y=mrt rie, determining 72 so that the term in 2? in the

resulting equation may vanish. We shall then, in either case,
obtain a simple equation in x.
The examples of this form in Diophantus are of the kind
@r*+ Bri + (x? + Dx + 2 =3,
where we can assume y= + ar?+ £r + ¢, determining 4 so that in
the resulting equation the coefficient of x® or of x may vanish;
when we again have a simple equation.
Ex. from Diophantus (1v. 28):
ort — 423+ 612 — 120+ 1 =)2
Diophantus assumes y = 324? — 6x + 1, and the equation reduces to
322% —362?=0, whence r=4§.
DIOPHANTUS’ METHODS OF SOLUTION 89

Diophantus is guided in his choice of signs in the expression
+ ax? + kx + ¢ by the necessity for obtaining a “rational ” result.

Far more difficult to solve are those equations in which, the
left-hand expression being bi-quadratic, the odd powers of x are
wanting, ze. the equations Ax*+ Cx? +E =3* and Ax*+ E =77
even when A or E is a square, or both are so. These cases
Diophantus treats more imperfectly.

3. Equation Aei+ Cit E=32
Only very special cases of this form occur. The type is
art — xt + = 52,

which is written
2

e
2 = 2
Cr: = ray h
Here p' is assumed to be ax or ¢/x, and in either case we have

a rational value for z.
Exx. from Diophantus:

Dt Vy. 27.
250 —g + 5 (v. 27)

This is assumed to be equal to 2522
a = 28 + 22 pp (v. 28.)

where 3? is assumed to be equal to on

4. Equation Art Ewmqr

The case occurring in Diophantus is x4 + 97 =» (V. 29). Dio-
phantus tries one assumption, y = 22 — 10, and finds that this gives
2* =, which leads to no rational result. Instead, however, of
investigating in what cases this equation can be solved, he simply
drops the equation x*+ 97 =? and seeks, by altering his original
assumptions, to obtain, in place of it, another equation of the same
type which can be solved in rational numbers. In this case, by
altering his assumptions, he is able to replace the refractory equa-
tion by a new one, x4+ 337 =? and at the same time to find a
suitable substitution for y, namely y=x?— 25, which brings out
a rational result, namely x=12 This is a good example of his
characteristic artifice of “ Back- a ” as Nesselmann calls it.

5. Equation of sixth di in the special form
— Az Br + et= 2,

1 “Methode der Zuriickrechnung und Nebenaufgabe.”
go INTRODUCTION

It is only necessary to put y =x*+¢, and we have
— Ar: + B= 2c2?,
B
=r

which gives a rational solution if B/(4 + 2¢) is a square.

or: - 512

6. If, however, this last condition does not hold, as in the case
occurring IV. 18, #*— 164° + x + 64 =»? Diophantus employs his
usual artifice of “back-reckoning,” which enables him to replace
the equation by another, 1% — 1281" + x + 4096 =? where the
condition is satisfied, and, by assuming y = + 64, x is found to
be

Second Class. Equation of the form

Are + Bai... + Kz + L=7~
Except for such simple cases as 42? = 3, Ax*= 373 where it is only
necessary to assume y = 7x, the only cases occurring in Diophantus
are of the forms
Ax + Br + C=?
A+ B+ (x + D =7"

1. Equation d+ Br 4 C= yp

There are only two examples of this form. First, in VI. 1 the
expression x® — 4x + 4 is to be made a cube, being already a square.
Diophantus naturally assumes x — 2 =a cube number, say 8, and
x = 10.

Secondly, in VI. 17 a peculiar case occurs. A cube is'to be
found which exceeds a square by 2. Diophantus assumes (x — 1)?
for the cube and (x + 1)* for the square, and thus obtains the
equation

Bp lir—=1=my:tart}

or 4 r=41v2+ 4,

previously mentioned (pp. 66-7), which is satisfied by x=4.
The question arises whether it was accidentally or not that this
cubic took so simple a form. Were + —1, x41 assumed with
knowledge and intention? Since 27 and 25 are, as Fermat
observes?, the only integral numbers which satisfy the conditions,
it would seem that Diophantus so chose his assumptions as to lead
back to a known result, while apparently making them arbitrarily
with no particular reference to the end desired. Had this not

! Note on VI. 17, Oeuvres, 1. pp. 333-4, II. p. 434. The fact was proved by Euler
(Algebra, Part 11. Arts. 188, 193). See note on VL. 17 infra for the proof.
DIOPHANTUS' METHODS OF SOLUTION 91

been so, we should probably have found him, here as elsewhere
in the work, first leading us on a false tack and then showing us
how we can correct our assumptions. The fact that he here
makes the right assumptions to begin with makes us suspect that
the solution is not based on a general principle but is empirical
merely.

2. Equation A+ Br + Cx + D =35
If A or D is a cube number, this equation is easy of solution.
For, hrst, if A =#, we have only to write y=ar+ i, and we

obtain a simple equation in x.

Secondly, if D = d?, we put y -5 xr +d.

If the equation is #%2* +4 B22 + Ce + d= 4°, we can use either
assumption, or we may put y =ax + 4, obtaining a simple equation
as before.

Apparently Diophantus used the last assumption only; for
in IV. 27 he rejects as impossible the equation

=H: 8x —~1=19%
because y = 2x —1 gives a negative value x= —-2, whereas either

TTY
of the other assumptions gives a rational value.

(2) Double-equations.

There are a few examples in which, of two functions of x, one
is to be made a square, and the other a cube, by one and the same
rational value of x. The cases are for the most part very simple;
e.g. in VI. 19 we have to solve

4x +2 =p"
2v + I = 2?

thus =22" and z=2,
A rather more complicated case is VI. 21, where we have the
double-equation
21 tL 2v=2
2+ yi
Diophantus assumes y = mx, whence x = 2/(72* — 2), and we have

( 2 J+ 2 Ys 2 3
2 ——- —— = gf
mi: — 2 mE: — 2 m:— 2

27mt i
(mt = 2)

! There is a special case in which C and 2 vanish, 443+ Ba®=3% Here y is put
equal to 7x, and x=B[(m?- 4). Cf. 1v. 6, 28 (2).

 

or =.
92 INTRODUCTION

To make 2m* a cube, we need only make 2m a cube or put
m=4. This gives x=1.
The general case
Ax + Bx* + (x= 2°
ox + cx =
would, of course, be much more difficult; for, putting y=mx, we
have

x=c/(m* - 0b),

and we have to solve

et {i 5) volt 3) +C(5)- 2
or Cem? +c (Bec —26C) m* + be (6C — Be) + Ac =,

of which equation the above corresponding equation is a very
particular case.

Summary of the preceding investigation.

1. Diophantus solves completely equations of the first degree,
but takes pains to secure beforehand that the solution shall be
positive. He shows remarkable address in reducing a number of
simultaneous equations of the first degree to a single equation in
one variable.

2. For determinate equations of the second degree he has
a general method or rule of solution. He takes, however, in the
Arithmetica, no account of more than one root, even where both
roots are positive rational numbers. But, his object being simply
to obtain some solution in rational numbers, we need not be
surprised at his ignoring one of two roots, even though he knew
of its existence.

3. No equations of a degree higher than the second are solved
in the book except a particular case of a cubic.

4. Indeterminate equations of the first degree are not treated
of in the work. Of indeterminate equations of the second degree,
as Ax? + Bx + C=? only those cases are fully dealt with in which
A or C vanishes, while the methods employed only enable us to
solve equations of the more general forms

Ax*+ C=yp* and Ax’ + Bxr+ C="
when 4,0r Cor }B2 - AC is positive and a square number, or (in
the case of 4x* + C= »?) when one solution is already known.
DIOPHANTUS METHODS OF SOLUTION 93

5. For double-equations of the second degree Diophantus has
a definite method when the coefficient of x? in both expressions
vanishes ; the applicability of this method is, however, subject to
conditions, and it has to be supplemented in one or two cases by
another artifice. Of more complicated cases we find only a few
examples under conditions favourable for solution by his method.

6. Diophantus’ treatment of indeterminate equations of degrees
higher than the second depends upon the particular conditions of
the problems, and his methods lack generality.

7. More wonderful than his actual treatment of equations are
the clever artifices by which he contrives to avoid such equations
as he cannot theoretically solve, eg. by his device of “back-
reckoning,” instances of which would have been out of place in
this chapter and can only be studied in the problems themselves.

I shall not attempt to class as “methods” certain headings
in Nesselmann’s classification of the problems, such as (a) “ Solution
by mere reflection,” (4) “ Solution in general expressions,” of which
there are few instances definitely so described by Diophantus, or
(¢) “Arbitrary determinations and assumptions.” The most that
can be done by way of describing these “methods” is to quote
a few characteristic instances. This is what Nesselmann has
done, and he regrets at the end of his chapter on “ Methods of
Solution” that it must of necessity be so incomplete. To under-
stand and appreciate the various artifices of Diophantus it is in
fact necessary to read the problems themselves in their entirety.

With regard to the “ Use of the right-angled triangle,” all that
can be said of a general character is that only “rational” right-
angled triangles (those namely in which the three sides can all be
represented by rational numbers) are used in Diophantus, and
accordingly the introduction of the “right-angled triangle” is
merely a convenient way of indicating the problem of finding
two square numbers, the sum of which is also a square number.
The general form used by Diophantus (except in one case, VI. 19,
gv.) for the sides of a right-angled triangle is (a®+ 8°), (a®— 6%),
2ab, which expressions clearly satisfy the condition

(a® + 02) = (@* — 0°) + (2ab).
The expression of the sides of a right-angled triangle in this form
Diophantus calls “forming a right-angled triangle from the
numbers a and 4.” His right-angled triangles are of course
formed from particular numbers. “Forming a right-angled
94 INTRODUCTION

triangle from 7, 2” means taking a right-angled triangle with sides
(72+ 2°), (72 — 2), 2.7.2, or 53, 45, 28.

II. METHOD OF LIMITS.

As Diophantus often has to find a series of numbers in
ascending or descending order of magnitude, and as he does not
admit negative solutions, it is often necessary for him to reject
a solution which he has found by a straightforward method
because it does not satisfy the necessary condition; he is then
very frequently obliged to find solutions which lie within certain
limits in place of those rejected.

1. A very simple case is the following: Required to find
a value of x such that some power of it, 2%, shall lie between two
given numbers. Let the given numbers be , 4. Then Diophantus’
method is to multiply both @ and & by 2%, 37, and so on, successively,
until some zth power is seen which lies between the two products.
Thus suppose that ¢" lies between ap" and 4p"; then we can put
x=c[p, in which case the condition is satisfied, for (¢/p)* lies
between a and &.

Exx. In 1v. 31 (2) Diophantus has to find a square between
1} and 2. He multiplies both by a square, 64; this gives 80 and
128, and 100 is clearly a square which lies between them; there-
fore (12)? or 23 satisfies the prescribed condition.

Here, of course, Diophantus might have multiplied by any
other square, as 16. In that case the limits would have become
20 and 32; between these lies the square 25, which gives the same
square 23 as that before found. :

In VI. 21 a sixth power (‘“cube-cube”) is sought which shall
lie between 8 and 16. The sixth powers of the first four natural
numbers are 1, 64, 729, 4096. Multiply 8 and 16 by 2° or 64, and
we have as limits 512 and 1024, between which 729 lies. There-
fore 720 is a sixth power satisfying the given condition. To
multiply by 729 in this case would not give us a solution.

2. Sometimes a value of x has to be found which will give
some function of x a value intermediate between the values of two
other functions of x.

Ex. 1. In Iv. 25 it is necessary to find a value of x such that
8/(x* + x) shall lie between x and x + 1.
The first condition gives 8 > 2® + 2
DIOPHANTUS' METHODS OF SOLUTION 95

Diophantus accordingly assumes that
S3=lz tL iP=2ttat iv tol,
which is greater than 3 + x2.

Thus r=} satisfies one condition. It is also seen to satisfy
= 8 : :
the second condition, or i <x+ 1. Diophantus, however, says

nothing about the second condition being satisfied ; his method is,
therefore, here imperfect.

Ex. 2. In Vv. 30 a value of x has to be found which shall make
x > { (x*— 60) but <1 (2*—60),

that is, x2 —60> 5x
x2 —60< 8x)’

Hence, says Diophantus, x is not less than 11 and not greater
than 12. We have already spoken (p. 60 sqq.) of his treatment
of such cases.

Next, the problem in question requires that 22—60 shall be
a square. Assume then that

x? —,60 = (xr — m)},
and we have x = (1m? + 60)/2m.

Since, says Diophantus, x is greater than 11 and less than 12,
it follows that
m?® +60 > 221 but < 24m;

and #2 must therefore lie between 19 and 21 (cf. p. 62 above).
He puts 72 =20, and so finds x =11%.

III. METHOD Or APPROXIMATION TO LIMITS.

We come, lastly, to a very distinctive method called by
Diophantus maptgdrns or mapioornros daywyy. The object of this
is to solve such problems as that of finding two, or three, square
numbers the sum of which is a given number, while each of them
approximates as closely as possible to one and the same number.

This method can be best exhibited by giving Diophantus’ two
instances, in the first of which #zo such squares, and in the second
three, are required. In cases like this the principles cannot be
so well indicated with general symbols as with concrete numbers,
which have the advantage that their properties are immediately
96 INTRODUCTION

obvious, and the separate expression of conditions is rendered
unnecessary.
Ex. 1. Divide 13 into two squares each of which >6 (Vv. 9).
Take half of 13, or 6}, and find what small fraction 1/#? added
. to it will make it a square: thus

I I
63 +5, Or 26 + 7 must be a square.

Diophantus assumes

I IN
26+ 7 = ( +2), or 267+ 1= + 1)}
=i 74 (37 +1)

whence y = 10 and 1/3? = 1}, or 1/2*=4}5; and
6} + 535 = a square, (33)

[The assumption of (57+ 1)* is not arbitrary, for assume
2692+ 1 = (py +1)%, and y is then 2p/(26 —p?); since 1/y should be
a small proper fraction, 5 is the most suitable and the smallest
possible value for p, inasmuch as 26 —p*< 2p or p* + 2p + 1 > 27.]

It is now necessary, says Diophantus, to divide 13 into two
squares the sides of which are both as near as possible to 53.

Now the sides of the two squares into which 13 is naturally
decomposed are 3 and 2, and

3is > & by
2is < $4 by ii.
But, if 3 — 4%, 2+ 1} were taken as the sides of two squares, the
sum of the squares would be
2.2601

zgyp=tt,

 

which is > 13.

Accordingly Diophantus puts 3 — gx, 24 11x for the sides of
the required squares, where therefore x is not exactly J; but
near it.

Thus (G—-9r)*+ (2+ 11x) =13,
and Diophantus obtains Z= of,

The sides of the required squares are 257, 228

[It is of course a necessary condition that the original number,
here 13, shall be a number capable of being expressed as the sum
of two squares.]
DIOPHANTUS’ METHODS OF SOLUTION 97

Ex. 2, Divide 10 into three squares such that each of them
is >3(v, 17)

[The original number, here 10, must of course be expressible
as the sum of three squares.]

Take one-third of 10, or 3}, and find what small fraction of the
form 1/2? added to 3} will make a square; ze. we have to make
30 + 2 a square, or 309° + I a square, where 3/x = 1/y.

Diophantus assumes

307+ 1 ={5p 4 1},

whence y = 2 and therefore 1/2? = 4%; and 34 +, =12L, a square.

[As before, if we assume 309* = (py + 1)%, y =2p/(30—2%); and,
since 1/y must be a small proper fraction, 30 —p? should be < 2p,
or p*+2p+1>31. Accordingly Diophantus chooses 5 for p as
being the smallest possible integral value.]

We have now, says Diophantus, to make wan of the sides
of our required squares as near as may be to 4

Now 10= 9+ 1= 3+ (+E)
and 3, 2, # are the sides of three squares the sum of which is 10.
Bringing (3, £, £) and 4! to a common denominator, we have

(33, 45, 2%) and 52.

And 3>42 by 23,
p<it by B,
te

If now we took 3— 85, 348%, 4431 for sides of squares, the

sum of the squares would be 3 (3; A or 383 which is > 10.

Accordingly Diophantus assumes as the sides of the three
required squares

3-357, $+ 37% $5 31%
where x must therefore be not exactly 4; but near it.

Solving (3-339) + (+372) +4 + 312) =10,

or 10—116x + 355522 = 10,
we find x = Ale

the required sides are therefore

and the required squares

JT AB04L 1658944
LEAEAL, JEpdaR0, IEDRON
98 INTRODUCTION

Other instances of the application of the method will be found
in V. 10, 12, 13, 14, where, however, the squares are not required to
be nearly equal, but each of them is subject to limits which may
be the same or different; e.g. sometimes each square is merely
required to be less than a given number (10, say), sometimes the
squares lie respectively between different pairs of numbers, some-
times they are respectively greater than different numbers, while
they are always subject to the condition that their sum is a given
number.

As it only lies within the scope of this Introduction to explain
what we actually find in Diophantus, I cannot do more than give
a reference to such investigations as those of Poselger in his
“ Beitrige zur unbestimmten Analysis” published in the Abkand-
lungen der Koniglichen Akademie dev Wissenschaften zu Berlin aus
dem Jahre 1832, Berlin, 1834. One section of this paper Poselger
entitles “ Anniherungs-methoden nach Diophantus,” and obtains
in it, on Diophantus’ principles, a method of approximation to the
value of a surd which will furnish the same results as the method
of continued fractions, with the difference that the “ Diophantine
method” is actually quicker than the method of continued frac-
tions, so that it may serve to expedite the latter.
CHAPTER V
THE PORISMS AND OTHER ‘ASSUMPTIONS IN DIOPHANTUS

I HAVE already mentioned (in Chapter I.) the three explicit refer-
ences made by Diophantus to “ The Porisms ” and the possibility
that, if these formed a separate work, it may have been from that
work that Diophantus took a number of other propositions relating
to properties of numbers which he enunciates or tacitly takes for
granted in the Arithmetica.

I begin with the three propositions for which he expressly
refers to % The Porisms.”

LPorism 1. In V. 3 he says, “We have it in the Porisms that,
‘If each of two numbers and their product when severally added to
the same given number produce squares, the squares with which
they are so connected are squares of two consecutive numbers.’”

That is to say, if r+a=m? y+a=n* and if xy +a is also a
square, then m~n=1.

~The theorem is not correctly enunciated, for it would appear
that mz ~»n=1 is not the only condition under which the three
expressions may be simultaneously squares.

For suppose

Xxta=m, y+a=w, xyta=p>.
By means of the first two equations we have
xy +a=mnt—a(m*+n—1) + a’
In order that
mnt —a(m? +n —1)+a® -
may be a square certain conditions must be satisfied. One suffi-

cient condition is
m+n: —1=2mn,

or m~n=+1I,
which is Diophantus’ condition.
! Literally ¢“(the numbers) arise from two consecutive squares” (yeybvaow dmwd ddo

TeTpaydywy TOV Kata TO €£Fs).
I=—2
100 INTRODUCTION

But we may also regard
min? — a (m+n — 1) + a =p?
as an indeterminate equation in 7 of which we know one solution,
namely m =n + 1.
Other solutions are then found by substituting z+ (z £ 1) for
m, whence we obtain the equation
(—a)z2+2{n*(nt1)—a(nti1)}z
+(—a)y (nt 1P—a(m—1)+a*=p%
or (—a)z2+2(m—a)y (nt 1) z+ {n(n t1)—al*=p%
which is easy to solve in Diophantus’ manner, since the absolute
term is a square.

But in the problem V. 3 #4ree numbers are required, such that
each of them, and the product of each pair, when severally added
to a given number, produce squares. Thus if the third number be
z, three additional conditions have to be satisfied, namely

2ta=1 zxta=v, zta=u’
The two last conditions are satisfied, if 7 + 1 = 2, by putting
2=2(2 +7) — I =4m?+ 4m + 1 — 4a,
when xz + a= {m(2m+1)- 2a}
and yz+a={(m+1)(2m+1)— 2a}? -
and perhaps this means of satisfying the conditions may have
affected the formulation of the Porism

The problem V. 4 immediately following assumes the truth of
the same Porism with — a substituted for + a.

Porism 2. In V. 5 Diophantus says, “ Again we have it in the
Porisms that, ‘ Given any two consecutive squares, we can find in
addition a third number, namely the number greater by 2 than the
double of the sum of the two squares, which makes the greatest of
three numbers such that the product of any pair of them added to
either the sum of that pair or the remaining number gives a square.”

That is, the three numbers

mt, (m+ 1)% 4(m*+m + 1)

! Euler has a paper describing and illustrating a general method of finding such
‘“porisms ” the effect of which is to secure that, when some conditions are satisfied, the
rest are simultaneously satisfied (“‘De problematibus indeterminatis quae videntur plus
quam determinata” in Novi Commentarii Acad. Petropol. 1756-357, Vol. VI. (1761),
P- 85 sqq. = Commentationes arithmeticae collectae, 1. pp. 245—259). This particular
porism of Diophantus appears as a particular case in § 13 of the paper.
THE PORISMS AND OTHER ASSUMPTIONS 101

have the property that the product of any two plus either the sum
of those two or the remaining number gives a square. In fact, if
X, VY, Z denote the numbers respectively,

XY+ XH Y=(m+ m+1)Y, XY Z=( 2+ m+2),
YZ + Y+Z=0Cm+3m+3), YZ+X=0@m+ 3m+ 2),
ZX + Z + X=0m+ m+2)’, ZX + Y=0@nr2+ m+1)

Porism 3 occurs in Vv. 16. Unfortunately the text is defective
and Tannery has had to supply three words! ; but there can be no
doubt that the correct statement of the Porism here in question is
“The difference of any two cubes is also the sum of two cubes,”
ze. can be transformed into the sum of two cubes, or two cubes can
be found the sum of which is equal to the difference between any
two given cubes. Diophantus contents himself with the enuncia-
tion of the proposition and does not show how to prove it or how
he effected the transformation in practice. The subject of the
transformation of sums and differences of cubes was investigated
by Vieta, Bachet and Fermat.

Vieta (Zetetica, Iv. 18-20) has three problems on the subject.

(1) Given two cubes, to find in rational numbers two other
cubes such that their sum is equal to the difference of the given
cubes’.

As a solution of a® — 4* =x? + 33, he finds
«ald — 207) f(a = 7)
ZiE 2 T 25h

A

1 Exouer 8 év Tols Iloplopaciy dtu ““ wdvrwy dbo kiBwy 7) bmepoxh kiBwy <Bdbo aivleud
o>."
2 The solution given by Vieta is obtainable thus. The given cubes being a3, 43, where
a>b, we assume x — b, a — £x as the sides of the required cubes.
Thus (x=0)3+ (a- kx)3=a3- 43,
whence x3 (1 — £3) + 3x2 (ak? — b) + 3x (6% — a?) =o.
This reduces to a simple equation if we assume
b2-a%k=o0, or k=0b%a?,
in which case :
radia) i220
eR TN
and the sides of the cubes are therefore
6(2a3-8%)  a(a®-28%)
a+e ad+68
Vieta’s second problem is similarly solved by taking a+ x, 4x—4 as the sides of the
requized cubes, and the third problem by taking x - 4, £x — a as the sides of the required
cubes respectively.

 

 
102 INTRODUCTION

(2) Given two cubes, to find in rational numbers two
others such that their difference is equal to the sum of the given

cubes.
Solving a® + & = x*—3?, we find that

a(a®+ 26°) _b(2a® +H)
a © oT

(3) Given two cubes, to find in rational numbers two cubes
such that their difference is equal to the difference of the given

cubes.
For the equation a® — # = x* — 3%, Vieta finds

be -0) _alal—a)
zr BLE = a+ 5
as a solution

In the solution of (1) x» is clearly negative if 26° > a?; therefore,
in order that the result may be “rational,” #* must be > 2’. But
for a “rational ” result in (3) we must, on the contrary, have @® < 28°.
Fermat was apparently the first to notice that, in consequence, the
processes in (I) and (3) exactly supplement each other, so that by
employing them successively we can effect the transformation
required in (1) even when a? is not > 242 : ;

The process (2) is always possible; therefore, by a suitable
combination of the three processes, the transformation of a sum
of two cubes into a difference of two cubes, or of a difference of
two cubes into a sum or a difference of two other cubes is always

! Vieta’s formulae for these transformations give any number of very special solutions
(in integers and fractions) of the indeterminate equation x3 + 33+ 23=13, including solutions
in which one of the first three cubes is negative. These special solutions are based on
the assumption that the values of two of the unknowns are given to begin with. Euler
observed, however, that the method does not give all the possible values of the other two
even in that case. Given the cubes 33 and 43, the method furnishes the solution

65\3 8 : ;

33+ 43+ (£5) = (+2) , but not the simpler solution 33+43+353=63. Euler ac-
cordingly attacked the problem of solving the equation a3+ 33+ 23=23 more generally.

He began with assuming only one, instead of two, of the cubes to be given, and, on that
assumption, found a solution much more general than that of Vieta. Next he gave a
more general solution still, on the assumption that ose of the cubes are given to begin
with. Lastly he proceeded to the problem 7v find all the sets of three integral cubes the
sum of whick is a cube and showed how to obtain a very large number of such sets
including sets in which one of the cubes is negative (Novi Commentarii Acad. Petropol.
1756-57, Vol. VL. (1761), p. 155 sq.= Commentationes arithmeticae, 1. pp. 193—20%).
The problem of solving 43+ 33=23+123 in integers in any number of ways had occupied
Frénicle, who gave a number of solutions (Oeuvres de Fermat, 111. pp. 420, 535) ; but the
method by which he discovered them does not appear,
THE PORISMS AND OTHER ASSUMPTIONS 103

practicable’. Fermat showed also how, by a repeated use of the
several processes as required, we can transform a sum of two cubes
into a sum of two other cubes, the latter sum into the sum of two
others and so on ad infinitum?

Besides the “Porisms” there are many other propositions
assumed or implied by Diophantus which are not definitely called

1 Fermat (note on 1v. 2) illustrates by the following case :
Given two cubes 125 and 64, to transform their difference into the sum of two other

cubes.
Here a=35, 6=4, and so 263>a3; therefore we must first apply the third process

by which we obtain
(5y- 0)
5 5 af sgn,
reel 63

3 3
As (32) >2 (2) , we can, by the first process, turn the difference between the

(2m
cubes ( —*= ) and ( =) into the sum of two cubes.
63 63

“In fact,” says Fermat, ‘‘if the three processes are used in turn and continued
ad infinitum, we shall get a succession ad infinitum of two cubes satisfying the same
condition ; for from the two cubes last found, the sum of which is equal to the difference
of the two given cubes, we can, by the second process, find two more cubes the difference
of which is equal to the sum of the two cubes last found, that is, to the difference
between the two original cubes; from the new difference between two cubes we can
obtain a new sum of two cubes, and so on ad infinitum.”

As a last illustration, to show how a difference between cubes can be transformed into
the difference between two other cubes even where the condition for process (3) is not
satisfied, Fermat takes the case of 8 — 1, z.e. the case where

a=2, b=1 and a3>28,.

First use process (1) and we have

en 44
3 3
Then use process (2), and

(2 3 (2) 12653 (Ey
—- + ( = =| — ge A omens .
3 3 183 183
2 Suppose it required to solve the fourth problem of transforming the sum of two cubes
into the sum of two other cubes.

Let it be required so to transform 23+ 13 or g.
First transform the sum into a difference of two cubes by process (2). This gives

Bp fie (2) i (2).
7 7

The latter two cubes satisfy the condition for process (3) and, applying that process,

we get
(5Y- (7) = Gant) - Go)
7 7 90391 90391
The cubes last found satisfy the condition for process (1), and accordingly the difference

between the said last cubes, and therefore the sum of the original cubes, is at last trans-
formed into the sum of two other cubes.

 

 
104 INTRODUCTION

porisms, though some of them are of the same character as the
three porisms above described.
Of these we may distinguish two classes.

1. The first class of theorems or facts assumed without ex-
planation by Diophantus are more or less of the nature of identical
JSormulae. Some are quite simple, e.g. the facts that the expressions
(3(@+6)}?—ab and a*(a+1)*+a*+ (a+ 1) are respectively
squares, that the expression a(a*— a) + a+ (a®* — a) is always a
cube, and the like.

Others are more difficult and betoken a certain facility in work-
ing with quasi-algebraical expressions. Examples of this kind are
the following :

(0) If X=a%+ 2a, V=(a+1)x+2(a+1),or, in other words,
2X +ti=s(art1), 2 ti={(at1)r+1]° then XV +1 05 a
square [IV. 20]. As a matter of fact,

XY +i1={a(@+1)x+(2a+1)}2
(B) 8 times a triangular number plus 1 gives a square [IV. 38].

In fact, 8 ry + T={(2v+4 1)

ly) WH Xto=w Yia=(m+1), and Z=2(X + VY)~1,
then the expressions YZ +a, ZX +a, XY +a are all squares.
(The upper signs refer to the assumption in Vv. 3, the lower to that
in V. 4.)

In fact, YZ ta={(m+1)(2m+ 1) T 2a}?
ZX ta={m(iam+ 1) FT 24)%,
XY ta={m@m+1)7F al

O) f X=mt2, Y=(m4+1)"+ 2, Z=2{m*+(m +1) +1} + 2,
then the six expressions

YZ-(Y+Z), Z£2X—-(Z+X), XY-(X+7)
YZ-X, ZX-Y, XyY-Z
are all squares [V. 6].
In fact,
YZ —(Y+2)=(@m*+3m+3), YZ—-X=(02m+3m+4), etc.

2. The second class is much more important, consisting of a
number of propositions in the Theory of Numbers which we find
THE PORISMS AND OTHER ASSUMPTIONS 105

first stated or assumed in the Arithmetica. It was, in general, in
explanation or extension of these that Fermat's most famous notes
were written. How far Diophantus possessed scientific proofs of
the propositions which he assumes, as distinct from a merely
empirical knowledge of them, must remain to a great extent
matter of speculation.

(@) Theorems in Diophantus vespecting the composition of num-
bers as the sum of two squares.

(1) Any square number can be resolved into two squares in
any number of ways, IL 8.

(2) Any number which is the sum of two squares can be
resolved into two other squares in any number of ways, II. 9.

N.B. It is implied throughout that the squares may be frac-
tional as well as integral.

(3) If there are two whole numbers eack of which ts the sum of
two squares, their product can be resolved into the sum of two squares
in two ways, 111 109.

The object of 111. 19 is to find four rational right-angled triangles
having the same hypotenuse. The method is this. Form two
right-angled triangles from (a, 4) and (¢, &) respectively, ze. let
the sides of the triangles be respectively

a+ 0, a — 0, 2a,
and CAdd —d2 2d.

Multiplying all the sides in each by the hypotenuse of the other,
we have two triangles with the same hypotenuse, namely

(a® + 0%) (c* + d?), (a* — 6*) (2 + d?), 2ab (3 + &2),
and (@® + 0) (2 + a2), (a® + B*) (2 — d?), 2cd (a® + B°).

Two other triangles having the same hypotenuse are obtained
by using the theorem enunciated. In fact,

(@® + 02) (+ d*) = (ac + bd)? + (ad F bc)?

and the triangles are formed from ac+ bd, ad F bc, being the
triangles

(a + 0*)( + d?), gqabcd + (a* — 0*)(c* — d?), 2 (ac + bd) (ad — bc),
(a® + 62) (c+ d*), 4abcd — (a* — 0°) (c* — d*), 2 (ac — bd) (ad + bc).
106 INTRODUCTION

In the case taken by Diophantus

A+ =22+12=5,
C+ai=312"=13,
and the four triangles are respectively
(65, 52, 39), (65, 60, 25), (65,63, 16), (65, 56, 33).

(If certain relations! hold between a, §, ¢, , this method fails.
Diophantus has provided against them by taking two triangles “in
the smallest numbers” (¥70 éAayloTwv apifudv), namely 3,4, 5 and
5.12, 13.)

Upon this problem 111. 19 Fermat has a long and important
note which begins as follows? :

“[1] A prime number of the form 4z + 1 is the hypotenuse of
a right-angled triangle in one way only, its square is so in two
ways, its cube in three, its biquadrate in four ways, and so on ad
infinitum.

“[2] The same prime number 47+ 1 and its square are the
sum of two squares in one way only, its cube and its biquadrate
in two ways, its fifth and sixth powers in three ways, and so on ad
infinitum.

“[3] If a prime number which is the sum of two squares be
multiplied into another prime number which is also the sum of
two squares, the product will be the sum of two squares in two
ways ; if the first prime be multiplied into the square of the second

1 (1) We must not have a/6=c|d or a/b=d]c, for in either case one of the perpendiculars
of one of the resulting triangles vanishes, making that triangle illusory. Nor (2) must
cd be equal to (a+0b)/(a-b) or to (a—b)/(a+8b), for in the first case ac—bd=ad+ bc,
and in the second case ac+bd=ad— bc, so that one of the sums of squares equal to
(a? +0?) (2+ 4d?) is the sum of two equal squares, and consequently the triangle formed
from the sides of these equal squares is illusory, one of its perpendicular sides vanishing.

2 G. Vacca (in Bibliotheca Mathematica, 113. 19o1, pp. 358-9) points out that Fermat
seems to have been anticipated, in the matter of these theorems, by Albert Girard, who
has the following note on Diophantus Vv. g (Ocuvres mathématiques de Simon Stevin par
Albert Girard, 1634, p. 156, col. 1):

*¢ ALB. GIR. Determinaison d’un nombre qui se peut diviser en deux quarrez entiers.

I. Tout nombre quarré.

II. Tout nombre premier qui excede un nombre quaternaire de I’unité.

III. Le produict de ceux qui sont tels.

IV. Et le double d’un chacun d’iceux.

Laquelle determinaison n’estant faicte n'y de I’Autheur n’y des interpretes, servira tant
en la presente et suivante comme en plusieurs autres.”

Now Girard died on g December, 1632; and the Theorems of Fermat above
quoted are apparently mentioned by him for the first time in his letter to Mersenne of
25 December, 1640 (Oeuvres de Fermat, 11. p. 213). Was the passage of Girard known
to Fermat ? :
THE PORISMS AND OTHER ASSUMPTIONS 107

prime, the product will be the sum of two squares in three ways ;
if the first prime be multiplied into the cube of the second, the
product will be the sum of two squares in four ways, and so on
ad infinitum.”

It is not probable that Diophantus was aware that prime num-
bers of the form 47 + 1 and numbers arising from the multiplication
of such numbers are the only classes of numbers which are always
the sum of two squares; this was first proved by Euler? But it
is remarkable that Diophantus should have selected the first two
prime numbers of the form 42+ 1, namely 5 and 13, which are
both sums of two squares, as the hypotenuses of his first two right-
angled triangles and then made their product, 65, the hypotenuse
of other right-angled triangles, that product having precisely the
property of being, as in Fermat's [3], the sum of two squares in
two ways. Diophantus may therefore have had an inkling, whether
obtained empirically or otherwise, of some of the properties enunci-
ated by Fermat.

(4) Still more remarkable is a condition of possibility of solution
prefixed to the problem Vv. 9. The object of this problem is “to
divide 1 into two parts such that, if a given number is added to
either part, the result will be a square.” Unfortunately, the text
of the added condition is uncertain. There is no doubt about the
first few words, « Ze given number must not be odd,” i.e. No number
of the form 4n + 3: [or 4n — 1] can be the sum of two squares.

~The text, however, of the latter half of the condition is corrupt.
The true condition is given by Fermat thus: “ Zhe given number
must not be odd, and the double of it increased by one, when divided
by the greatest square which measures it, must not be divisible by a
prime number of the form 4n—1" (Note upon Vv. 9; also in a
letter to Roberval) There is room for any number of conjectures
as to what may have been Diophantus’ words.

1 For a fuller account of this note see the Supplement, section I.

? Novi Commentarii Acad. Petropol. 1752-3, Vol. IV. (1758), pp. 3-40, and 1754-5,
Vol. v. (1760), pp. 3-58 = Commentationes arithmeticae, 1. pp. 155-173 and pp. 210-233 ;
cf. Legendre, Zaklentheorie, tr. Maser, I. p. 208; Weber and Wellstein’s Encyclopddie
der Elementar- Mathematik, 1,. pp. 285 sqq.

3 Qeuvres de Fermat, 11. pp. 203-4. See the Supplement, section I.

4 Bachet’s text has det 89 Tov Sidbuevor wire mepioodv elvar, pire 6 durhaciwy abTod
Yu a. pelfova Expy pépos 8.  perpeirar vmwd Tob a%%, oF,

He also says that a Vatican MS. reads wire 6 durhaciwr alrod dpifudy povdda a.
pelfova Exy pépos Téraprov, 1) perpeirar Ymd Tod wpdTov dpibod.

Neither does Xylander help us much. He frankly tells us that he cannot understand
108 INTRODUCTION

There would seem to be no doubt that in Diophantus’ condition
there was something about “double the number” (ze. a number of
the form 47), also about “greater by unity” and “a prime number.”
It seems, then, not unlikely that, if Diophantus did not succeed in
giving the complete sufficient and necessary condition stated by
Fermat, he made an approximation to it; and he certainly knew
that no number of the form 47+ 3 could be the sum of two
squares’.

(6) On numbers which are the sum of three squares.

In the problem Vv. 11 a condition is stated by Diophantus re-
specting the form of a number which, added to three parts of unity,
makes each of them square. If a be this number, clearly 3a+ 1
must be divisible into three squares.

Respecting the number a Diophantus says, “It must not be 2
or any multiple of 8 increased by 2.”

That is, a number of the form 24n+ 7 cannot be the sum of three
squares. Now the factor 3 of 24 is irrelevant here, for with respect
to three this number is of the form 32 + 1, and this, so far as 3 is
concerned, might be a square or the sum of two or three squares.
Hence we may neglect the factor 3 in 242.

We must therefore credit Diophantus with the knowledge of

 

the passage. ‘‘Imitari statueram bonos grammaticos hoc loco; quorum (ut aiunt) est
multa nescire. Ego vero nescio heic non multa, sed paene omnia. Quid enim (ut
reliqua taceam) est uxre 6 durhaciwy advTod ap uo a etc., quae causae huius mposdiopiouod,
quae processus? immo qui processus, quae operatio, quae solutio ?”’

Nesselmann discusses an attempt made by Schulz to correct the text, and himself
suggests wire Tov durhaciova avTob dplfudy povdde melfova Exew, 8s perpelratr Uwé Twos
mpwrov dptfuol. But this ignores uépos Térapror and is not satisfactory.

Hankel, however (Gesck. d. Math. p. 169), says: ‘Ich zweifele nicht, dass die
von den Msscr. arg entstellte Determination so zu lesen ist: Ael 0% Tov Otdduevor pie
mwepiooov elvar, wife Tov durhaciova alrol dpfudr movdd a uelfova perpeicfar wb TOU
mpwrov dptfuod, 8s dv uovddt a@ pelfwy Expy pépos Téraprov.” This correction seems a
decidedly probable one. Here the words uépos Téraprov find a place; and, secondly,
the repetition of worddt a umelfwv might well confuse a copyist. Tov for 700 is of course
natural enough; Nesselmann reads Twos for Tov.

Tannery, improving on Hankel, reads Aet 8%) Tov 8dbuevor wife mepioadr elvar, uhre
T7ov Surhdotov avrod kal movddt mig pelfova werpetcfar Vmd Tov wpdTou dpifuod < od 6
Movdde ulg pelfwy > Exp pépos Téraprov t, ‘‘ the given number must not be odd, and twice
it plus 1 must not be measured by any prime number which, when increased by 1, is
divisible by 4.”

1 A discussion of the text and a suggestion as to the considerations which may have
led to the formulation of the condition will be found in Jacobi, ¢‘ Ueber die Kenntnisse
des Diophantus von der Zusammensetzung der Zahlen” (Berliner Monatsberichte, 1847 ;
Gesammelte Werke, V11., 1891, pp. 332-344).
THE PORISMS AND OTHER ASSUMPTIONS 109

the fact that no number of the form 8n + 7 can be the sum of
three squares’.

This condition is true, but does not include @// the numbers
which cannot be the sum of three squares, for it is not true that
all numbers which are not of the form 8z + 7 are made up of three
squares. Even Bachet remarked that the number # might not be
of the form 32749, or a number of the form 96x + 28 cannot be
the sum of three squares.

Fermat gives the conditions to which @ must be subject thus

Write down two geometrical series (common ratio of each 4),
the first and second series beginning respectively with 1, 8,

I. 4 16 64 250 1024 4000

3.32. 128 3512 2043 8102 32763:
then @ must not be (1) any number obtained by taking twice any
term of the upper series and adding all the preceding terms, or
(2) the number found by adding to the numbers so obtained any

multiple of the corresponding term of the second series.
Thus a must not be

8k+2.1 = 184-0
328+2.4+1 = 324-+0
1282 42.16+4+1 = 128% + 37,

SI2R+ 2.044 16+ 4+ 1=512/ + 149,

and so on, where £=0 or any integer.

That is,since I + 4 +42 + ... + 4" =1(4" — 1), a cannot be either

2.44 )=4 (7. 44 = 1)

or 22. 40410 LL -D=3(240. 447 4 1);
therefore 3a + 1 cannot be of the form 4" (24% + 7) or 4" (8% + 7).

Again, there are other problems, eg. V. 10 and Vv. 20, in which,
though conditions are necessary for the possibility of solution, none
are mentioned ; but suitable assumptions are tacitly made, without
explanation. It does not follow, from the omission to state the
conditions, that Diophantus was ignorant of even the minutest

points connected with them; as, however, we have no definite
statements, we must be content to remain in doubt.

1 Legendre proved (Zaklentheorie, tr. Maser, 1. p. 386), that numbers of this form are
the only odd numbers which are not divisible into three squares.
2 Note on Diophantus v. 11.
110 INTRODUCTION

(¢) Composition of numbers as the sum of four squares.

Every number is either a square or the sum of two, three or four
squares. This well-known theorem, enunciated by Fermat!, and
proved by Lagrange? (who followed up results obtained by Euler)
shows at once that any number can be divided into four squares
either integral or fractional, since any square number can be divided
into two other squares, integral or fractional. We have now.to look
for indications in the Arithmetica as to how far Diophantus was
acquainted with the properties of numbers as the sum of four squares.
Unfortunately, it is impossible to decide this question with anything
like certainty. There are three problems, IV. 29, 30 and V. 14, in
which it is required to divide a number into four squares, and from
the absence of mention of any condition to which the number must
conform, considering that in both cases where a number is to
be divided into #kree or two squares, V. 11 and V. 9, he does
state a condition, we should probably be right in inferring that
Diophantus was aware, at least empirically, that any number could
be divided into four squares. That he was able to prove the
theorem scientifically it would be rash to assert. But we may
at least be certain that Diophantus came as near to the proof of
it as did Bachet, who takes all the natural numbers up to 120
and finds by trial that all of them can actually be expressed
as squares, or as the sum of two, three or four squares in whole
numbers. So much we may be sure that Diophantus could do, and
hence he might have empirically satisfied himself that it is possible
to divide any number into four squares, integral or fractional, even
if he could not give a rigorous mathematical demonstration of the

general theorem.

1 See note on Diophantus IV. 29; cf. also section I. of the Supplement.

2 ¢« Démonstration d'un théoréme d’arithmétique ”’ in Nouveaux Mémoires de I’ Acad.
royale des sciences de Berlin, année 1770, Berlin 1772, pp. 123-133 = Oeuvres de Lagrange,
IIL. pp. 187-201; cf. Wertheim’s account of the proof in his Diophantus, pp. 324-330.
CHAPTER VI

THE PLACE OF DIOPHANTUS

IN algebra, as in geometry, the Greeks learnt the beginnings
from the Egyptians. Familiarity on the part of the Greeks with
Egyptian methods of calculation is well attested. Thus (1) Psellus
in the letter published by Tannery! speaks of “the method
of arithmetical calculations used by the Egyptians, by which
problems “in analysis are handled” (§ kar Alyvmrriovs Tv
apilfudv wéfodos, O/ 1s olkovouelTar TA Kata THY AVANUTIKNY
mpofBMijpara); the details which he goes on to give respecting
the technical terms for different kinds of numbers, including the
powers of the unknown quantity, in use among the Egyptians
are doubtless taken from Anatolius. (2) The scholiast to Plato’s
Charmides 165 E may be drawing on the same source when he
says that “parts of Aoyioriks (the science of calculation) are the
so-called Greek and Egyptian methods in multiplications and
divisions, and the additions and subtractions of fractions....The
aim of it all is the service of common life and utility for contracts,
though it seems to deal with things of sense as if they were
perfect or abstract.” (3) Plato himself, in the Laws? says that
free-born boys should, as is the practice in Egypt, learn, side by
side with reading, simple mathematical calculations adapted to their
age, which should be put into a form such as to give amusement
and pleasure as well as instruction; eg. there should be different
distributions of such things as apples, garlands, etc., different
arrangements of numbers of boys in contests of boxing or wrestling,
illustrations by bowls of different metals, gold, copper, silver, etc,
and simple problems of calculation of mixtures; all of which are '
useful in military and civil life and “in any case make men more
useful to themselves and more wide-awake.”

1 Dioph. II. pp. 37-42. 2 Laws, Vil. 819 A-C.
112 INTRODUCTION

The Egyptian calculations here in point (apart from their
method of writing and calculating in fractions, which differed
from that of the Greeks in that the Greeks worked with ordinary
fractions, whereas the Egyptians separated fractions into sums
of submultiples, with the exception of ¢ which was not decomposed)
are the Aawu-calculations. Hau, meaning a /eap, is the term used
to denote the unknown quantity, and the calculations in terms
of it are equivalent to the solutions of simple equations with one
unknown quantity’. Examples from the Papyrus Rhind? corre-
spond to the following equations:

x +x =19,
fr+dr +r +x =33,
(+i) -1(x +3) = 10.

Before leaving the Egyptians, it is right to mention their
anticipation, though in an elementary form, of a favourite method
of Diophantus, that of the “false supposition” or “regula falsi”
as it is sometimes called. An arbitrary assumption is made as to
the value of the unknown, and the value is afterwards corrected
by a comparison of the result of substituting the wrong value in
the original expression with the actual fact. Two instances
mentioned by Cantor® may be given. The first, taken from the
Papyrus Rhind, is the problem of dividing 100 loaves among five
persons in numbers forming an arithmetical progression and such
that one-seventh of the sum of the first three shares is equal to
the sum of the other two. If a+4+4d, a+ 3d, a+ 2d, a+d, a
are the shares, we have

3a +9d=7 (2a +d),
or d = sia.
Ahmes merely says, without explanation, “ make the difference,
as it is, 54,” and then, assuming @=1, writes the series 23, 173,
12, 64, 1. The addition of these gives 60, and 100 is 1% times 60.
Ahmes says simply “multiply 1% times” and thus gets the correct
values 38%, 29%, 20, 103 ¢, 13. The second instance (taken from
the Berlin Papyrus 6619) is the solution of the equations
x+y} = 100,

x:y=1:4 or y=ixr

! For a complete account of the subject the reader is referred to Moritz Cantor’s
Geschichte der Mathematik, 15. Chapter 11., especially pp. 74-81.

* Eisenlohr, Ein mathematisches Handbuch der alten Agypter (Papyrus Rhind des

British Museum), Leipzig, 1877.
3 Geschichte der Math. 14. pp. 78-9 and p. 95.
THE PLACE OF DIOPHANTUS 113

x is first assumed to be 1, and x%+ 3? is then found to be 25/16.
In order to make 100, 25/16 has to be multiplied by 64 or 8%. The
true value of x is therefore 8 times 1, or 8.

The simple equations solved in the Papyrus Rhind are just the
kind of equations of which we find numerous examples in the
arithmetical epigrams included in the Greek Anthology. Most
of these appear under the name of Metrodorus, a grammarian,
who probably lived about the time of the Emperors Anastasius I.
(491-518 A.D.) and Justin I. (518-527 A.D.). They were obviously
only collected by Metrodorus, from ancient as well as more recent
sources; none of them can with certainty be attributed to Metro-
dorus himself. Many of the epigrams (46 in number) lead to
simple equations, with one unknown, of the type of the /au-
equations, and several of them are problems of dividing a number
of apples or nuts among a certain number of persons, that is
to say, the very type of problem alluded to by Plato. For
example, a number of apples has to be determined such that, if
four persons out of six receive one-third, one-eighth, one-fourth
and one-fifth respectively of the total number of apples, while the
fifth person receives ten apples, there remains one apple as the
share of the sixth person, ze.

rt ix+ir+ir+ 10+ 1=x

We are reminded of Plato’s allusion to problems about bowls
(praia) of different metals by two problems (Anthol. Palat. XIV.
12 and 50) in which the weights of bowls have to be found. We
can now understand the allusions of Proclus! and the scholiast
on Charmides 165 E to uphitar and ¢ealirar apifuoi, the adjectives
being respectively formed from wihov, an apple, and ¢udry, a
bowl. It is clear from Plato’s allusions that the origin of such
simple algebraical problems dates back, at least, to the fifth
century B.C. ?

I have not thought it worth while to reproduce at length the
problems contained in the Anthology? but the following is a
classification of them. (1) Twenty-three are simple equations
containing one unknown and of the type shown above; one of
these is the epigram on the age of Diophantus and incidents
in his life (XIv. 126). (2) Twelve more are easy simultaneous

1 Proclus, Comment. on Eucl. [., ed. Friedlein, p. 40, 5.

2 They are printed in Greek, with the scholia, in Tannery’s edition of Diophantus
(11. pp. 43-72 and x), and they are included in Wertheim’s German translation of

Diophantus, pp. 331-343.
H. D. 8
114 INTRODUCTION

equations containing two unknowns, and of the same sort as
Diophantus I. 1-6; or, of course, they can be reduced to a simple
equation with one unknown by means of an easy elimination.
One other (XIV. 51) gives simultaneous equations in three un-
knowns

x= wi Poh d=104+
HY 3’ 3 2 2

and one (XIV. 49) gives four equations in four unknowns
X+y=40, x+2=48%, x+u=36 x+y+2+u= 060.

With these may be compared Diophantus I. 16-21. (3) Six more
are problems of the usual type about the filling of vessels by pipes:
e.g. (XIV. 130) one pipe fills the vessel in one day, a second in two,
and a third in three; how long will all three running together
take to fill it? Another about brickmakers (XIV. 136) is of the
same sort,

The Anthology contains (4) two indeterminate equations of
the first degree which can be solved in positive integers in an
infinite number of ways (XIV. 48 and 144); the first is a distribution
of apples satisfying the equation x — 37 =y, where » is not less
than 2, and the original number of apples is 3x; the second leads
to the following three equations between four unknown quantities :

Xp =n,
x =27h,
X= 3,

the general solution of which is = 44, y = £, x, = 3%, y, = 2£. These
very equations, made however determinate by assuming that
x+y =x,+3 =100, are solved in Diophantus I. 12.

We mentioned above the problem in the Anthology (XIV. 49)

leading to the following four simultaneous linear equations with
four unknown quantities,

r+y=a,
Z+2m)
r+ u=c

r+ y+e+u=d.

The general solution of any number of simultaneous linear
equations of this type with the same number of unknown quantities
was given by Thymaridas, apparently of Paros, and an early
Pythagorean. He gave a rule, épodos, or method of attack (as
THE PLACE OF DIOPHANTUS 115

Iamblichus?, our informant, calls it) which must have been widely
known, inasmuch as it was called by the name of the éwdvOnpua,
“flower” or “bloom,” of Thymaridas. The rule is stated in general
terms, but, though no symbols are used, the content is pure
algebra. Thymaridas, too, distinguishes between what he calls
aopiorov, the undefined or unknown quantity, and the @piouévo,
the definite or known, therein anticipating the very phrase of
Diophantus, mAffos povddwyv aocpiorov, “an undefined number of
units,” by which he describes his @pifuos or x. Thymaridas’ rule,
though obscurely expressed, states in effect that, if there are #
equations between # unknown quantities x, x, 2,... x,_, of the

following form,
r+r =a,

X+Xp=0an,,
Xt tt... E, =,
then the solution is given by
a (ay +a, +... Fan) —s
n—2
Iamblichus goes on to show that other types of equations can

be reduced to this, so that the rule does not leave us in the lurch
(00 mapérker) in those cases either. Thus we can reduce to
Thymaridas’ form the indeterminate problem represented by the
following three linear equations between four unknown quantities :

r+y=a (z+ un),

x+z2=>0(u+y),

r+u=c(y+2).
From the first equation we obtain

: r+y+eztu=(a+1)(s+ wn),
from which it follows that, if x, », 2, « are all to be integers,
x+y + z+un must contain a+ 1 as a factor. Similarly it must
contain 6+ 1 and ¢+ 1 as factors.
Suppose now that vr + y+e+u=(a+1)(0+1) (c+ 1). There-

fore, by means of the first equation, we get

@+2) (147) =@+ G+ 1) (+1)

1 Tamblichus, Zn WNicomachi arithmeticam introductionem (ed. Pistelli), pp. 62,
18-68, 26.

8—2
116 INTRODUCTION

or x+y=a(b+1)(c+ 1).
Similarly x+2=0(c+1)(a+1),
x+u=c(a+1)(6+1),
and the equations are in the form to which Thymaridas’ rule is
applicable.
Hence, by that rule,

mtr ner ~(at (G+ 1)(c+ I)
2

 

In order to ensure that x may always be integral, it is only
necessary to assume
r4+y+etu=2(a+1)(0+1)(c+1)
The factor 2 is of course determined by the number of un-

knowns. If there are 7 unknowns, the factor to be put in place
ofzisn—2.

Iamblichus has the particular case corresponding to a= 2,
b=3,c=4. He goes on to apply the method to the equations

k
pty = (tu)

m
Bk =u ty),

?

Stitt + 2),

for the case where 2//=%,m/n= 4, pig =1.

Enough has been said to show that Diophantus was not the
inventor of Algebra. Nor was he the first to solve indeterminate
problems of the second degree.

Take, first, the problem of dividing a square number into two
squares (Diophantus II. 8), or of finding a right-angled triangle
with sides in rational numbers. This problem was, as wé learn
from Proclus?, attributed to Pythagoras, who was credited with

the discovery of a general formula for finding such triangles which
may be shown thus:

2 + (2) B= (= i ,
2 2
where 7 is an odd number. Plato again is credited, according
to the same authority, with another formula of the same sort,
(ny + (Fr —1P=(n+ 1)

1 Comment. on Euclid, Book I. pp. 428, 75qq.

 
THE PLACE OF DIOPHANTUS 117

Both these formulae are readily connected with the geometrical
proposition in Eucl. 11. 5, the algebraical equivalent of which may

be stated as
[a
ey = ab.
2 2

The content of Euclid Book 11. is beyond doubt Pythagorean, and
this way of stating the proposition quoted could not have escaped
the Pythagoreans. If we put 1 for & and the square of any odd
number for @, we have the Pythagorean formula; and, if we put
a=2n% 6=2, we obtain Plato's formula. Euclid finds a more
general formula in Book X. (Lemma following X. 28). Starting
with numbers 2 =c¢ + 0 and v =c¢ — 4, so that
Uh = — &,
Euclid points out that, in order that »v may be a square, » and »
must be “similar plane numbers” or numbers of the form mp?
mng®. Substituting we have
mnp® + wy fs -— ry
2 2

 

nePnpg? = (

But the problem of finding right-angled triangles in rational
numbers was not the only indeterminate problem of the second
degree solved by the Pythagoreans. They solved the equation

202 —9p=+ 1
in such a way as to prove that there are an infinite number of
solutions of that equation in integral numbers. The Pythagoreans
used for this purpose the system of “side-” and “diagonal-”
numbers? afterwards fully described by Theon of Smyrna? We
begin with unity as both the first “side” and the first “ diagonal”;
thus
asl, d=1
We then form (a,, 4,), (a;, d;), etc., in accordance with the following
law,
atl + dy, do 20d
a; =ay+d,, dy=2a,+d,;
and so on. Theon states, with reference to these numbers, the
general proposition that
di=2a, 41,

and observes that (1) the signs alternate as successive &’s and a's

1 See Proclus, /n Platonis rempublicam commentarii (Teubner, Leipzig), Vol. 11.
c.. 27 P- 27, 11-18.
2 Theon of Smyrna, ed. Hiller, pp. 43, 44
118 INTRODUCTION

are taken, d?— 2a? being equal to — 1, 4? — 2a) equal to +1,
dy — 2a equal to — 1 and so on, (2) the sum of the squares of all
the @’s will be double of the sum of the squares of a// the a’s. For
the purpose of (2) the number of successive terms in each series,
if finite, must of course be even. The algebraical proof is easy.

dl—2a2 = an + dn) — 2 (Wp + dp)
= 00, =n
=e (dn? — 2a’)
= + (dns — 2any"),
and so on, while &2— 24,2=—1. Proclus tells us that the property

was proved by means of the theorems of Eucl. IL 9, 10, which
are indeed equivalent to
(22+9Y—2(a+F = 22" ~~.

Diophantus does not particularly mention the indeterminate
equation 22°—-1=97 still less does he mention “side-” and
“diagonal-” numbers. But from the Lemma to VI. 15 (quoted
above, p. 69) it is clear that he knew how to find any number of
solutions when one is known. Thus, seeing that z=1, y=1 is
one solution, he would put

2 (1 +x)*— 1 =a square
= (px — 1) say,
whence x= (4+ 2p)/(p* — 2).

Take the value p=2, and we have r=4, or x+1=5; and
2.50 —1=40=7. Putting +35 in place of », we find a still
higher value, and so on.

In a recent paper Heiberg has published and translated, and
Zeuthen has commented on, still further Greek examples of in-
determinate analysis’. They come from the Constantinople MS.
(probably of 12th c.) from which Schone edited the Metrica of
Heron. The first two of the thirteen problems had been published
before (though in a less complete form)?; the others are new.

The first bids us find two rectangles such that the perimeter
of the second is three times that of the first, and the area of the
first is three times that of the second (the first of the two con-
ditions is, by some acgident, omitted in the text). The number 3

L Bibliotheca Mathematica, Villy, 1907-8, pp. 118-134.

2 Hultsch’s Heron, Geeponica, 78, 79. The two problems are discussed by Cantor,
Agrimensoren, p. 62, and Tannery, Mém. de la soc. des sc. de Bordeaux, 1V,, 1882.
THE PLACE OF DIOPHANTUS I19

is of course only an illustration, and the problem is equivalent to
the solution of the equations

u+v=n(x+y)

y=n. uv |

the solution given in the text is equivalent to
X=2f=1, y=¢

u=n(4*—2), v=

Zeuthen suggests that this solution may have been arrived at

thus. As the problem is indeterminate, it would be natural to

make trial of some hypothesis, e.g. to put v=2. It would follow

from the first equation in (1) that # is a multiple of #, say nz. We

have then :
r+y=1+a2
xy =z,
whence xy =n (x+y)— 1,
or (x — n*) (y—n®) =n? (1® — 1).

An obvious solution of this is
x—-1=t—-1, y—-1=n

The second problem is equivalent to the solution of the

equations
xt+y=u+t+v 1);
a, SR ;
and the solution given in the text is
EHP mp Um Lie dit sesninasvere (2),
u=n—1, v=n{—1) 5)
zee b, pm) IF ed 3).

In this case trial may have been made of the assumption

v= nx, y= nu,

when the first equation in (1) would give
(n—1)x=m—1)u,
a solution of which is x=n*—1, u=n— 1.

The fifth problem is of interest in one respect. We are asked
to find a right-angled triangle (in rational numbers) with area
of 5 feet. We are told to multiply 5 by some square containing 6
as a factor, eg. 36. This makes 180, and this is the area of the

( triangle (9, 40,41). Dividing each side by 6, we have the triangle
required. The author, then, is aware of the fact that the area
of a right-angled triangle with sides in whole numbers is divisible
120 INTRODUCTION

by 6. If we take the Euclidean formula for a right-angled triangle,

thus making the sides

772% — 12° mi? + 12>
a. um, a. > a. tind

 

 

where a is any number, and #¢, » are numbers which are both odd
or both even, the area is

2 (172 — 12) (12 + n)

4

and, as a matter of fact, the numerator mn (m — un) (m +n) is
divisible by 24, as was proved later (for another purpose) by
Leonardo of Pisal. There is no sign that Diophantus was aware
of the proposition; this however may be due to the fact that he
does not trouble as to whether his solutions are integral, but is
satisfied with rational results.

The last four problems (numbered 10 to 13) are of great
interest. They are different particular cases of one problem, that
of finding a rational right-angled triangle such that the numerical
sum of its area and all its three sides is a given number. The
author’s solution depends on the following formulae, where a, &
are the perpendiculars, and ¢ the hypotenuse, of a right-angled
triangle, S its area, 7 the radius of its inscribed circle, and
s=%(@+o0+0):

S=rs=tab, rts=a+td, c=s—7,

J

(The proof of these formulae by means of the usual figure, that
used by Heron to prove his formula for the area of a triangle
in terms of its sides, is easy.)

Solving the first two equations, in order to find a and 4, we
have

)

al _7+sF J+ 5F — 8rs)
0) 2

which formula is actually used by the author for finding a and é.
The method employed is to take the sum of the area and the three
sides S + 2s, separated into its two obvious factors s (» + 2), to put
s (r+2)=4 (the given number), and then to separate A into
suitable factors to which s and » + 2 may be equated. They must
obviously be such that s#, the area, is divisible by 6. To take the
first example where A is equal to 280: the possible factors are

1 Seritti, ed. B, Boncompagni, 11. (1862), p. 264. Cf. Cantor, Gesck. d. Math. 11,,
p- 4°.
THE PLACE OF DIOPHANTUS 121

2X T40, 4% 70, 57356, 7 x40, 8x35, 10x 28,14 X20, ‘The
suitable factors in this case are » +2 = §, 5 = 35, because 7 is then
equal to 6, and 7s is a multiple of 6.

The author then says that

Orsi 33Y 3.5.35) 1-1 0
2 2 2

pail

21,
2

and c=35—6=20.

The triangle is therefore (20, 21, 29) in this case. The
triangles found in the other cases, by the same method, are
(9, 40, 41), (8, 15, 17) and (9, 12, 15).

Unfortunately there is no guide to the date of the problems
just given. The form, however, cannot be that in which the
discoverer or discoverers of the methods indicated originally
explained those methods. The probability is that the original
formulation of the most important of the problems belongs to
the period between Euclid and Diophantus. This supposition best
agrees with the fact that the problems include nothing taken from
the great collection in the Arithmetica. On the other hand, it is
strange that none of the seven problems above mentioned is found
in Diophantus. The five of them which relate to rational right-
angled triangles might well have been included by him; thus he
finds rational triangles such that the area plus or minus one of the
perpendiculars is a given number, but not the rational triangle
which has a given area; and he finds rational triangles such that
the area plus or minus the sum of two sides is a given number,
but not the rational triangle such that the sum of the area and
the three sides is a given number. The omitted problems might,
it is true, have come in the lost Books; but, on the other hand,
Book VI. is the place where we should have expected to find
them. Nor do we find in the above problems any trace of
Diophantus’ peculiar methods.

Lastly, the famous Cattle-Problem attributed to Archimedes!’
has to be added to the indeterminate problems propounded before
Diophantus’ time. According to the heading prefixed to the
epigram, it was communicated by Archimedes to the mathe-
maticians at Alexandria in a letter to Eratosthenes. The scholiast

1 Archimedes, ed. Heiberg, Vol. 11. p. 430 sqq.
T22 INTRODUCTION

on Charmides 165E also refers to the problem “called by Archi-
medes the Cattle-Problem.” Krumbiegel, who discussed the
arguments for and against the attribution to Archimedes, con-
cluded apparently that, while the epigram can hardly have been
written by Archimedes in its present form, it is possible, nay
probable, that the problem was in substance originated by
Archimedes’. Hultsch? has a most attractive suggestion as to
the occasion of it. It is known that Apollonius in his exvrokiov
had calculated an approximation to the value of 7 closer than
that of Archimedes, and he must therefore have worked out more
difficult multiplications than those contained in the Measurement
of a circle. Also the other work of Apollonius on the multipli-
cation of large numbers, which is partly preserved in Pappus, was
inspired by the Sand-reckoner of Archimedes; and, though we
need not exactly regard the treatise of Apollonius as polemical,
yet it did in fact constitute a criticism of the earlier book. That
Archimedes should then reply with a problem involving such a
manipulation of immense numbers as would be difficult even for
Apollonius is not altogether outside the bounds of possibility. And
there is an unmistakable vein of satire in the opening words of
the epigram, “ Compute the number of the oxen of the Sun, giving
thy mind thereto, if thou hast a share of wisdom,” in the tran-
sition from the first part to the second, where it is said that
ability to solve the first part would entitle one to be regarded
as “not unknowing nor unskilled in numbers, but still not yet
to be counted among the wise,” and again in the last lines.
Hultsch concludes that in any case the problem is not much
later than the time of Archimedes and dates from the beginning
of the second century B.C. at the latest.

I have reproduced elsewhere? from Amthor, details regarding
the solution of the problem, and I need do little more than state
here its algebraical equivalent. Eight unknown quantities have
to be found, namely, the numbers of bulls and cows respectively
of each of four colours (I use large letters for the bulls and small
letters for the cows). The first part of the problem connects the
eight unknowns by seven simple equations; the second part adds
two more conditions.

L Zeitschrift fiir Math. wu. Physik (Hist. litt. Abtheilung), xxv. (1880), p. I21 sq.
Awmthor added (p. 153 sq.) a discussion of the problem itself.

2 Art. Archimedes in Pauly-Wissowa’s Real- Encyclopidie, 11. 1, pp. 534, 535

3 The Works of Archimedes, pp. 319-326.
THE PLACE OF DIOPHANTUS 123

First part of Problem.
(I) Wat) X+Y, .iieieiererees (1),
Xeld dt DZ + V0 iver {2),
Zl HWY .........c00 3)
(11 wad + D(X +2) iin ld),
wm (Bb DWZ +2) i de 5),
2 = LEVY LL (6),
7 2B ANH) ..cinreerieennn. (7).

Second part.

Wt .X =nsquares. ois. tienes es (8),
Y + Z = a triangular number ............ (9).

The solution of the first part gives
W=1036064327 w= 72063607,
X= 746051470, 2=43032467,
Y= 41493877, y=75439213%,
Z= 7%:8000%, 2735153207,
where 7 is an integer. The solution given by the scholiast! corre-
sponds to 7 = 80.

The complete problem would not be unmanageable but for the
condition (8). If for this were substituted the requirement that
W + X shall be merely a product of two unequal factors (“ Wurm’s
problem ”), the solution in the least possible numbers is

W =1217263415886, w = 846192410280,

X = 876035035422, x= 574579625058,
Y= 487233469701, y= 6386887080909,
Z = 8640054790380, =z=412838131360.

But, if we include condition (8) and first of all find a solution
satisfying the conditions (1) to (8), we have then, in order to
satisfy condition (9), to solve the equation

g(g+1)/2 =51285802909803 . £*
=3.7.11.20.353.4657°. £°

If we multiply by 8, and put

20+ 1=1¢ 2.0657 Emy,
we have the equation
lm I=. 3.7. 11,200.38 1%
or r= 4729494 1° = I.
! Archimedes, ed. Heiberg, Vol. II. pp. 454, 455.
124 INTRODUCTION

The value of W would be a number containing 206545
digits.

Such are the very few and scattered particulars which we
possess of problems similar to those of Diophantus solved or
propounded before his time. They show indeed that the kind of
problem was not invented by him, but on the other hand they
show little or no trace of anything like his characteristic alge-
braical methods. In the circumstances, and in default of discovery
of fresh documents, the question how much of his work represents
original contributions of his own to the subject must remain a
matter of pure speculation. It is pretty obvious that one man
could not have been the author of all the problems contained in
the six Books. There are also inequalities in the work; some
problems are very inferior in interest to the remainder, and some
solutions may be assumed to be reproduced from other writers
of less calibre, since they reveal none of the mastery of the subject
which Diophantus possessed. Again, it seems probable that the
problem V. 30, which is exceptionally in epigrammatic form, was
taken from someone else. The Arithmetica was no doubt a
collection, much in the same sense as Euclid’s Elements were. And
this may be one reason why so little trace remains of earlier
labours in the same field. It is well known that Euclid’s Elements
so entirely superseded the works of the earlier writers of Elements
(Hippocrates of Chios, Leon and Theudius) and of the great
contributors to the body of the Elements, Theaetetus and Eudoxus,
that those works have disappeared almost entirely. So no doubt
would Diophantus’ work supersede, and have the effect of con-
signing to oblivion, any earlier collections of problems of the
same kind. But, if it was a compilation, we cannot doubt that
it was a compilation in the best sense, therein resembling Euclid’s
Elements; it was a compilation by one who was a master of the
subject, who took account of and assimilated all the best that had
been written upon it, arranged the whole of the available material
in due and progressive order, but also added much of his own, not
only in the form of new problems but also (and even more) in the
mode of treatment, the development of more general methods, and
SO on. :

It is perhaps desirable to add a few words on the previous
history of the theory of polygonal numbers. The theory certainly
goes back to Pythagoras and the earliest Pythagoreans. The
triangle came first, being obtained by first taking 1, then adding
THE PLACE OF DIOPHANTUS 125

2 to it, then 3 to the sum; each successive number would be
represented by the proper number of dots, and, when each number
was represented by that number of dots arranged symmetrically
under the row representing the preceding number, the triangular
form would be apparent to the eye, thus:

Next came the Pythagorean discovery of the fact that a similar
successive addition of odd numbers produced
successive square numbers, the odd numbers
being on that account called gnomons, and again
the process was shown by dots arranged to re-
present squares. The accompanying figure shows
the successive squares and gnomons.

Following triangles and squares came the figured numbers
in which the “gnomons,” or the numbers added to make one
number of a given form into the next larger of the same form,
were numbers in arithmetical progression starting from 1, but with
common difference 3, 4, §, etc., instead of 1, 2. Thus, if the
common difference is 3, so that the successive numbers added to
I are 4, 7, 10, etc, the number is a pentagonal number, if the
common difference is 4 and the gnomons 5, 9, 13, etc., the number
is a hexagonal number, and so on. Hence the law that the
common difference of the gnomons in the case of a z-gon is
7n— 2.

Perhaps these facts had already been arrived at by Philippus
of Opus (4th c. B.C), who is said to have written a work on
polygonal numbers’. Next Speusippus, nephew and successor of
Plato, wrote on Pythagorean Numbers, and a fragment of his
book survives’, in which linear numbers, polygonal numbers,
triangles and pyramids are spoken of: a fact which leaves no
room for doubt as to the Pythagorean origin of all these con-
ceptions? :

Hypsicles, who wrote about 170 B.C, is twice mentioned by
Diophantus as the author of a “definition” of a polygonal number,

 

1 Bioypdeoi, Vitarum scriptores Graeci minores, ed. Westermann, 1845, p. 448.

* Theologumena arithmeticae (ed. Ast), 1817, pp. 61, 62; the passage is translated with
notes by Tannery, Pour histoire de la science hellene, pp. 386-390.

% Cantor, Geschichte der Mathematik, 14, p. 249.
126 INTRODUCTION

which is even quoted verbatim. The definition does not mention
any polygonal number beyond the pentagonal; but indeed this
was unnecessary : the facts about the triangle, the square and the
pentagon were sufficient to enable Hypsicles to pass to a general
conclusion. The definition amounts to saying that the nth a-gon
(1 counting as the first) is

In{2+@—1)(a-2))

Theon of Smyrna? Nicomachus? and Iamblichus* all devote
some space to polygonal numbers. The first two, who flourished
about 100 A.D., were earlier than Diophantus, and are accordingly
of interest here. Besides a description of the successive polygonal
numbers, Theon gives the theorem that two successive triangular
numbers added together give a square. That is,

p=n0yn wnt) ,
2 2

The fact is of course clear if we divide a square
into two triangles as in the figure.

Nicomachus gave various rules for transforming triangles into
squares, squares into pentagons, etc.

 

1. If we put two consecutive triangles together we get a square
(as in Theon’s theorem).

2. A pentagon is obtained from a square by adding to it a
triangle - the side of which is 1 less than that of the square;
similarly a hexagon from a pentagon by adding a triangle the side
of which is 1 less than that of the pentagon; and so on.

In fact,

Ini2+(n—1)(@a—2)+3(n—1)n=%nl2+n—-1){(a+1)—2}]
Next Nicomachus sets out the first triangles, squares, pentagons,
hexagons and heptagons in a diagram thus :

Triangles IL. 3 6 10 15 21 “28. 36: 4% 5%
Squares E 4 0916.28 36 4p G4 S81 100
Pentagcons ... Xt: 5: 12: 22 35 51 70 02 M17 i145
Hexacons ... Ui 6 15 28 45 66. 01 :120 133 190
Heptacons... 1 7 18 34 %5 31 112. 148 130: 235

and observes that

1 Dioph. I. pp. 470-472.

2 Expositio rerum mathematicarum ad legendum Platonem utilium, ed. Hiller,
PP. 31-40.

8 Introductio arithmetica, ed. Hoche, 11. 8-12, pp. 87-99.

tn Nicopachi arithmeticam introd., ed. Pistelli, pp. 58-61, 68-72.
THE PLACE OF DIOPHANTUS 127

3. Each polygon is equal to the polygon immediately above
it in the diagram plus the triangle with 1 less in its side, ze. the
triangle in the preceding column.

4. The vertical columns are arithmetical progressions, the
common difference of which is the triangle in the preceding
column.

But Plutarch, a contemporary of Nicomachus, mentioned
another method of transforming triangles into squares: Every
triangular number taken eight times and then increased by 1 gives
a sguare’, That is,

7 Fiza 1=(2n+ 1)

Diophantus generalised this proposition into his theorem for
transforming any polygonal number into a square.

If P be a polygonal number, a the number of angles,

8P (a — 2) + (a — 4)* = a square,

He deduces rules for finding a polygonal number when the
side and the number of angles are given, and for finding the side
when the number and the number of its angles are given. These
fine results and the fragment of the difficult problem of finding
the number of ways in which any given number can be a polygonal
number no doubt represent part of the original contributions by’
Diophantus to the theory of that class of numbers.

1 Plat. guaest. V. 2, 4, 1003 F.
 
THE ARITHMETICA

BOOK 1

PRELIMINARY

Dedication.

“ Knowing, my most esteemed friend Dionysius, that you are
anxious to learn how to investigate problems in numbers, I have
tried, beginning from the foundations on which the science is
built up, to set forth to you the nature and power subsisting in
numbers.

“Perhaps the subject will appear rather difficult, inasmuch as
it is not yet familiar (beginners are, as a rule, too ready to despair
of success); but you, with the impulse of your enthusiasm and
the benefit of my teaching, will find it easy to master; for
eagerness to learn, when seconded by instruction, ensures rapid
progress.”

After the remark that “all numbers are made up of some
multitude of units, so that it is manifest that their formation is
‘subject to no limit,” Diophantus proceeds to define what he calls
the different “species” of numbers, and to describe the abbreviative
signs used to denote them. These “species” are, in the first
place, the various powers of the unknown quantity from the second
to the sixth inclusive, the unknown quantity itself, and units.

Definitions.

A square (=x?) is Swaps (“power”), and its sign is a 4 with V’
superposed, thus 4”.

A cube (=x?) is k¥Bos, and its sign K.

A square-square (= x*) is SuvapoSivauss’, and its sign is 474.

A square-cube (= x°) is SuvapdruBos, and its sign AK".

A cube-cube (= z°) is kvBoruBos, and its sign K'K.

1 The term duramodivaues was already used by Heron (Aetrica, ed. Schone, p. 48,
11, 19) for the fourth power of a side of a triangle.

H. DD. 9
130 THE ARITHMETICA

«It is,” Diophantus observes, “from the addition, subtraction
or multiplication of these numbers or from the ratios which they
bear to one another or to their own sides respectively that most
arithmetical problems are formed” ; and “each of these numbers...
is recognised as an element in arithmetical inquiry.”

“ But the number which has none of these characteristics, but
merely has in it an indeterminate multitude of units (whiijfos
povdadwy dépiaTov) is called aplbucs, ‘number, and its sign is
¢ [=2)

“ And there is also another sign denoting that which is in-
variable in determinate numbers, namely the unit, the sign being

M with o superposed, thus MY

Next follow the definitions of the reciprocals, the names of
which are derived from the names of the corresponding species
themselves.

Thus

from apifucs [x] we derive the term dpibuoativ [= 1/2]

.  Svvaus [x7] a Svvapootov [= 1/2?)
«080s [27] > 2 kvBoaTov [= 1/x%]

,» Ouvapodivaus [x] 5 dvvapobvvapoatov [= I1/x4]
ww  SupapbevBos [2°] % SvvaporvBoaTov [= 1/x°%]

»  kvuBokuBos [1°] ’ kvBokvBooTov [= 1/xF],

and each of these has the same sign as the corresponding original
species, but with a distinguishing mark which Tannery writes in
the form X above the line to the right.

Thus d¥%= 1/2" just as y¥==1.

Sign of Subtraction (minus).

“A minus multiplied by a minus makes a plus’; a minus
multiplied by a plus makes a minus; and the sign of a minus is a
truncated V turned upside down, thus A.”

Diophantus proceeds: “It is well that one who is beginning
this study should have acquired practice in the addition, subtraction
and multiplication of the various species. He should know how
to add positive and negative terms with different coefficients to

! The literal rendering would be ‘“ A wanting multiplied by a wanting makes a
forthcoming.” The word corresponding to minus is Nets (*“ wanting ”’): when it is
used exactly as our mzznus is, it is in the dative Nelyet, but there is some doubt whether
Diophantus himself used this form (cf. p. 44 above). For the probable explanation of
the sign, see pp. 42-44. The word for “forthcoming” is Umapés, from dmwdpyw, to exist.
Negative terms are Neiwovra edn, and positive drdpyxovra.
BOOK 1 131

other terms? themselves either positive or likewise partly positive
and partly negative, and how to subtract from a combination of
positive and negative terms other terms either positive or likewise
partly positive and partly negative.

“Next, if a problem leads to an equation in which certain
terms are equal to terms of the same species but with different
coefficients, it will be necessary to subtract like from like on both
sides, until one term is found equal to one term. If by chance
there are on either side or on both sides any negative terms, it will
be necessary to add the negative terms on both sides, until the
terms on both sides are positive, and then again to subtract like
from like until one term only is left on each side.

“This should be the object aimed at in framing the hypotheses
of propositions, that is to say, to reduce the equations, if possible,
until one term is left equal to one term; but [ will show you later
how, in the case also where two terms are left equal to one term, such
a problem is solved.”

Diophantus concludes by explaining that, in arranging the
mass of material at his disposal, he tried to distinguish, so far as
possible, the different types of problems, and, especially in the
elementary portion at the beginning, to make the more simple lead
up to the more complex, in due order, such an arrangement being
calculated to make the beginner's course easier and to fix what
he learns in his memory. The treatise, he adds, has been divided
into thirteen Books.

PROBLEMS

1. To divide a given number into two having a given
difference. :
Given number 100, given difference 40.
Lesser number required x. Therefore
2x + 40 = 100,
x = 30.
The required numbers are 70, 30.
2. To divide a given number into two having a given ratio.
Given number 60, given ratio 3: 1.
Two numbers x, 3x. Therefore r= 135. .
The numbers are 45, 15.

1 los, “species,” is the word used by Diophantus throughout.
132 THE ARITHMETICA

3. To divide a given number into two numbers such that one
is a given ratio of the other plus a given difference’.
Given number 80, ratio 3: I, difference 4.
Lesser number x. Therefore the larger is 3x + 4, and
4r+ 4=80, so that z=190.
The numbers are 61, 19.

4. To find two numbers in a given ratio and such that their
difference is also given.
Given ratio 5 : 1, given difference 20.
Numbers 5+, ». - Therefore 4x = 20, x= 35, and
the numbers are 25, 5.

5. To divide a given number into two numbers such that given
fractions (not the same) of each number when added together
produce a given number.

Necessary condition. The latter given number must be such
that it lies between the numbers arising when the given fractions
respectively are taken of the first given number.

First given number 100, given fractions 4 and 1, given
sum of fractions 30.
Second part 52. Therefore first part = 3 (30 — 7).
Hence 90+ 2v=100, and z=13,
The required parts are 75, 25.

6. To divide a given number into two numbers such that a
given fraction of the first exceeds a given fraction of the other
by a given number.

Necessary condition. The latter number must be less than that
which arises when that fraction of the first number is taken which
exceeds the other fraction.

Given number 100, given fractions 1 and % respectively,
given excess 20.
Second part 6x. Therefore the first part is 4 (x + 20).
Hence 107 + 30= 100, = 2, and ee
the parts are 88, 12. oy 2 je

! Literally “to divide an assigned number into two in a given ratio and difference (év
Noyw kal bmepox7 77 6ofeloy).” The phrase means the same, though it is not so clear, as
Euclid’s expression (Data, Def. 11 and passim) do0évre pelfwy 7) év Noyw. According to
Euclid’s definition a magnitude is greater than a magnitude ‘by a given amount (more)
than in a (certain) ratio” when the remainder of the first magnitude, after subtracting
the given amount, has the said ratio to the second magnitude. This means that, if x, y
are the magnitudes, & the given amount, and #4 the ratio, x — d=4y or x=4y +d.
BOOK 1 133

7. From the same (required) number to subtract two given
numbers so as to make the remainders have to one another a
given ratio. :

Given numbers 100, 20, given ratio 3: I.
Required number x. Therefore x — 20'= 3 (x — 100), and
% = 140.

8. To two given numbers to add the same (required) number so
as to make the resulting numbers have to one another a given ratio.

Necessary condition. The given ratio must be less than the
ratio which the greater of the given numbers has to the lesser.
Given numbers 100, 20, given ratio 3:1.
Required number x. Therefore 3x + 60 =x + 100, and
x = 20.

9. From two given numbers to subtract the same (required)
number so as to make the remainders have to one another a given
ratio.

Necessary condition. The given ratio must be greater than the
ratio which the greater of the given numbers has to the lesser.

Given numbers 20, 100, given ratio 6: I.
Required number x. Therefore 120 — 62 = 100 — x, and
y=g.

10. Given two numbers, to add to the lesser and to subtract
from the greater the same (required) number so as to make the
sum in the first case have to the difference in the second case
a given ratio.

Given numbers 20, 100, given ratio 4 : I.
Required number x. Therefore (20 +x) =4 (100 — x), and
x =706,

11. Given two numbers, to add the first to, and subtract the
second from, the same (required) number, so as to make the
resulting numbers have to one another a given ratio.

Given numbers 20, 100, given ratio 3 : I.
Required number x. Therefore 32 — 300 = x + 20, and
x = 160.

12. To divide a given number twice into two numbers such
that the first of the first pair may have to the first of the second
pair a given ratio, and also the second of the second pair to the
second of the first pair another given ratio.
134 THE ARITHMETICA

Given number 100, ratio of greater of first parts to lesser
of second 2:1, and ratio of greater of second parts
to lesser of first parts 3: I.
x lesser of second parts.
The parts then are
2c 00 — bx
Sa and J 2
Therefore 300 — 5x = 100, x = 40, and
the parts are (80, 20), (60, 40).

13. To divide a given number thrice into two numbers such that
one of the first pair has to one of the second pair a given ratio,
the second of the second pair to one of the third pair another
given ratio, and the second of the third pair to the second of the
first pair another given ratio.

Given number 100, ratio of greater of first parts to lesser
of second 3:1, of greater of second to lesser of
third 2:1, and of greater of third to lesser of
first 4:1

x lesser of third parts.

Therefore greater of second parts = 2x, lesser of second
= 100 — 2x, greater of first = 300 — 6.

Hence lesser of first = 6x —200, so that greater of third
= 24x — 800.

Therefore 252 — S00 = 100, x = 36, and
the respective divisions are (84, 16), (72, 28), (64, 36).

14. To find two numbers such that their product has to their
sum a given ratio. [One is arbitrarily assumed.]

Necessary condition. The assumed value of one of the two
must be greater than the number representing the ratiol.
Ratio 3: 1, » one of the numbers, 12 the other (5 3).
Therefore 12v= 32+ 36, +=4, and
the numbers are 4, 12.

15. To find two numbers such that each after receiving from
the other a given number may bear to the remainder a given

ratio.
Let the first receive 30 from the second, the ratio being

then 2: 1, and the second 50 from the first, the ratio
being then 3:1; take x + 30 for the second.

1 Literally ‘“‘the number homonymous with the given ratio.”
Zibe Sper
BOOK "1 135

Therefore the first = 2x — 30, and
(x + 80) = 3 (2x — 80).
Thus »=64, and
the numbers are 98, 94.

16. To find three numbers such that the sums of pairs are
given numbers.
Necessary condition. Half the sum of the three given numbers
must be greater than any one of them singly.
Let (1) + (2) = 20, (2) + (3) = 30, (3) + (1) = 4.
x the sum of the three. Therefore the numbers are
x —30, X—40, x-— 20.
The sum x = 3x — 90, and » = 45.
The numbers are 15,5, 25.

17. To find four numbers such that the sums of all sets of three
are given numbers.
Necessary condition. One-third of the sum of the four must be
greater than any one singly.
Sums of threes 22, 24, 27,20 respectively.
x the sum of all four. Therefore the numbers are
222, x—24, X27, Z=30
Therefore 42-93 =2, 2 = 31, and
the numbers are 9, 7, 4, 11.

13. To And three numbers such that the sum of any pair
exceeds the third by a given number.
Given excesses 20, 30, 40.
2x the sum of all three.
We have (1) + (2) = (3) + 20.
Adding (3) to each side, we have: twice (3) + 20= 2x, and
(3)==zx-10.
Similarly the numbers (1) and (2) are x— 15, x— 20
respectively.
Therefore 3x — 45 = 2x, x=45, and
the numbers are 30, 25, 35.
[Otherwise thus’. As before, if the third number (3) is z,
(1) + (2)=x + 20,
Next, if we add the equations
(1) +{2)—- (3) = ol
(2) +3) ~{1)= 30}

I Tannery attributes the alternative solution of I. 18 (as of I. 19) to an old scholiast.
136 THE ARITHMETICA

we have (2)=1 (20 + 30) = 25.

Hence (1)=x—35.
Lastly GY +(1)-(2)=40,
or 2x —§ — 25 = 40.
Therefore X= 75,

The numbers are 30, 25, 35.)

19. To find four numbers such that the sum of any three
exceeds the fourth by a given number.
Necessary condition. Half the sum of the four given differences
must be greater than any one of them.
Given differences 20, 30, 40, 50.
2x the sum of the required numbers. Therefore the
numbers are
x—15, x—20, ¥— 25, x — IO.
Therefore 4x — 70= 2x, and x = 35.
‘The numbers are 20, 15, 10, 25.

[Otherwise thus’. If the fourth number (4) is x,
(1) + (2) + (3) = x + 20.

Put (2) + (3) equal to half the sum of the two excesses 20
and 30, z.e. 25 [this is equivalent to adding the two
equations

(1)+ (2) + (3)= (4) = 20,
(2) (3): (4) — {1} = 30}.

It follows by subtraction that (1) =x — 3.

Next we add the equations beginning with (2) and (3)
respectively, and we obtain

(3) + (4) = 1 (30 + 40) = 35,

so that {D=135 ~~

It follows that (2)=x- 10.

Lastly, since (4) + (1) + (2) = (3) = 50,

3¥—15—(35 —2)=350, and x= 25.
The numbers are accordingly 20, 15, 10, 25.]

20. To divide a given number into three numbers such that the
sum of each extreme and the mean has to the other extreme a
given ratio.

Given number 100; and let (1) +(2)=3.(3) and (2) + (3)
=4.(1)

1 Tannery attributes the alternative solution of 1. 1g (as of I. 18) to an old scholiast.
BOOK 1 137

x the third number. Thus the sum of the first and second
= 3%, and the sum of the three = 4x = 100.
Hence x = 25, and the sum of the first two = 75.
Let » be the first’. Therefore sum of second and third
= 4p, 57 =100 and y= 20.
The required parts are 20, 55, 25.

21. To find three numbers such that the greatest exceeds the
middle number by a given fraction of the least, the middle exceeds
the least by a given fraction of the greatest, but the least exceeds
a given fraction of the middle number by a given number.

Necessary condition. The middle number must exceed the
least by such a fraction of the greatest that, if its denominator? be
multiplied into the excess of the middle number over the least, the
coefficient of x in the product is greater than the coefficient of
x in the expression for the middle number resulting from the
assumptions made?

Suppose greatest exceeds middle by i of least, middle
exceeds least by 1 of greatest, and least exceeds
1 of middle by 10. [Diophantus assumes the three
given fractions or submultiples to be one and the

same. |

x + 10 the least. Therefore middle = 3x, and greatest
= 6x — 30.

Hence, lastly, 6x— 30 — 3x =1 (x + 10),

or x4 10=9xr—90, and x= 124.

The numbers are 45, 373, 223.

1 As already remarked (p. 52), Diophantus does not use a second symbol for the
second unknown, but makes dptfués do duty for the second as well as for the first.

2 ¢ Denominator,” literally the “number homonymous with the fraction,” Z.e. the
denominator on the assumption that the fraction is, or is expressed as, a submultiple.

3 Wertheim points out that this condition has reference, not to the general solution of
the problem, but to the general applicability of the particular procedure which Diophantus
adopts in his solution. Suppose X, V, Z required such that X — ¥Y=2Z[m, Y—- Z=X|n,
Z-a=Y[p. Diophantus assumes Z=x+a, whence V=px,  X=n (px—-x-a). The
condition states that 2p —7>p. If we solve for x by substituting the values of X, ¥, Z
in the first equation, we in fact obtain

mi{(np-n-p)x—na}=x+a,
or x (mnp—mn—mp—1)=a (mn+1).

In order that the value of x may be positive, we must have mnp>mn+mp+1,
that is,

I
7, p+ —
BRR
or (if m, n, p are positive integers) np>n+p.
138 THE ARITHMETICA

[Another solution’.

Necessary condition. The given fraction of the greatest must
be such that, when it is added to the least, the coefficient of x in
the sum is less than the coefficient of x in the expression for the
middle number resulting from the assumptions made?

Let the least number be x + 10, as before, and the given
fraction !; the middle number is therefore 3x.
Next, greatest = middle + } (least) = 34x + 33.
Lastly, 3x=x+ 10+} (34r + 3%)
=24r+ 114.
Therefore x= 12}, and
the numbers are, as before, 45, 374, 224.]

22. To find three numbers such that, if each give to the next
following a given fraction of itself, in order, the results after each
has given and taken may be equal.

Let first give 1 of itself to second, second } of itself to
third, third 1 of itself to first.

Assume first to be a number of x's divisible by 3, say
3x, and second to be a number of units divisible by
4, say 4.

Therefore second after giving and taking becomes x + 3.

Hence the first also after giving and taking must become
x43; it must therefore have taken x +3 —2zx, or
3—x; 3-2 must therefore be 1 of third, or this

= 15 = 57.
Lastly, 15-5r—(3—2)+1=x+3,
or 13-4r=2-+3 and a=2,

The numbers are6, 4, 5.

23. To find four numbers such that, if each give to the next
following a given fraction of itself, the results may all be equal.
Let first give § of itself to second, second I of itself
to third, third } of itself to fourth, and fourth } of
itself to first.
Assume first to be a number of #’s divisible by 3, say 3,
and second to be a number of units divisible by 4,

say 4.

! Tannery attributes this alternative solution, like the others of the same kind, to an
ancient scholiast.

2 Wertheim observes that the scholiast’s necessary condition comes to the same thing
as Diophantus’ own.
BOOK I 139

‘The second after giving and taking becomes x + 3.
Therefore first after giving x to second and receiving
4 of fourth =x + 3; therefore fourth
=06(r+3—22)=18 67
But fourth after giving 3 —x to first and receiving 1 of
third =x + 3 ; therefore third = 30r — 60.
Lastly, third after giving 6x— 12 to fourth and receiving
1 from second =2 + 3.
That is, 247-47 =x +3, and #=§2.
‘The numbers are therefore A880, 4, 330, 114,
or, after multiplying by the common de-
nominator, 150, 92, 120, 114,

24. To find three numbers such that, if each receives a given
fraction of the sum of the other two, the results are all equal.

Let first receive § of (second + third), second 1 of
(third + first), and third 1 of (first + second).

Assume first =x, and for convenience’ sake (Tod mwpoyeipov
évexev) take for sum of second and third a number of
units divisible by 3, say 3.

Then sum of the three=x+ 3,

and first + 1 (second + third) = x + 1.

Therefore second + 1 (third + first) =x +1;

hence 3 times second + sum of all = 4x + 4,

and therefore second = x + 1.
Lastly, third + { (first + second) =x + 1,
or 4 times third + sum of all = 5x + 3,
and third =x 4 §.
Therefore r+E+H+ E+) =2+3,
and r=13,

The numbers, after multiplying by the common
denominator, are 13, 17, 19.

25. To find four numbers such that, if each receives a given
fraction of the sum of the remaining three, the four results are

equal.

Let first receive } of the rest, second } of the rest,
third 1 of rest, and fourth } of rest.

Assume first to be x and sum of rest a number of units
divisible by 3, say 3.

Then sum of all=2x +3.

‘Now first +1 (second + third + fourth) =x + 1.
140 THE ARITHMETICA

Therefore second + # (third + fourth + first) =x +1,

whence 3 times second + sum of all = 4x + 4,
and therefore second =x + §.

Similarly third =x + §,

and fourth =x + §.

Adding, we have 4r+43i=x+3,

and : x =4L

The numbers, after multiplying. by a common
denominator, are 47, 77, 92, 101,

26. Given two numbers, to find a third number which, when
multiplied into the given numbers respectively, makes one product
a square and the other the side of that square.

Given numbers 200, 5; required number x.
Therefore 200xr = (5x)? and
x =8.

27. To find two numbers such that their sum and product are
given numbers.

Necessary condition. The square of half the sum must exceed
the front by a square number. éoTe 8¢ TodTO TAaouaTikoVL.
Given sum 20, given product g6.
2x the difference of the required numbers.
Therefore the numbers are 10+, 10—ux.
Hence 100 — 22 = g6.
Therefore £=2, and
the required numbers are 12, 8.

28. To find two numbers such that their sum and the sum of
their squares are given numbers.

Necessary condition. Double the sum of their squares must
exceed the square of their sum by a square. éori 8¢ kat ToiTo
TAATUATIKOV

! There has been controversy as to the meaning of this difficult phrase. Xylander,
Bachet, Cossali, Schulz, Nesselmann, all discuss it. Xylander translated it by ‘“effictum
aliunde.” Bachet of course rejects this, and, while leaving the word untranslated,
maintains that it has an active rather than a passive signification ; it is, he says, not
something ‘made up” (effictum) but something ‘‘a quo aliud quippiam effingi et
plasmari potest,” ‘‘ from which something else can be made up,” and this he interprets as
meaning that from the conditions to which the term is applied, combined with the
solutions of the. respective problems in which it occurs, the rules for solving mixed
quadratics can be evolved. Of the two views I think Xylander’s is nearer the mark.
mhacuatikdy should apparently mean ‘of the nature of a w\doua,” just as dpauarikéy
means something connected with or suitable for a drama; and mAdoua means something
BOOK 1 141

Given sum 20, given sum of squares 208.
Difference 2x.
Therefore the numbers are 10+, 10 — 1.
Thus 200 +222=208, and r=2.

The required numbers are 12, 8.

29. To find two numbers such that their sum and the difference
of their squares are given numbers.
Given sum 20, given difference of squares 8o.
Difference 2».
The numbers are therefore 10 +x, 10 — x.
Hence (10+ 2x) — (10—x)*= 80,
or 40r=80, and x=2,
The required numbers are 12, 8.

30. To find two numbers such that their difference and product
are given numbers. :

Necessary condition. Four times the product together with
the square of the difference must give a square. €éore 8¢ kal TodTO
TAAT UATLKOV.

Given difference 4, given product gb.
2x the sum of the required numbers.
Therefore the numbers are x+2, r—2; accordingly
Zi—4:=00, and r= 10.
The required numbers are x2, 8.

31. To find two numbers in a given ratio and such that the
sum of their squares also has to their sum a given ratio.
Given ratios 3:1 and 5:1 respectively.
Lesser number ».
Therefore 102° = 5.4%, whence z= 2, and
the numbers are 2, 6.

32. To find two numbers in a given ratio and such that the
sum of their squares also has to their difference a given ratio.
Given ratios 3:1 and 10:1.
Lesser number x, which is then found from the equation
1022 = 10. 2x. :
Hence v= 2, and
the numbers are 2, 6.

 

“formed” or “moulded.” Hence the expression would seem to mean ‘“this is of the
nature of a formula,” with the implication that the formula is not difficult to make up
or discover. Nesselmann, like Xylander, gives it much this meaning, translating it ‘das
ldsst sich aber bewerkstelligen.” Tannery translates mhaopuarikér by ‘‘formativum.”
142 THE ARITHMETICA

33. To find two numbers in a given ratio and such that the
difference of their squares also has to their sum a given ratio.
: Given ratios 3:1 and 6: I.
Lesser number zx, which is found to be 3.
The numbers are 3, g.

34. To find two numbers in a given ratio and such that the
difference of their squares also has to their difference a given
ratio.

Given ratios 3:1 and 12: 1.
Lesser number x, which is found to be 3.
The numbers are 3, g.

Similarly by the same method can be found two numbers in
a given ratio and (1) such that their product is to their sum in a
given ratio, or (2) such that their product is to their difference in a
given ratio.

35. To find two numbers in a given ratio and such that the
square of the lesser also has to the greater a given ratio.
Given ratios 3:1 and 6: I respectively.
Lesser number x, which is found to be 18.
The numbers are 18, 54.

36. To find two numbers in a given ratio and such that the
square of the lesser also has to the lesser itself a given ratio.
Given ratios 3:1 and 6: I.
Lesser number x, which is found to be 6.
The numbers are 6, 18.

37. To find two numbers in a given ratio and such that the
square of the lesser also has to the sum of both a given ratio.
Givenratios 3:1 and 2:1.
Lesser number x, which is found to be 8.
The numbers are 8, 24.

38. To find two numbers in a given ratio and such that the
square of the lesser also has to the difference between them a
given ratio.

Given ratios 3:1 and 6:1.
Lesser number x, which is found to be 12.
The numbers are 12, 36.
BOOK 11 143

Similarly can be found two numbers in a given ratio and

(1) such that the square of the greater also has to the
lesser a given ratio, or

(2) such that the square of the greater also has to the
greater itself a given ratio, or

(3) such that the square of the greater also has to the sum
or difference of the two a given ratio.

39. Given two numbers, to find a third such that the sums of
the several pairs multiplied by the corresponding third number
give three numbers in arithmetical progression.

Given numbers 3, 5.
Required number x.
The three products are therefore 3x + 15, 5x +15, 82
Now 3x +15 must be either the middle or the least of
the three, and 35x15 either the greatest or the
middle.
(1) 5x + 15 greatest, 3x + 15 least.
Therefore 52+ 13+ 32x +15=2.32, and
cs
serie
(2) s5x+ 15 greatest, 3x + 15 middle.
Therefore (54+ 15)— (3x + 15)=3x+ 15 — 8x, and
=
=
(3) 8x greatest, 3x + 15 least,
Therefore 8x + 3x + 15 =2 (54 + 15), and
y==1g,

x

BOOK 11

[The first five problems of this Book are mere repetitions of problems in
Book I. They probably found their way into the text from some ancient
commentary. In each case the ratio of one required number to the other
is assumed to be 2 : 1. The enunciations only are here given.]

I. To find two numbers such that their sum is to the sum of
their squares in a given ratio [cf. I. 31].

2. To find two numbers such that their difference is to the
difference of their squares in a given ratio [cf. I. 34].
144 THE ARITHMETICA

3. To find two numbers such that their product is to their sum
or their difference in a given ratio [cf. I. 34].

4. To find two numbers such that the sum of their squares is to
their difference in a given ratio [cf. I. 32].

5. To find two numbers such that the difference of their squares
is to their sum in a given ratio [cf. I. 33].

61. To find two numbers having a given difference and such
that the difference of their squares exceeds their difference by a
given number.

Necessary condition. The square of their difference must be
less than the sum of the said difference and the given excess
of the difference of the squares over the difference of the
numbers.

Difference of numbers 2, the other given number 20.
Lesser number 2. Therefore x + 2 is the greater, and
4x + 4 = 22.
Therefore x = 4%, and
the numbers are 44, 63.

7, To find two numbers such that the difference of their
squares is greater by a given number than a given ratio of
their difference® [Difference assumed.)

Necessary condition. The given ratio being 3:1, the square of
the difference of the numbers must be less than the sum of three
times that difference and the given number.

Given number 10, difference of required numbers 2.
Lesser number ». Therefore the greater is x + 2, and
4r+4=3.2+ 10
Therefore x= 3, and
the numbers are 3, 5.

8. To divide a given square number into two squares”.

1 The problems 11. 6, 7 also are considered by Tannery to be interpolated from some

ancient commentary.

2 Here we have the identical phrase used in Euclid’s Data (cf. note on p. 132 above) :
the difference of the squares is 77s Umepoxis alr@v 800évTi dplbug pelfwr 7 év Noy,
literally *‘ greater than their difference by a given number (more) than in a (given) ratio,”
by which is meant ‘‘greater by a given number than a given proportion or fraction
of their difference.”

3 It is to this proposition that Fermat appended his famous note in which he
enunciates what is known as the ‘great theorem” of Fermat. The text of the note is
as follows:

““On the other hand it is impossible to separate a cube into two cubes, or a
BOOK 11 145

Given square number 16.

2? one of the required squares. Therefore 16 — #* must
be equal to a square.

Take a square of the form! (mx —4), m being any
integer and 4 the number which is the square root
of 16, og. take (ax—4Y, and equate it to 16-27%

Therefore 44°— 16x + 16=16—x2,

or 522=16%, and r=214, -

256

The required squares are therefore ry a

9. To divide a given number which is the sum of two squares
into two other squares?

 

biquadrate into two biquadrates, or generally any power except a square into two powers
with the same exponent. 1 have discovered a truly marvellous proof of this, which
however the margin is not large enough to contain.”

Did Fermat really possess a proof of the general proposition that x™ 4 y™=2™ cannot
be solved in rational numbers where 7 is any number >2? As Wertheim says, one
is tempted to doubt this, seeing that, in spite of the labours of Euler, Lejeune-Dirichlet,
Kummer and others, a general proof has not even yet been discovered. Euler proved
the theorem for m=3 and m=4, Dirichlet for 7=35, and Kummer, by means of the
higher theory of numbers, produced a proof which only excludes certain particular
values of 72, which values are rare, at all events among the smaller values of 72; thus
there is no value of 7 below 100 for which Kummer's proof does not serve. (I take
these facts from Weber and Wellstein’s Zncyclopidie der Elementar-Mathematik, 1,
p. 284, where a proof of the formula for 7=4 is given.)

It appears that the Gottingen Academy of Sciences has recently awarded a prize
to Dr A. Wieferich, of Miinster, for a proof that the equation xP+yP=2P cannot be
solved in terms of positive integers not multiples of p, if 2” —2 is not divisible by p%
“This surprisingly simple result represents the first advance, since the time of Kummer,
in the proof of the last Fermat theorem” (Bulletin of the American Mathematical Society,
February 1910).

Fermat says (‘‘Relation des nouvelles découvertes en la science des nombres,”
August 1659, Oeuvres, 11. p. 433) that he proved that zo cube is divisible into two cubes by
a variety of his method of infinite diminution (descente infinie or indéfinie) different from
that which he employed for other negative or positive theorems; as to the other cases, see
Supplement, sections I., II.

1 Diophantus’ words are: “I form the square from any number of dapifuol minus
as many units as there are in the side of 16.” It is implied throughout that » must
be so chosen that the result may be rational in Diophantus’ sense, Z.e. rational and
positive. ?

2 Diophantus’ solution is substantially the same as Euler's (A4/gebra, tr. Hewlett,
Part 11. Art. 219), though the latter is expressed more generally.

Required to find x, y such that

Raf
IfxZ /,theny =r

Put therefore x=f+ps, y=g-gz:

H, D. I0
146 THE ARITHMETICA

Given number 13 = 22+ 3%

As the roots of these squares are 2, 3, take (x + 2)" as the
first square and (mx — 3)* as the second (where 7 is
an integer), say (2x — 3).

Therefore (x2 +42 + 4) + (4x2 +9 — 12x) = 13,

or 522+ 13 — 8x =13.

Therefore x =8, and

the required squares are par

25° 25
10. To find two square numbers having a given difference.

Given difference 60.
Side of one number xr, side of the other x plus any
number the square of which is not greater than 60,

say 3
Therefore +3) ~2'=60;
x=28%}, and

the required squares are 72%, 132%.

11. To add the same (required) number to two given numbers
so as to make each of them a square.

(1) Given numbers 2, 3; required number x.
Therefore * 1 2 | must both be squares.
z+r3

This is called a double-equation (SumhoicoTns).
To solve it, take the difference between the two expressions
and resolve it into two factors'; in this case let us say
4 i
Then take either
(a) the square of half the difference between these factors
and equate it to the lesser expression,

or (b) the square of half the sum and equate it to the

 

greater.
hence 2fpz + p22% — 2gyz + ¢%% =o,
289 — 2 fp
d =r" 48
an z Tq
spon fle?-A) fete (47)
SO that ma IE

in which we may substitute all possible numbers for p, g.
1 Here, as always, the factors chosen must be suitable factors, z.e. such as will lead to
a ‘“‘rational ”’ result, in Diophantus’ sense.
BOOK II 147

In this case (a) the square of half the difference is 225.

Therefore x + 2=%2p and x = 2 the squares being 225 289,
Taking (&) the square of half the sum, we have x + 3 =282

which gives the same result.

(2) To avoid a double-equation?,

first find a number which when added to 2, or to 3,
gives a square. :

Take eg. the number 22 — 2, which when added to 2 gives

"a square.

Therefore, since this same number added to 3 gives a

square,
2 + 1 =a square ={(r — 4}, say,

the number of units in the expression. (in this case 4)
being so taken that the solution may give 2? > 2.

Therefore x = 42, and

the required number is Zz as before.

12. To subtract the same (required) number from two given
numbers so as to make both remainders squares.
Given numbers 9, 21.
Assuming 9g —2*® as the required number, we satisfy one
condition, and the other requires that 12 4 2* shall be
a square.
Assume as the side of this square x minus some number
the square of which > 12, say 4.
Therefore F—4r=12+2°
and r=4,
The required number is then 8%.
[Diophantus does not reduce to lowest terms, but says
x =4 and then subtracts $ from 9 or £8,

1 This is the same procedure as that of Euler, who does not use double-equations.
Euler (4/gebra, tr. Hewlett, Part 11. Art. 214) solves the problem

Tia
x+7=22

Suppose x+4=p%;
therefore x=p2—-4, and x+7=2%+3.

Suppose that P+3=(2+9)%;
therefore 2=(3-9%)[2q.

Thus x=(9— 22% + ¢%)/49%

or, if we take a fraction 7/s instead of ¢,
x= (gst — 22722 + 24) [ 4222.
10—2
148 THE ARITHMETICA

13. From the same (required) number to subtract two given
numbers so as to make both remainders squares.
Given numbers 6, 7.

{1) Let xz be the required number.
Therefore rr i are both squares.

The difference is 1, which is the product of, say, 2 and %;

and, by the rule for solving a double equation,
I21

r—7=5% and x= —.

(2) To avoid a double-equation, seek a number which exceeds
a square by 6, say x? +6.
Therefore #2 — 1 must also be a square = (x — 2)?, say.
Therefore x = 3%, and

. : 21
the required number is =.

14. To divide a given number into two parts and to find a
square which when added to each of the two parts gives a square
number.

Given number 20.
Take two numbers! such that the sum of their squares
< 20, say 2,3.

1 Diophantus implies here that the two numbers chosen 722s¢ be such that the sum of
their squares <20. Tannery pointed out (Bibliotheca Mathematica, 1887, p. 103) that
this is not so and that the condition actually necessary to ensure a real solution in
Diophantus’ sense is something different. We have to solve the equations

x+y=a, 2+x=u 2+y=12
We assume #=z+m, v=2z2+n, and, eliminating x, y, we obtain
_a— (m+n?)

2 (m+n)

In order that z may be positive, we must have m2+n2<a; but z need not be positive
in order to satisfy the above equations. What is really required is that x, y shall both be
positive.
Now from the above we derive
x—y=(u?—- 22) =2z (m—n) +m? — nt
(772 — n) (a + 2mn)
m+n ;

Solving for x, y, we have
_m (a+ mn-n?) _ un (a+mn—m?)
on m+n ;

I ’

m+n

If, of the two assumed numbers, 7 > n, the condition necessary to secure that x, y shall
both be positive is a + mn > m2.
BOOK II 149

Add x to each and square.
We then have
riLarta
x*+6x+9 1

hd

6xr+9
are the same square.
Let then x? be the required square, and we have only to

and, | are respectively subtracted, the remainders

47 + 4 :
make rd 4! the required parts of 20.
Thus 10% + 13 =.20,
and = Jo.

The required parts are then gs. 132Y, and
q p I0 I0

the required square is 22.
I00

15. To divide a given number into two parts and to find a
square which, when each part is respectively subtracted from it,
gives a square.

Given number 20.
Take (x + m)* for the required square’, where 72* is not
greater than 20,
eg. take (x + 2)>
This leaves a square if either 44 + 4
or 1: 2243
Let these then be the parts of 20.

| is subtracted.

1 Here again the implied condition, namely that 72 is not greater than 20, is not
necessary ; the condition necessary for a real solution is something different.

The equations to be solved are x+y=qa, 22-x=u?, 22-y=202%

Diophantus here puts (+)? for 22, so that, if x=2m¢ + m?, the second equation is
satisfied. Now (&+)%—y must also be a square, and if this square is equal to (£4 m — 2)2,
say, we must have

y=ant+amn— nl
Therefore, since x+y=a,

2 (m+n) E+m?+2mn—nl=a,

a—mi+n—amn

whence t= oh

’

and it follows that
_m(a—mn+n?) _ n(a—mn+m?)
m+n :

oF m+n Ne
*
If m>n, it is necessary, in order that x, y may both be positive, that a+ 72> nn,

which is the true condition for a real solution,
150 THE ARITHMETICA

Therefore 6x + 7 = 20, and x = 18,

The required parts are therefore z 4), and
the required square is =
16. To find two numbers in a given ratio and such that each
when added to an assigned square gives a square. ;
Given square 9, given ratio 3: I
If we take a square of side (mx + 3) and subtract 9
from it, the remainder may be taken as one of the
numbers required.
Take, e.g, (x + 3)? — 9, or x? + 6x, for the lesser number.
Therefore 322+ 18x is the greater number, and 322+ 18x +9
must be made a square = (2x — 3)? say.
Therefore z= 30, and
the required numbers are 1080, 3240.

17. To find three numbers such that, if each give to the next
following a given fraction of itself and a given number besides,
the results after each has given and taken may be equal

First gives to second { of itself + 6, second to third } of
itself + 7, third to first + of itself + 8.

Let first and second be 5x, 6x respectively.

When second has taken x + 6 from first it becomes 7x + 6,
and when it has given x+7 to third it becomes
Or-1,

But first when it has given x +6 to second becomes
4x —6; and this too when it has taken } of third
+ 8 must become 6x —1.

Therefore } of third + 8 = 2x + 5, and

third = gr — 21.

Next, third after receiving & of second + 7 and giving 4 + of
itself + 8 must become 6x — 1.

Therefore 132 — 19=6x—1, and r=1¢

18,

The required numbers are 2, m 22,

1 Tannery is of opinion that the problems II. 17 and 18 have crept into the text
from an ancient commentary to Book I. to which they would more appropriately belong.
Cf. 1. 22, 23.
BOOK 11 I51

18. To divide a given number into three parts satisfying the
conditions of the preceding problem
Given number 8o.
Let first give to second 1 of itself + 6, second to third
1 of itself + 7, and third to first } of itself + 8.
[What follows in the text is not a solution of the problem
but an alternative solution of the preceding. The
first two numbers are assumed to be 5x and 12, and
the numbers found are 27° 2s8 217
TOA g 10
19. To find three squares such that the difference between the
greatest and the middle has to the difference between the middle
and the least a given ratio.
Given ratio 3: I.
Assume the least square = 2? the middle = 2* + 2x + 1.
Therefore the greatest = 2? + 8x + 4 = square = (x + 3)? say.
Thus x= 2}, and
the squares are 30%, 12%, 61.

20. To find two numbers such that the square of either added
to the other gives a square?

1 Though the solution is not given in the text, it is easily obtained from the general
solution of the preceding problem, which again, at least with our notation, is easy.

Let us assume, with Wertheim, that the numbers required in 11. 17 are sx, 6y, 72.
Then by the conditions of the problem

4x—6+2+8=5y-7+x+6=62-8+y+7,

from which two equations we can find x, z in terms of y.

In fact x=(26y —18)[19 and z=(17y - 3)/19,
and the general solution is

5(26y -18)[19, 6y, 7 (17y-3)/19.

[mn his solution Diophantus assumes x=y, whence r=] .

Now, to solve II. 18, we have only to equate the sum of the three expressions to 8o,
and so find y.

We have 1631
y(5.26+6.19+7.17)—5.18-%7.3=80.19, y=
and the required numbers are
0140 510756: 081
363° 363’ 363°
2 Euler (Algebra, Part 11. Art. 239) solves this problem more generally thus.
Required to find x, y such that 2%2+y and y2+x are squares.
If we begin by supposing x2+y=2% so that y=p?- x% and then substitute the value
of in terms of x in the second expression, we must have
pt — 29%? + xt 4 x =square.
But, as this is difficult to solve, let us suppose instead that
Ayy=(p-f=pl- pasa,
152 THE ARITHMETICA

Assume for the numbers x, 2x + 1, which by their form
satisfy one condition.
The other condition gives
42 + 5x + I =square = (2x — 2)? say.
Therefore x= 2, and
the numbers are 2 9
21. To find two numbers such that the square of either minus
the other number gives a square.
x +1, 2x +1 are assumed, satisfying one condition.
The other condition gives
44° 4+ 3x = square = gx? say.
Therefore x = 2, and ;
II
the numbers are Pa
22. To find two numbers such that the square of either added
to the sum of both gives a square.
: Assume x, x + 1 for the numbers. Thus one condition is
=X satisfied. :
It remains that
2+ 4r + 2 = square = (x — 2)? say.
Therefore x=1, and
the numbers are L 5,
4 4

[Diophantus has £, 12.]

23. To find two numbers such that the square of either minus
the sum of both gives a square.
Assume x, x +1 for the numbers, thus satisfying one
condition.
Then 2*—2x—1=square=(r— 3) say.
Therefore x= 24, and
the numbers are 2%, 33.

 

and that Ha=(g-3)2=¢%-2qy +32
It follows that y+ 2px=7p%
x+29y=g%
whence x= Wine? = ral .
429 — 1 407 ~ 1
Suppose, for example, p=2, ¢=3, and we have zl, = f and so on. We

must of course choose p, ¢ such that x, y are both positive. Diophantus’ solution is
obtained by putting p= -1, g=3.
BOOK II 153

24. To find two numbers such that either added to the square
of their sum gives a square.
Since 22+ 322, 22+ 82% are both squares, let the numbers
be 32% 822 and their sum =
Therefore 1212* = 2? whence 1122=x, and r=}.

The numbers are therefore 3-, 2.
T2121

25. To find two numbers such that the square of their sum
minus either number gives a square.
If we subtract 7 or 12 from 16, we get a square.
Assume then 122? 72 for the numbers, and 162* for the
square of their sum.
Hence 194° = 4x, and x=.
The numbers are 12, 112
361’ 361
26. To find two numbers such that their preduct added to
either gives a square, and the sides of the two squares added
together produce a given number.
Let the given number be 6.
Since x (4x —1) +x is a square, let x, 4x — 1 be the numbers.
Therefore 42> + 3x — 1 is a square, and the side of this
square must be 6 — 2x [since 2x is the side of the
first square and the sum of the sides of the square
is 6].
Since 422+ 3x — 1 =(6— 22),
we have x = 37, and

the numbers are 2 121

27

27. To find two numbers such that their product minus either
gives a square, and the sides of the two squares so arising when
added together produce a given number.

Let the given number be 5.
Assume 4x + 1, x for the numbers, so that one condition
is satisfied. :

Also 428 — 3x — 1 =(5— 2x)
Therefore x = ¢, and

the numbers are 20 © 1a

7
154 THE ARITHMETICA

28. To find two square numbers such that their product added
to either gives a square.
Let the numbers! be 22, 2
292+ 1%)
PIR + 2
To make the first expression a square we make 2? +1 a
square, putting
2+ 1 = (x — 2) say.
Therefore x =$, and 2* =f.
We have now to make % (#*+ 1) a square [and y must be
different from x].
Put 95* +9 = (37 — 4)’, say,
and y=.
9 49

Therefore the numbers are % oo

Therefore of are both squares.

29. To find two square numbers such that their product minus
either gives a square.
Let 22, 72 be the numbers.

: i
Then Td 5 are both squares.

22 2

A solution of #2 — 1 = (a square) is ¥?=2§.
We have now to solve
2% 9* — 2% = a square.
Put Li A say.
Therefore y = £7, and
the numbers are 2 i00
4’ 64
30. To find two numbers such that their product + their sum
gives a square.
Now mm? + n* + 2mn is a square.
Put 2, 3, say, for m, n respectively, and of course
224-3742, 2,3 is a square.
Assume then product of numbers = (22 + 3%) 2? or 1342 and
sSiM=2.2.370r 122°
The product being 134% let x, 132 be the numbers.
Therefore their sum 14x = 127% and x=].

The numbers are therefore 3 %.

! Diophantus does not use two unknowns, but assumes the numbers to be x2 and 1
until he has found x. Then he uses the same unknown (x) to find what he had first taken
to be unity, as explained above, p. 52. The same remark applies to the next problem.
BOOK II 155

31. To find two numbers such that their sum is a square and
their product + their sum gives a square.
2.2m .m =a square, and (2m)? + m* + 2.2m .m = a square.
Um=2, 21-2242. 4.2= 360014.
Let then the product of the numbers be (42 + 2?)x* or 20x?
and their sum 2.4. 222 or 16%? and take 2x, 10% for
the numbers.

Then 12x = 162% and x= §.
The numbers are : 2

32. To find three numbers such that the square of any one of
them added to the next following gives a square.

Let the first be x the second 2¢ + 1, and the third
2(2x+1)+1 or 4x + 3, so that two conditions are
satisfied. :

The last condition gives (4x + 3)’ + x = square = (4x — 4)%,
say.

Therefore x = ;'., and

51)

the numb e 1 25 190
numbers ar 52

33. To find three numbers such that the square of any one of
them minus the next following gives a square.
Assume x + 1, 2x + 1, 4x + 1 for the numbers, so that two
conditions are satisfied.
Lastly, 162° 7x = square = 257%, say,
and xr=1.

The numbers are 8 2 37,
9°..9°.9
34. To find three numbers such that the square of any one
added to the sum of all three gives a square.
{3m —n)}*+ mn is a square. Take a number separable
into two factors (i, 7) in three ways, say 12, which is
the product of (1, 12), (2,6) and (3,4).
The values then of { (2 — n) are 54, 2, 1.
Take 54x, 2x, 4x for the Ba, and for their sum 122%
Therefore 3z=122°, and z=1

The numbers are It 4 1
gia an

8

i)

[Diophantus says £, and 22, § 2.]
156 THE ARITHMETICA

35. To find three numbers such that the square of any one

minus the sum of all three gives a square.
{4 (m + n)}* — mn is a square. Take, as before, a number

divisible into factors in three ways, as 12.
Let then 61x, 4x, 34x be the numbers, and their sum 1222

Therefore 14x = 122% and r= 1.

The numbers are “=, = gt
BOO 111

1. To find three numbers such that, if the square of any one
of them be subtracted from the sum of all three, the remainder
is a squarel.

Take two squares 22, 42%; the sum is 522
If then we take 52° as the sum of the three numbers, and
x, 2x as two of them, we satisfy two conditions.
Next divide 5, which is the sum of two squares, into two
other squares #, 12L [IL 9], and assume 2x for the
third number.
17

Therefore x + 2x + r= 52% and x =.
17 34. 34
25 2 125
[Diophantus writes 85% for x and 8%, 11%, 3% for the numbers.]

The numbers are

2. To find three numbers such that the square of the sum of
all three added to any one of them gives a square.
Let the square of the sum of all three be 22 and the
numbers 3r2, 222, 1522
Hence 26% =x, x =, 2

8 5
the numbers are 5 &6 66

3. To find three numbers such that the square of the sum of

all three minus any one of them gives a square.
Sum of all three 4x, its square 1627, the numbers 72?

122%, 1542
Then 3412 = 4x, x=, and
the numbers are 2, 48 6
289’ 289° 289°

! The fact that the problems 111. 1—4 are very like II. 34, 35 makes Tannery suspect
that they have found their way into the text from some ancient commentary.
BOOK III 157

4. To find three numbers such that, if the square of their sum
be subtracted from any one of them, the remainder is a square.
Sum », numbers 22%, 52° 107%
Then 172% =x, r= +, ani

I0
the numbers are a

oy 289’ 289°
5. To find three numbers such that their sum is a square and
the sum of any pair exceeds the third by a square.
Let the sum. of the three be (r+ 1); let first + second
= third + 1, so that third = J#* + 25 let second + third
= first +27 so that ust =v 4, *
Therefore second = {x2 +4.  ~
It remains that first + third = second + a square.
Therefore 2x = square = 16, say, and x = 8.
The numbers are 8%, 324, 40.
Otherwise thus’.

First find three squares such that their sum is a square.
Find eg. what square number +4 +9 gives a square,
that is, 36;

4, 36, 9 are therefore squares with the required property.
Next find three numbers such that the sum of each pair =
the third + a given number ; in this case suppose
first 4+ second — third = 4,
second + third — first = 9,

third + first — second = 36.
This problem has already been solved [I. 18].
The numbers are respectively 20, 6%, 224.

1 We should naturally suppose that this alternative solution, like others, was inter-
polated. But we are reluctant to think so because the solution is so elegant that it
can hardly be attributed to a scholiast. If the solution is not genuine, we have here
an illustration of the truth that, however ingenious they are, Diophantus’ solutions are not
always the best imaginable (Loria, Ze sczenze esalte nell’ antica Grecia, Libro v. pp. 138-9).
In this case the more elegant solution is the alternative one. Generalised, it is as follows.
We have to find x, y, z so that

—x+y+2=a square
x—-y+z=a use| y
x+y — z=a square

and also : x+y +z=a square,
We have only to equate the first three expressions to squares @?, 42, 2 such that
a?+02+c2=a square, 4% say, since the sum of the first three expressions is itself
X+y+z.

The solution is then

x= (2+, y=- (+a), 2=1 (a+).
158 THE ARITHMETICA

6. To find three numbers such that their sum is a square and
the sum of any pair is a square.
Let the sum of all three be x2 + 2x + 1, sum of first and
second 22, and therefore the third 2x + 1; let sum of
second and third be (x — 1)%
Therefore the first = 4x, and the second =x? — 4x.
But first + third = square,
that is, 6x + I = square = 121, say.
Therefore x = 20, and
the numbers are 80, 320, 41.

[An alternative solution, obviously interpolated, is practically
identical with the above except that it takes the square 36 as
the value of 6x +1, so that x=3% and the numbers are 140
_ 840 385 456 1

36 36° 36
7. To find three numbers in A.P. such that the sum of any
pair gives a square.
First find three square numbers in A.P. and such that half
their sum is greater than any one of them. Let
x% (x + 1) be the first and second of these ; therefore
the third is 22+ 4x + 2 = (x — 8)? say.
Therefore x=4§2 or 81;
and we may take as the numbers 961, 1681, 2401.
We have now to find three numbers such that the sums
of pairs are the numbers just found.
The sum of the three = 803 = 25211, and
the three numbers are 120%, 840%, 15601.

8. Given one number, to find three others such that the sum
of any pair of them added to the given number gives a square, and
also the sum of the three added to the given number gives a
square.

Given number 3.
Suppose first required number + second = x*+ 4x +1,
second + third =x2+6x+6,
sum of all three =2%+ 8x + 13.
Therefore third =4x + 12, second =x2 +22 —6, first = 22+ 7.
Also first + third + 3 = a square,
that is, 6x + 22 = square = 100, suppose.
Hence x=13, and
the numbers are 33, 189, 64.
BOOK III 159

9. Given one number, to find three others such that the sum
of any pair of them minus the given number gives a square, and
also the sum of the three minus the given number gives a square.

Given number 3.
Suppose first of required numbers + second = 22 + 3,

second + third =2?%+ 2x44,

sum of the three =x? + 4x + 7.
Therefore third = 4x + 4, second = 22 — 2x, first =2x+ 3.
Lastly, first + third — 3 = 6x + 4 = a square = 64, say.
Therefore x = 10,and

(23, 80, 44) is a solution.

10. To find three numbers such that the product of any pair
of them added to a given number gives a square.

Let the given number be 12. Take a square (say 235)
and subtract 12. Take the difference (13) for the
product of the first and second numbers, and let these :
numbers be 13x, I1/x respectively.

Again subtract 12 from another square, say 16, and let the
difference (4) be the product of the second and third
numbers.

Therefore the third number = 4x.

The third condition gives 52+? + 12 =a square; NOW

S2= 4.13, and 13 1s not a square; but, if it were a
square, the equation could easily be solved. 1

Thus we must find two numbers to replace 13 and 4 such
that their product is a square, while either + 12 is _
also a square. : gat:

Now the product is a square if both are squares; hence we

must find two squares such that either + 12 = a square.
= “This is easy? and, as we said, it makes the equation easy
to solve.”

The squares 4, satisfy the condition.

1 The equation 5242+ 12 =#2 can in reality be solved as it stands, by virtue of the fact
that it has one obvious solution, namely x=1. Another solution is found by substituting
y+1 for x, and so on. Cf. pp. 69, 70 above. The value x=1 itself gives (13, 1, 4) as
a solution of the problem.

2 The method is indicated in II. 34. We have to find two pairs of squares differing.
by 12. (a) If we put 12=6. 2, we have

1 2 I 2
; HED) +.12= {6+a)} ,
and 16, 4 are squares differing by 12, or 4 is a square which when added to 12 gives a

) 2
square. (6) If we put 12=4.3, wé find fa 3) or Lt be a square which when

added to 12 gives a square.
160 THE ARITHMETICA

Retracing our steps, we now put 4%, 1/x and z/4 for the
numbers, and we have to solve the equation
x24 12 = square = (x + 3)? say.
Therefore x = §, and

(= 2, 5) is a solution,

11. To find three numbers such that the product of any pair
minus a given number gives a square.

Given number. 10.

Put product of first and second = a square + 10 =4 + IO,
say, and let first = 14, second = 1/x.

Let product of second and third = a square + 10= 19, say ;
therefore third = 19x.

By the third condition, 266x* — 10 must be a square; but
266 is not a square?

Therefore, as in the preceding problem, we must find two
squares each of which exceeds a square by 10.

The squares 301, 121 satisfy these conditions?

Putting now 30ix, 1/x, 124x for the numbers, we have,
by the third condition, 3708%+? — 10 = square [for
370 Diophantus writes 3704 1;

therefore 5929x%— 160 = square = (77x — 2)} say.

Therefore x = $4, and

the numbers are 222% 71 2

A Ey

1 Euler (Algebra, Part 11. Art. 232) has an elegant solution of this problem in whole
numbers. Let it be required to find x, y, z such that xy +a, yz +a, zx +a are all Speen
Suppose ay +a=p?% and make z=x+y+g¢;
therefore xzta=xt+xy+gxt+a=x2+qx+p?,
and yita=xy+2+gy+a=y +r +22;
and the right hand expressions are both squares if g= +2, so that z=x+y +25.

We can therefore take any value for p such that g>>a, split p?—a into factors,
take those factors respectively for the values of x and y, and so find z.

E.g. suppose a=12 and p2=25, so that xy=13; let x=1, y=13, and we have
z=14*10=24 Or 4, and (1, 13, 4), (1, 13, 24) are solutions.

2 As a matter of fact, the equation 266x2? — ro=#2 can be solved as it stands, since it
has one obvious solution, namely x=1. (Cf. pp. 69, 70 above and note on preceding
problem, p. 159.) The value x=r1 gives (14, 1, 19) as a solution of the problem.

8 Tannery brackets the passage in the text in which these squares are found, on

the ground that, as the solution was not given in the corresponding place of 111. 10, there
was no necessity to give it here. 10 and 1 being factors of 10,

I 2 I 2
{Z0-n)} + 10= {3 ton) 1
thus 30% is a square which exceeds a square by ro. Similarly [366+2) or 12} is such
= 2

a square. The latter is found in the text by putting 72 — 10=square = (7 -- 2)2, whence
m=3%, and m?=12}.

 
BOOK 111 161

12. To find three numbers such that the product of any two
added to the third gives a square.

Take a square and subtract part of it fo the third number ;
let x*4+ 6x +9 be one of the sums, and 9 the third
number.

Therefore product of first and second = »* 4-62; let first
= % so that second =x + 6,

By the two remaining conditions

10% + 54
1Ior+ 6

Therefore we have to find two squares differing by 48;
“this is easy and can be done in an infinite number
of ways.”

The squares 16, 64 satisfy the condition. Equating these
squares to the respective expressions, we obtain
x= 1, and

the numbers are 1, 7, 9. .

} are both squares.

13. To find three numbers such that the product of any two
minus the third gives a square.
First x, second x + 4; therefore product = x2 + 4%, and we
assume third = 4x.
Therefore, by the other conditions,
422+ 15%
4r*—-xr—4
The difference = 16x + 4 = 4 (4x + 1), and we put
tet 5)1* = 47% + 157.
Therefore x = $4, and
105 100

2
the numbers are 23, 195 , .
20°. 20 20

| are both squares. >

14. To find three numbers such that the product of any two
added to the square of the third gives a square

1 Wertheim gives a more general solution, as follows. If we take as the required
I I A) :
numbers X=- ax, V=ax+4?, Z=-0 two conditions are already satisfied, namely

4 4
XY+Z%=a square, and YZ+ X?=a square.
It only remains to satisfy the condition ZX + ¥2?=a square, or
433

ax? 4 oe ab?x + P= =a square.
Put PLICHR. ne 3 4b2x + b= (ax + £82)2,
166% (42 = 1)
nd ee NE TY
5 a(33-324)°

where Z remains undetermined.
HD. TT
162 THE ARITHMETICA

First x, second 4x + 4, third 1. Two conditions are thus
satisfied.
The third condition gives

#+ (47 + 4) = a square = (47 — 5), say.
Therefore x= %, and
the numbers (omitting the common on dninoter)

‘are g, 328, 73.

15. To find three numbers such that the product of any two
added to the sum of those two gives a square.

[Lemma.] The product of the squares of any two con-
secutive numbers added to the sum of the said
squares gives a square?

Let 4, 9 be two of the required numbers, x the third.

Therefore "®* 7 9 are both squares.
52+ 4

The difference = 3524 5=5(z+ 1).
Equating the square of half the sum of the factors to
10x + 9, we have
fE+6)f=10r+0.
Therefore x = 28, and (4, 9, 28) is a solution.

! The problem can of course be solved more elegantly, with our notation, thus. (The
same remark applies to the next problem, 111. 16.)
If x, y, =z are the required numbers, xy+x+y, etc. are to be squares. We may
therefore write the conditions in the form
(y+1) (241) =a square +1,
(2+ 1)(x + 1) =a square +1,
(x+1)(y+1)=a square +1.
Assuming a? 42, ¢? for the respective squares, and putting £=x+1, n=p+1, f=z+1,
we have to solve
ni=a’+1,
$E=02+1,
f=c2+1.
[This is practically the same problem as that in the Lemma to Dioph. v. 8.]
Multiplying the second and third equations and dividing by the first, we have

£={(2+1) (2+ 1)/(a®+1)},
with similar expressions for 7, {

x, y, z are these expressions minus 1 respectively. a2, 42, must of course be so
chosen that the resulting values of £, », ¢ may be rational. Cf. Euler, Commentationes
arithmeticae, 11. p. 577.

? In fact, a? (a+ 1)2+a%+(a+1)2={a (a+ 1) +1}2
BOOK III 163

Otherwise thus.
Assume first number to be x, second 3.
Therefore 4x + 3 = square = 25 say, whence x = 54, and 5%,
3 satisfy one condition.

1 This alternative solution would appear to be undoubtedly genuine.
Diophantus has solved the equations
Yit+y+z=1s
zx+ztx=v2
xy + x+y =w?
Fermat shows how to solve the corresponding problem with fo» numbers instead of
three. He uses for this purpose Diophantus’ solution of Vv. 5, namely the problem
of finding x2, 32 22, such that
J22 + x2=12, 22x24 p2= 2, AYR 2=p2
P2242 422 =22, 24224 x2=22 xP a? +y a

Diophantus finds (2. , 198) as a solution of the latter problem. Fermat takes
these as the first three of the four numbers which are to satisfy the condition that the
product of any two plus the sum of those two gives a square, and assumes x for the

fourth. Three relations out of six are already satisfied, and the other three require

 

190 pl or 105% 20
9 9 9 9

to be made squares: a ‘‘triple-equation” to be solved by Fermat’s method. (See the

Supplement, section Vv.)
Fermat does not give the solution, but I had the curiosity to work it out in order to

verify to what enormous numbers the method of the triple-equation leads, even in such

comparatively simple cases.
- We may of course neglect the denominator g and solve the equations

34x + 23=122,
73% + 64=12
205% + 196 =?
The method gives
_ _ 459818598496844787200
631629004828419699201 ’
the denominator being equal to (25132230399)2
Verifying the correctness of the solution, we find that, in fact,
: : 2
Hove ( 2505136897 ) ;
25132230399
73 rm fala )
64 25132230399
205 i =(Lzbarbol yf,
196 25132230399
Strictly speaking, as the value found for x is negative, we ought to substitute y — 4
for it (where — A is the value found) in the three equations and start afresh. The
portentous numbers which would thus arise must be left to the imagination.

11—2
164

THE ARITHMETICA

Let the third be #, while 54, 3 are the first two.

Therefore grt 3

63x + 54

but, since the coefficients in one expression are respectively

greater than those in the other, but neither of the ratios

of corresponding coefficients is that of a square lo a

square, our suppositions will not serve the purpose; we
cannot solve by our method.

Hence (to replace 54, 3) we must find two numbers such
that their product + their sum =a square, and the
ratio of the numbers increased by I respectively is
the ratio of a square to a square.

Let these be y and 4y + 3, which satisfy the latter con-
dition; and, in order that the other may be satisfied,
we must have

47* + 8y + 3 = square = (2y — 3)? say.

Therefore r=

Assume now 3, 41, x for the three numbers.

53% +45

Br +1

or, if we multiply by 25 and 100 respectively,

130% + 105
130+ 30

The difference is 75 = 3.25, and the usual method of
solution gives x = {%.

3.42 7

h x = hs
The numbersiare 2, +. .%,

| must both be squares;

Therefore i are both squares,

| are both squares.

16. To find three numbers such that the product of any two

minus the sum of those two gives a square.

Put x for the first, and any number for the second; we
then fall into the same Seay as in the last
problem.

We have to find two numbers such that

(a) their product minus their sum =a square, and

(6) when each is diminished by 1, the remainders
have the ratio of squares.
Now 47 + I, y + 1 satisfy the latter condition.
The former (@) requires that
ie — I = square = (2p — 2)’, say,
which gives y =
Assume then 12 a , x for the numbers.
BOOK III 165

Therefore ° 2 are both squares,
Re

or, if we multiply by 4, 16 respectively,

AE 2 are both squares.
The difference is 12 =2.6, and the usual method gives
L=3.
The numbers are 33 , 38 = 3=23.

17. To find two numbers such that their product added to -
both or to either gives a square,
Assume x, 4x — 1 for the numbers, since.
x(4x— 1) + x= 44% a square.
422+ 3r — 1
422 4 4 —
The difference is x=4x. 1, and we find

xX = Sr.
The numbers are 2s. 2
224° 224

Therefore also if are both squares.

18. To find two numbers such that their product szznus either,
or minus the sum of both; gives a square.

1 With this problem should be compared that in paragraph 42 of Part 1. of the
Inventum Novum of Jacobus de Billy (Oeuvres de Fermat, 111. pp. 351-2), where three
conditions correspond to those of the above problem, and there is a fourth in addition.
The problem is to find £, 3 (£>7) such that

£=5y
7—%y
£tn=2%n
E-n=4

Suppose n=x, £{=1-x; the first two conditions are thus satisfied. The other

two give :

are all squares.

x2 — x4 1 =122,
x2—3x+1=22%
Separating the difference 2x into the factors 2x, 1, we put, as usual,

[erie
he =x*—-x+1,

whence x= 2 , and the numbers are : y 3
To find another value of x by means of the value thus found, we put y+? in place of

x in the double-equation, whence

1

» = ot 2 =u?
2 2.
-2y+—
ry fis

Multiplying the lower expression by 49, we can solve in the usual way. Our expressions
166 THE ARITHMETICA

Assume x + 1, 4x for the numbers, since
4x (xr + 1) — 4x = a square.

; 2 —
Therefore also ih il : | are both squares.
The difference is 4x = 4x. 1, and we find
x =17.

The numbers are 2%, 5.

19. To find four numbers such that the square of their sum
plus or minus any one singly gives a square.
Since, in any right-angled triangle,
(sq. on hypotenuse) + (twice product of perps.) =a square,
we must seek four right-angled triangles [in rational
numbers] having the same hypotenuse,
or we must find a square which is divisible into two
squares in four different ways; and “we saw how to
divide a square into two squares in an infinite
number of ways.” [IL 3.]
Take right-angled triangles in the smallest numbers,
(3. 4, 3) and (3, 12, 13); and multiply the sides of

 

are now y2 — Sos 2 and 495% — LW 2 19 , and the difference between them is 48y% — 1105.

The solution next mentioned by Billy was clearly obtained by separating this
difference into factors such that, when the square of half their i is equated to

ry - 25 L0 , the absolute terms cancel out. The factors are —— ey 7 , and we put
’y P
4 Gx = 4
220 21 fl an 49
mo) yh = yt
i( 7 2) y= ” 64
This gives y= —- a , whence x= a , and the numbers are

48647065 22715927
71362992’ 71362992" :
A solution in smaller numbers is obtained by separating 48y*— 110y into factors such

 

that the terms in #2 in the resulting equation cancel out. The factors are 6y, gl , and

 

 

we put
(- 3 Y Brel] 5,
47959 47950 3 _ £196;
wheiice EI 0416’ and, om lily T0416" 3° 10416"
This Se give a negative value for 1—x; but, owing to the yay of the

 

         

Sn i : 86 2 ;
original double-equation in x, since x=51%03 qatisfies it, so does x=. y26 ;
10416 a

 

 

10416 41449
51865 an 51865
Cf. note > IV. 23.

numbers are : a solution also mentioned by De Billy.
BOOK III 167

the first by the hypotenuse of the second and wice
versa.

This gives the triangles (39, 52, 65) and (25, 60, 65); thus
65? is split up into two squares in zo ways.

Again, 65 is “naturally” divided into two squares in two
ways, namely into 72+ 4® and 82+ 1%, “which is due
to the fact that 65 is the product of 13 and 5, each of
which numbers is the sum of two squares.”

Form now a right-angled triangle! from 7, 4. The sides
are (P= 45, 2.7.4, 7244) or (33, 56,09).

Similarly, forming a right-angled triangle from 8, 1, we
obtain (2.8.1, 8 — 13,824 .1%) or 16, 63, 63.

Thus 652 is split into two squares in four ways.

Assume now as the sum of the numbers 65x and

as first number 2.30. 524% = 40562%, :

» Second ,, 2.29.602%=30007,

wdhird 2.33. 5604 = 36001",

» fourth, 2.16.6322= 20164,
the coefficients of #2 being four times the areas of the
four right-angled triangles respectively.

The sum 127682 = 65x, and x = Ff5.

The numbers are

17136600 12675000 15615600 8517600
163021824’ 163021824’ 163021824’ 163021824"

20. To divide a given number into two parts and to find a
square which, when either of the parts is subtracted from it, gives
a square’

Given number 10, required square x* + 2x + I.
Put for one of the parts 2x + 1, and for the other 4x.
The conditions are therefore satisfied if

6x + I = IO.
Therefore x =13;
the parts are (4, 6) and the square 6}.

1 If there are two numbers p, ¢, to “form a right-angled triangle” from them means
to take the numbers p%+¢2%, 22 —¢?, 229. These are the sides of a right-angled triangle,
since

(22 +7) = (22 - 9)* + (apg).

2 This problem and the next are the same as II. 15, 14 respectively. It may therefore
be doubted whether the solutions here given are genuine, especially as interpolations
from ancient commentaries occur most at the beginning and end of Books,
168 THE ARITHMETICA

21. To divide a given number into two parts and to find a
square which, when added to either of the parts, gives. a square.

Given number 20, required square 2*+ 2x + I.

If to the square there be added either 2x4 3 or 42+ 8,
the result is a square.

‘Take 2x + 3, 4x +8 as the parts of 20, and 62+ 11 = 20,
whence x = 14}.

Therefore the parts are (6, 14) and the square 64.

BOOK 1V

1. To divide a given number into two cubes such that the sum
of their sides is a given number.
Given number 370, given sum of sides 10.
Sides of cubes § + x, § — x, satisfying one condition.
Therefore 30424 280= 370, 2 = 2,
and the cubes are 75, 37, or 343, 27.

2. To find two numbers such that their difference is a given
number, and also the difference of their cubes is a given number.
Difference 6, difference of cubes 504.
Numbers x + 3, x — 3.
Therefore 132° +54 = 304, 2° = 25, and v= 5.
The sides of the cubes are 8, 2 and the cubes 512, 8.

3. To multiply one and the same number into a square and
its side respectively so as to make the latter product a cube and
the former product the side of the cube.

Let the square be 2% Its side being x, let the number

be 8/x.
Hence the products are 8x, 8, and
(S23 =3.
Therefore 2 = 8x, x=1}, and the number to be multiplied
is 32.

The square is I and its side 1.
16 4
1 It will be observed that Diophantus chooses, as his given numbers, numbers such
as will make the resulting “pure” quadratic equation give a ‘rational ” value for x. If
the given numbers are 2a, 26, respectively, we assume é+x, &—x as the sides of the
cubes, and we have
203+ 66x = 2a,
so that x?=(a-4%/36; x is therefore ‘‘irrational” unless (@—4?)/36 is a square. In
Diophantus’ hypothesis a is taken as 185, and 4 as 5, and the condition is satisfied. He
shows therefore incidentally that he knew how to find two numbers a, 4 such that
(a—03)/36 is a square (Loria, Ze sciernze esatte nell’ antica Grecia, Libro Vv. pp. 129-30).
A similar remark applies to the next problem, 1v. 2.
BOOK IV 169
/

4. To add the same number to a square and its side/ re-
spectively and make them the same! [7.e. make the first product a
square of which the second product is the side].

Square 2? with side x. :

Let the number added to #4? be such as to make a square
say 322

‘Therefore 347 + z= side of 42="2¢, and z=1.

The square is > its side 4 and the number :

5. To add the same number to a square and its sid and make
them the opposite?
Square 2? the number a square number of times 2?
minus x, say 41° — x.
Hence 52° —x=side of 422 = 2x, and x=}.

The square is 9 its side 3 and the number 5

25
6. To add the same square number to a cube and a square
and make them the same.

Let the cube be #* and the square any square number of
{imes #7, say g2%

We want now a square which when added to gx? makes
a square. Take two factors of 9, say 9 and 1, sub-
tract 1 from 9, take half the difference and square.
This gives 16.

~ “Therefore 162* is the square to be added.
Next, 2* + 162° =a cube = 84% say; and x = 18.

The cube is therefore 4096 the square 2% and
343 1 49

the added square number 1098

! In this and the following enunciations I have kept closely to the Greek partly
for the purpose of showing Diophantus’ mode of expression and partly for the brevity
gained thereby.

In Prop. 4 to ‘make them the same” means what I have put in brackets ; to ‘make
them the opposite” in Prop. 5 means to make the first product a side of which the second
product is the square.

? Nesselmann solves the problem generally, thus (Notes in Zeitschrift fiir Math. u.
Physik, XXXVI (1892), Hist. litt. Abt. p. 162).

x24+y=4/(x+y); therefore x*+ 242 +y2=u+y, or y2— (1 — 242) y=x — x4,
Solving for y, we obtain, as one of the solutions,

y=ioat N (a=).
2 4

; ; I 2
To make the expression under the radical a square we put trrmat= (mat) 4
4 2

m1 m4 m3 —m— 1

whence x= ——, y= ———————
Zeit? (72+ 1)?

Diophantus’ solution corresponds to 7 =2,
170 THE ARITHMETICA

7. To add the same square number to a cube and a square
respectively and make them the opposite.
For brevity call the cube (1), the second square (2) and
the added square (3).
Now, since (2) + (3) = a cube, suppose (2) + (3) = (1).
Since a*4 8% + 2ab is a square, suppose (1)= (a? + b%),
(3) = 2ab, so that the condition that (1) + (3) = square
is satisfied.
But (3) is a square, and, in order that 2aé may be a square,
we put a=2,0= 2%
Suppose then (1)=22+(2xF=352% (3)=2.x.2x=42";
therefore (2) = 2? by subtraction.
But 522 is a cube; therefore x =,
and the cube (1)= 125, the square (2)=25, the
square (3) = 100.
Otherwise thus.
Let (2) + (3) = (1).
Then, since (1) + (3) = a square, we have to find two squares
such that their sum + one of them = a square.
Let the first of these squares be #2, the second 4.
Therefore 22* + 4 = square = (2x — 2), say; thus x =4,
and the squares are 16, 4.
Assume now (2) = 4x7, (3)= 162%
Therefore 20x? is a cube, so that x= 20;
the cube (1) is 8000, the square (2) is 1600, and the
added square (3) is 6400.

8. To add the same number to a cube and its side and make

them the same.
Added number x, cube 81% say. Therefore second sum
= 3x, and this must be the side of 84° + x.
That is, 32°-t2=272" and 1042 =z, or 1ga’=1.

1 Nesselmann (gg. cit. p. 163) gives a more general solution.
We have x3 +y= (x +y)3, whence 1 =3x2+ 3xy +52
Solving for y, we find

y= “ea f(s -2 2) => {-3r+Jls-329)

; 7 \? m 2 _ 612
Lastly, putting 4 — 322= € AN x) , we find x= rs » M(4=320) == er ;
— bmn = (m2 — 3n?)
3n2 +m?
always be positive, 72/7 must be >3+44/12; Diophantus’ solution corresponds to 7 =7,
n=1.

and y= If the positive sign be taken, then, in order that y may
BOOK IV 171

But 19 is not a square. Hence we must find, to replace
it, some square number. Now 19a* arises from
271% — 82% where 27 is the cube of 3, and 8 the cube
of 2. And the 3x comes from the assumed side 2x,
by increasing the coefficient by unity.

Thus we must find two consecutive numbers such that their
cubes differ by a square.

Let them be y, y+ 1.

Therefore 39% + 37 + 1 = square = (1 — 2y)% say, and y =7.

Going back to the beginning, we assume added number

: =z, side of cube =72.

The side of the new cube is then 8x, and

3432+ x =75122%

Therefore 22 = 115, and x = J.

The cube is 3% its side 7, and the added
2197 137%

XT
number 5

9. To add the same number to a cube and its side and make
them the opposite’.

Suppose the cube is 843, its side being 2x, and the added
number is 272% ~2x (The coefficients 8, 27 are
chosen as cube numbers.)

Therefore 354% — 22 =side of cube 2724%=32, or 352° =5.

This gives no rational value.

But 35=27+8, and 5=7342,

Therefore we have to find two cubes such that their sum
has to the sum of their sides the ratio of a square
to a square?

Let sum of sides =any number, 2 say, and side of first
cube = 2, so that the side of the other cube is 2 — 2. .

I Nesselmann (gg. c#t. p. 163) solves as follows. The equation being x + y= (3+ 2)3,
put y=2z— x3, and the equation becomes x+z—a3=23, or B+ z3=ux +2.
Dividing by x +2, we have x2 — xz+22=1.

Solving for x, we obtain x= {z2£4/(4— 322) }.

To mak 22 a squa te it to (7 if therefore z= 17" so that
make 4-32? a square, equate it to (— 2-2) ; therefore ST

2mn = (m? — 3n?)
Te
m? + gn?
corresponds to m=2, n=1.
% It will be observed that here and in the next problem Diophantus makes no use of
the fact that

,and y=z- 23. If the positive sign be taken, Diophantus’ solution

(B+) (x+y) =22 - xy +52
Cf. note on IV. 11 below.
172 THE ARITHMETICA

Therefore 8 — 122 + 65° must be twice a square.

That is, 4 — 65 + 32° = square = (2 — 42)? say; ¢=1%, and
the sides are 13, 1§.

Neglecting the denominator and the factor 2 in the
numerators, we take 5, 8 for the sides.

Starting afresh, we put for the cube 12524 and for the
number to be added 5124°— 5.x; we thus get

O37 —~5r=8v and +r =1.

The cube is 125, its side 5 and the added number?%Z,
343 7 343

10. To find two cubes the sum of which is equal to the sum
of their sides.

1 et the sides be 2, 32

This gives 354% = 5x; but this equation gives an irrational
result.

We have therefore, as in the last problem, to find two
cubes the sum of which has to the sum of their sides
the ratio of a square to a square’.

These are found, as before, to be 53, 82

Assuming then 5.x, 8x as the sides of the required cubes,
we obtain the equation 6374°=13x, and x= 1.

The cubes are X23, 512

343’ 343°

1 Here, as in the last problem, Diophantus could have solved his auxiliary problem
of making (x4 3°)/(x +») a square by making x — xy +? a square in the same way as in
Lemma 1. to v. 7 he makes x2 + xy +3? a square.

The original problem, however, of solving

x3 +2 =x+y
can be more directly and generally solved thus. Dividing out by (x+y), we must have
22—zy+yi=1.
This can be solved by the method shown in the note to the preceding problem.
Alternatively, we may (with Wertheim) put x2- xy +32=(x+4y)% and at the same
time 1 = = (x + 4&y).
Thus we have to solve the equations

x (1+28)=y (1 La

 

xX+hy= CE
; . — kp? I+ 24
which give a it TI A
g Tw Ero

where # remains undetermined.

Diophantus’ solution is obtained by taking the positive sign and putting k=1 or by
4

taking the negative sign and putting 4= — 2,
2
BOOK IV : 173

11. To find two cubes such that their difference is equal to
the difference of their sides.

Assume 2x, 3x as the sides.

This gives 1924® =x, and x is irrational.

We have therefore to find two cubes such that their
difference has to the difference of their sides the
ratio of a square to a square’. Let them be (z+ 1);
23, so that the difference of the sides may be a square,
namely 1.

Therefore 32+ 32+ 1 =square = (I — 22)}, say, and z=7.

Starting afresh, assume 7x, 8x as the sides; therefore
1692° =x, and x = ;.

The sides of the two cubes are therefore % y

1 Nesselmann (Die Algebra der Griechen, pp. 447-8) comments on the fact that
Diophantus makes no use here of the formula (x*—3%)/(x — y) =x%*+ xy + ?%, although he
must of course have known it (it is indeed included in Euclid’s much more general
summation of a geometrical progression, 1X. 35). To solve the auxiliary problem in
1v. 11 he had only to solve the equation

2%? + xy +y*=a square,
which equation he does actually solve in his Lemma I. to Vv. 7.
The whole problem can be more simply and generally solved thus. We are to have

 

Se p=ry,
or. Xt+axy+yi=1.
Nesselmann’s method of solution (cf. note on IV. 9) gives x=t {-y=J(4-37},
i + (722 — 3722 ; ST .
and hence y= Line 5s X= ae fo 5 3) Diophantus’ solution is obtained by
m2 + 3n m+ 3n

putting 72 =1, »=2 and taking the lower sign.

Wertheim’s method (see note on preceding problem) gives in this case
Sir? Sei me

1 — A+ A? 1—- A+ A?
where 4 is undetermined.

If we take the negative sign and put 2= — 3, we obtain Diophantus’ solution.

Bachet in his notes to 1v. 10, 11 solves the problems represented by

3k y8=yp (x+y)

subject to the condition that m is either a square or the third part of a square. His method
corresponds to that of Diophantus. He does not divide out by x= y, and he reduces the
problem to the subsidiary one of finding £, 5 such that the ratio of £93 to £7 is the
ratio of a square to a square. His assumptions for the ‘‘ sides,” £, 7, are of the same kind
as those made by Diophantus; in the first problem he assumes x, 6 —~x and in the second
x, +2. In fact, it being given that (x3 +39)/(x+y) =a, Bachet assumes x+y =z and

thus obtains
2

3-324 2 =a,
which equation can easily be solved by Diophantus’ method if « is a square or triple of a
square.

Fermat observes that the dwopiomés of Bachet is incorrect because not general. It
should be added that the number (2) may also be the product of a square number into a
prime number of the form 3741, as 7, 13, 19, 37 etc. or into any number which has no
factors except 3 and prime numbers of the form 32+ 1, as 21, g1 ete.  ““ The proof and

the solution are to be obtained by my method.”
174 THE ARITHMETICA

12. To find two numbers such that the cube of the greater
+ the less =the cube of the less + the greater

Assume 2%, 3x for the numbers.

Therefore 272 +24=82%+ 32, or 1094°=2, and x is
irrational.

But 19 is the difference of two cubes, and 1 the difference
of their sides. Therefore, as in the last problem,
we have to find two cubes such that their difference
has to the difference of their sides the ratio of a
square to a square?

The sides of these cubes are found, as before, to be 7, 8.

Starting afresh, we assume 7x, 8x for the numbers; then
3437+ 8xr=5122°+ 7x, and x = {4.

The numbers are Zz 3.
13° 13
13. To find two numbers such that either, or their sum, or
their difference added to unity gives a square.

Take for the first number any square less I; let it be,
say, 9x*+ 6x. But the second + 1 =a square; and
first + second + 1 also =a square. Therefore we must
find a square such that the sum of that square and
gx*+4 6x = a square.

Take factors of the difference 9a®+ 6x, say 9x +6, x;
the square of half the difference between these factors
= 162 + 24% +9. :

Therefore, if we put for the second number this expres-
sion minus 1, or 164% + 24x + 8, three conditions are
satisfied.

The remaining condition gives difference + I = square,
or 722 + 18x + 9 =square = (3 — 3x), say.

Therefore x= 18, and (3024, 5624) is a solution.

14. To find three square numbers such that their sum is equal
to the sum of their differences.

Sum of differences = (greatest) — (middle) + (middle) —
(least) + (greatest) — least=twice difference of greatest
and least.

This is equal to the sum of all three, by hypothesis.

Let the least square be 1, the greatest 22+ 2x + 1;

1 This problem will be seen to be identical with the preceding problem.
2 See'note, p. 173.
BOOK: IV 175

therefore twice difference of greatest and least = sum of
the three =22" 44x,
But least + greatest =x* + 2x + 2, so that
middle = 2% + 2x — 2.
Hence 2% + 2x — 2 = square = (xr — 4)% say, and x= 3,
The squares are (2, = 1) or {196, 121, 25).
15. To find three numbers such that the sum of any two
multiplied into the third is a given number.
Let (first + second) x third = 35, (second + third) x first
= 27 and (third + first) x second =32.
Let the third be x,
Therefore (first + second) = 35/2.
Assume first = 10/z, second = 25/z; then

250 |]
fi + 10=27
250
x?
These equations are inconsistent; but they would not be if
25 — 10 were equal to 32 — 27 or 5.
Therefore we have to divide 35 into two parts (to replace

25 and 10) such that their difference is 5. The parts

are 15,20 {CL 1 1]
We take therefore 15/x for the first number, 20/x for the

second, and we have

 

r25=32)

 

 

300
5 LI ma
300

oF +2032

Therefore x= 5, and (3, 4, 5) is a solution.

1 As Loria says (Le scienze esatte nell’ antica Grecia, Libro Vv. p. 131), this method of
the ‘“ false hypothesis,” though somewhat indirect, would not be undeserving of a place
in a modern textbook.

Here again, as in 1v. 1, 2, Diophantus tacitly chooses, for his given numbers, numbers
which will make the resulting *‘ pure” quadratic equation give a rational value for x.

We may put the solution more generally thus. We have to solve the equations

(y+2)x=a, (z+x)y=b, (x+y)z=c.

Diophantus takes z for his principal unknown and, writing the third equation in the
form x+y=c[z, he assumes x=afz, y=f/z, where a, B have to be determined. One
equation connecting a, 8 is a+pB=c. Next, substituting the values of x, y in the first two
equations, we have

2 +a=a, z +B8=4,
176 THE ARITHMETICA

16. To find three numbers such that their sum is a square,
while the sum of the square of each added to the next following
number gives a square.

Let the middle number be any number of x's, say 4x;
we have therefore to find what square + 4x gives
a square. Split 4x into two factors, say 2x, 2, and
take the square of half their difference, (* —1)2 This
is the square required.

Thus the first number is x — I.

Again, 162 + third number =a square. Therefore, if we
subtract 164* from a square, we shall have the third
number. Take as the side of this square the side of
1622, or 4%, plus 1.

Therefore third number = (4x + 1)? — 1622 = 8x + I.

Now the sum of the three numbers =a square; therefore
132=a square = 16g), say’.

The numbers are then 139% — 1, 529% 104° + I.

Lastly, (third)®+ first = a square.

Therefore 108164 + 221° = a square,

or 108162 + 221 =a Suan = (104¥ + 1)? say.

Therefore y = 83 = 28,

and [pn 157300 317304
2704’ 2704 ’ 2704

 

 

) is a solution.

17. To find three numbers such that their sum is a square, while
the square on any one minus the next following also gives a square.
~ The solution is precisely similar to the last.

 

whence it follows that a — =a —4. From this condition and a+g=¢, we obtain

= (a-b+0), B=2(-a+b+d).

ol EY ,

al fer irglrt a1 A peti (-a+d+c)(atb-o)
=t=y/ 1 2(—ab:bo) ? = TRE-ira J

Now x, y, z must all be rational, and this is the case if
—atbtec=29r, a-b+c=2rp, atb-c=2py,
where p, ¢, » are any integers.
This gives a=p(g+r), b=qg@+p), c=r(p+9);
a fact which can hardly have been unknown to Diophantus, since his values a=27, 6=32,
¢=35 correspond to the values p=3, g=4, »=15 (Loria, loc. cit.).
! Diophantus uses the same unknown s for y as for x, writing actually kal ylverat 6

 

s AY, literally “and x becomes 1342.”
BOOK IV 197

The middle number is assumed to be 4x. The square
which exceeds this by a square is (x + 1)% and we
therefore take x + 1 for the first number.

For the third number we take 164? — (4x — 1)? or 8x — 1.

The sum of the numbers being a square,

13x = a square = 1697? say.

The numbers are then 1332 + 1, 5252 10492 — 1.

Lastly, since (third)? — first =a square,

10816y* — 2219* = a square,

or 108162 — 221 =a square = (104 — 1)? say.

Thus y =111,

Suid 170989 640692 1270568
10816 ° 10816 > 10816

 

) is a solution.

18. To find two numbers such that the cube of the first added
to the second gives a cube, and the square of the second added to
the first gives a square. :

First number x Therefore second is a cube number
minus x°, say 8 —2°, ;

Therefore 2%— 162% + 64 + x =a square = (2° + 8)? say,
whence 324°=2%, or 322°=1,

This gives an irrational result; x would however be
rational if 32 were a square.

But 32 comes from 4 times 8. We must therefore sub-
stitute for 8 in our assumptions a cube which when
multiplied by 4 gives a square. If 3° is the cube,
47" = a square = 16)* say; whence y= 4.

Thus we must assume x, 64 — 2° for the numbers.

Therefore 2°— 128.2% + 4096+ x = a square = (2* + 64)? say ;
whence 256.1% =x, and x = 4%.

The numbers are 1, #2143

16° 4096
19. To find three numbers indeterminately! such that the
product of any two increased by 1 is a square.
Take for the product of first and second some square
minus 1, say 2* + 2x; this satisfies one condition.

1.et second = #, so that first =24 2.

Now product of second and third + 1 =a square; let the
1 The expression is év 7¢ doploTe, which is defined at the end of the problem to mean

in terms of one unknown (and units), so that the conditions of the problem are satisfied
whatever value is given to the unknown.
7

H. D. 12
178

THE ARITHMETICA

square be (3x4 1)’ so that product of second and
third = 92% + 6x;

therefore third =gx + 6.

Also product of third and first + 1 = a square; therefore
ox + 24x + 13 = a square.

Now, if 13 were a square, and the coefficient of x were
twice the product of the side of this square and the
side of the coefficient of 2°, the problem would be
solved indeterminately.

But 13 comes from 2.6 + 1, the 2 in this from twice 1,
and the 6 from twice 3. Therefore we want two
coefficients (to replace 1, 3) such that the product
of their doubles + 1 =a square, or four times their
product + 1 = a square.

Now four times the product of any two numbers plus the
square of their difference gives a square. Thus the
requirement is satisfied by taking as coefficients any
two consecutive numbers, since the square of their
difference is 1. [The assumption of two consecutive
numbers for the coefficients simultaneously satisfies
the second of the two requirements indicated in the
italicised sentence above. ]

Beginning again, we take (+1)? —1 for the product of
first and second and (2x + 1)?— 1 for the product of
second and third.

Let the second be x, so that first =x 4-2, third=42x +4.

[Then product of first and third + 1 = 442 + 122 + 9, and
the third condition is satisfied.)

Thus the required indeterminate solution! is
(x +2, x, 4x + 4).

1 The result obtained by Diophantus really amounts to the more general solution

1.

a*x+2a, x, (a+1)%x+2(a+1).

With this solution should be compared that of Euler (4/gebra, Part 11. Art. 231). °
To determine x, y, z so that

xy+1, yz2+1, zx+1 are all squares.

Suppose swt 1=22 yz-t1=g%

so that x= -1)[z, y=(¢®-1)[z.
1) 7.

Therefore xy +1= Zara + 1 =a square,

oF (p%- 1) (¢%— 1) + 2%2=a square
=(z-7)? say; [Euler has (z+7)?%]
LLP = 2.

whence eelnti- 09)

27 ?

where any numbers may be substituted for g, ¢, 7.
BOOK IV 179
20. To find four numbers such that the product of any two
increased by unity is a square.
For the product of first and second take a square minus 1,
say (+ 1) —1=2% +221.
Let first = x, so that second =x +4 2.

 

For example, if »=pg¢ + 1, we shall have

 

+9)? 2 (pg+1)(p2-1 2(pg+1)(g2—1
= 2; and = Ls hnpetiBy 2 pt
II. But, if whole numbers are required, we put xy+ 1 =p? and assume z=x+y +4.
We then have xz+1=a+ay+gr+ 1 =x +gx +p?
and rei Eay ty opt my hopt sl

These expressions are both squares if g = = 22.
Thus a solution is obtained from xy=p%—- 1 combined with either
z=x+y+2p, or z=X+y—2p.

We take a certain value for p? split p2 - 1 into two factors, take these factors for the
values of x, y respectively, and so find z.

For example, let p=3, so that p?- 1=8; if we make x=2, y=4, we find z=either 12
or 0; and in this case x=12, y=4, z=12 is the solution.

If we put p2=(£+1)%, we have xy=£*+2¢; and if we put x=£+2, y=§, we have

s=ftetite(f+1)=48+4 or 0.

The solution is then (£+ 2, £, 4£+ 4), as in Diophantus.

Fermat in his note on this problem shows how to find three numbers satisfying not
only the conditions of the problem but three more also, namely that each of the numbers
shall itself when increased by 1 give a square, i.e. to solve the equations

p+1=22 E+1=4s2 Ip+1=4,
(+1=u2, n+1=v% (+iz=wl

Solve, he says, the present problem of Diophantus in such a way that the terms
independent of x in the first and third of the numbers obtained by his method shall be
such as when increased by 1 give a square. It is easy to find a value for a such that

I . J
2a+ 1 and 2 (a+ 1) +1 are both squares. Fermat takes the value 2a=23 , which satisfies

3
the conditions, and the general expressions for the three numbers sought are therefore
199 oil 2h
he ae
Each of these has, when increased by 1, to become a square, that is; we have to solve the
triple-equation

 

 

169 9 5
ie

x+1=172
7228 Spit en
ge B=

Fermat does not give the solution; but it is effected as follows.
Multiplying the third expression by 36 and the first by £18 36 (in order that the

absolute terms in the two may be equal), we have to solve

12—2
180 THE ARITHMETICA

For the product of first and third take (2x +1)*—1, or
424° + 4, the coefficient of x being the number next
following the coefficient (1) taken in the first case, °
for the reason shown in the last problem;

thus third number = 4x + 4.

Similarly take (34 + 1)? — 1, or 942+ 6x, for the product of
first and fourth; therefore fourth = gx +6.

And product of third and fourth + 1

= (4x + 4) (9x + 6) + 1 = 362° + 60x + 25,
which is a square.

 

x+1=2%

2
(222) x+121=u"?
7.12

/8z2\2
2) xti2r=w2
12
In order to solve by the method of the triple-equation, we make x+1 a square by

putting x=y%+2y.
Substitute this value in the other two expressions, and for convenience multiply each

by 144; this gives
1432 :
(42) bra) + (13202=a i

(85)2 (#2 + 2y) + (132)2=a square
The difference = (3% + 2y) (8s + 12) (8s = )

7
738 (12 Tat)
=50 py (BE pp),
TEST

The square of half the difference of the factors equated to the smaller expression gives

143 45? = (4° 2 2,
( 2 2) 2 ) (2+ 2p) + (132)%;

_ 324736

whence y= 85085

 

; and we find that

wag oymio S870

. 7239457225
It is easily verified that

BY a (C1210) 85\* _ [1837
(2) tv = {G0 i 2 x+vet=( 85085 Y

 

so that the value of x satisfies the three equations.
The numbers satisfying Fermat's six conditions are then
380 nel 10000! | _SOIOSIUANIG sj 7008, 00 humby
5184 36 171348100 7239457225 5184" 36 4008004
! This results from the fact that, if we have three numbers x, y, z such that

xy+1=(mx+1)2 and xz+r1={(m+1)x+1}2

then yet+1={m(m+1)x+(2m+1)}2
BOOK 1V 181

Lastly, product of second and fourth + 1 = 92? + 24x + 13;
therefore 9x* + + 13 =square = (3x — 4) say;
which gives r=.

All the conditions are now ride

2/33 08 108; ion?
and 5 ae ip IS the solution?
1 The remaining condition was: product of second and third + 1 =a square. That this
is satisfied also follows from the general property stated in the last note. In fact
(x +2) (4x+4) + 1=422+ 122 +0,
which is a square.
2 With this solution should be compared Euler’s solution (A4/gebra, Part 11. Art. 233)
of the problem of finding x, y, 2, 2 such that the six expressions
xy+a, yz+a, zx+a, xv+a, yvta, z2v+ta
are all squares. The solution follows the method adopted to solve the corresponding
problem with three unknowns x, y, z only. See note on III. 10 above.
If we begin by supposing xy+a=p?% and take z=x+4y+2p, the second and third

expressions become squares (vide note on III. 10, p. 160).
If we further suppose z=x+y-2p, the fourth and fifth expressions also beconie

squares (vide the same note).
Consequently we have only to secure that the sixth expression zz +a shall be a square ;

that is,
x24 2xy +32 — 49% + a=a square,

 

or (since xy +a =p? x? — 2xy + y% — 3a =a square.
Suppose that (x-y2-3a=(x-y-9¢)%;
therefore x -y=(g+3a)[2g,
7*+ 3a 2
or NY
Fo 29
9 ar 7*+3a
Consequently P=xyta=yi+ 27 yta.
If we put p=y+r, we have
ary +72= 72 Lon y+a,
2am
ii VE agr+3a’

from which p, x, and therefore z, » also, are found in terms of ¢, », where ¢, » may have
any values provided that x, y, 2, v are all positive.

Euler observed that this method is not suited for finding integral solutions, and,
pursuing the matter further, he gave the following very elegant solution of Diophantus’
actual problem (the case where a=1) in integers (‘‘Miscellanea analytica” in Com-
mentationes arithmeticae, 11. pp. 45-6).

Six conditions have to be satisfied. If x, y, z, v are the required numbers, let x=,
y =n, where m, n are any integers such that mz +1 =272.

Then put z= + n+ 2/, and three conditions are already satisfied, for

xy +1=mn+ 1=1[%, by hypothesis,
xz+1=m(m+n+2l)+1=(+m)
yi+1=n(m+n+20)+1=(/+n)

The three conditions remaining to be satisfied are

xy + 1=mv+1=a square,
yv+1=nv+1=a square,
2w+1=(m+n+2/)v+1=a square.

Let us make the continued product of these expressions a square.
182 THE ARITHMETICA

21. To find three numbers in proportion and such that the
difference of any two is a square.
Assume x for the least, x + 4 for the middle (in order
that the difference of middle and least may be a
square), x + 13 for the greatest (in order that differ-
ence of greatest and middle may be a square).

 

This product will be found to be
142 (m+n+l)o+{(m+n+i)?—1} 22+ mn (m+n+-2l)23
2
Let us equate this to dr +(m+n+l)v- Loa , in order that the terms in 2, 2? as well
as the absolute term may vanish ; therefore

mn (m+n+20)= ~ (ntti) +o,

whence [I= mnt kn (m+ n+ 20)
=(mn+1)(m+n+10)+imn
=P(m+n+l)+imn
=/(/+m) (+n),

and therefore v=4l (I+ m)({+n).

It is true that we have only made the product of the three expressions mz +1, no +1,
(m+n+2/)v+1 a square; but, as the value of # has turned out to be an integral number,
so that all three formulae are prime to one another, we may conclude that each of the
expressions is a square.

The solution is therefore

x=m, y=n, z=m+n+2,, v=4l (+m) (+n),
where mn+1=17».

In fact, while three of the conditions have been above shown to be satisfied, we find,
as regards the other three, that

xv+1=glm (+m) ({+n)+ 1=(22+ 20m — 1)%,
yo+1=4ln(l+m) (+n) +1=(20242/n—1)%
w+ 1=4l(m+n+20) (+m) (l+n)+1= (42+ 2lm+2ln—1)%

It is to be observed that / may be either positive or negative.

Ex. Letm=3, #=3, so that l= =x. :

If /= + 5, the solution is 3, 8, 21, 2080; if /= — 5, the solution is 3, 8, 1, 120.

Fermat shows how to solve this problem, alternatively, by means of the “triple-equation.”

Take #Aree numbers with the required property, e.g. 3, 1, 8. Let x be the fourth, and
we have then to satisfy the conditions

3xti=u2, xti=of, SBrti=wl

Put x=yp%+ 2, so as to make the second expression a square, and then substitute the .
value of x in the other two. We have then the double-equation

3(y2+2y)+1=02,
8 (+a) +1=w

The difference =5 (32+ 29) = 33 (y+ 2).

We put then (r+ 1P=8F+uy)+1,
whence y=10, and x=y5?+ 2 = 120, which value satisfies the triple-equation.

The four numbers are then 3, 1, 8, 120, which solution is identical with one of those
ottained by Euler as above. :
BOOK 1V 183

If now 13 were a square, we should have an indeterminate
solution satisfying three of the conditions.

We must therefore replace 13 by a square which is the
sum of two squares. Any rational right - angled
triangle will furnish what is wanted, say 3, 4, 5;

we therefore put for the numbers x, x + 9, x + 25.

The fourth condition gives

x(x + 25)=(xr+9)}, and xr=8§L,
Thus 5 HE 256 is a solution.
7e 7 57
22. To find three numbers such that their solid content! added
to any one of them gives a square.

Assume continued product 2? + 2, first number 1, second
number 4x + 9, so that two conditions are satisfied.

The third number is then (#* + 22)/(4x + 9).

This cannot be divided out unless 12: 4x=2x:9 or,
alternately, 22: 22= 42:9; but it could be done if 4
were half of g.

Now 4x comes from 6x — 2x, and the 6x in this from
twice 3x; the 9 comes from 3%

Therefore we have to find a number 7 to replace 3 such
that 2m — 2 = im: thus mw? = 4m — 4, whence? m = 2.

We put therefore for the second number (x + 2)? — (2? + 2%),
or 2x + 4; the third number is then

(* + 2x)/(2x + 4) or }x.

Lastly, the third condition requires

2+ 2x + x =a square = 44% say.
Therefore x =§,

and (x 3, 2) is a solution?,

1 6 €f avTQv aTepebs, *¢ the solid (number formed) from them ” =the continued product
of the three numbers. ] 3

2 Observe the solution of a mixed quadratic.

3 Fermat gives a solution which avoids the necessity for the auxiliary problem.

Let the solid content be x2 — 2.x, the first number 1, and the second number 2x; two
conditions are thus satisfied.

The third number is now a? — 2x divided by 2x. 1, or samt; and the third condition
gives

a?-3x_1=a square.
2

Now x must be greater than 2; we therefore put

2 wn
R-2%-1=(rc nlf,

where m is greater than 2,
184 THE ARITHMETICA

23. To find three numbers such that their solid content minus
any one gives a squarel.
First number x, solid content 22+ x; therefore product of
second and third =x + 1.
Let the second be 1, so that the third is x + 1.
The two remaining conditions require that
BLr—1 :
in | shall both be squares. [Double-equation.]
The difference =2=3 .2%, say;
thus (r+ }P=2*+xr—1,and r=11."
The numbers are (Z I, 2).
24. To divide a given number into two parts such that their
product is a cube minus its side.
Given number 6. First part x; therefore second =6 — z,
and 6x — 2? = a cube minus its side.
Form a cube from a side of the form mx — 1, say 2x — 1,
and equate 6x — 2? to this cube minus its side.
Therefore 84% — 1222 + 4x= 6x — x2

1 A remarkable problem of this kind (in respect of the apparent number of conditions
satisfied) is given by De Billy in the Znventum Novum, Part 1. paragraph 43 (Ocuvres
de Fermat, 111. p. 352): To find three numbers & 7, { (£, {, 7 being in ascending order
of magnitude) such that the following nine expressions may become squares :

(1) E-&n, (4) n-E-&n, (9) En-Enf,
(2) n-tme, (5) $-£-fng, (8) ni-1mt
(3) $-Z25, (6) n-¢-Gt, (9) »?-&ng.

Take x, 1, 1 —x as the values of §, u, { respectively. Then six conditions, namely,
(1), (3), (4), (6), (7), (8), are all automatically satisfied.
By conditions (2) and (9) alike,

I-x+x2=a square.
And, by (5), 1-3x+x?=a square.

Solving this double-equation in the usual way, we get x=3, and the numbers are

Another solution can be obtained by putting y +3 in place of x in the two expressions,

and so on. Cf. note on III. 18 above.

It would appear from a letter from Fermat to De Billy of 26 Aug. 1659 (Oeuvres, 11.
pp- 436-8) that this problem and the above single solution were De Billy’s own. De Billy
had supposed that this was the only solution, but Fermat observed that there were any
number, as the above double-equation has any number of solutions. Fermat gave

10416 47449
(Bue, © $1863

 

) as another solution.
BOOK IV 185

Now, if the coefficient of x were the same on both sides,
this would reduce to a simple equation, and x would
be rational.

In order that this may be the case, we must put for 2
in our assumption, where 3m —m =6 (the 6 being
the given number in the original hypothesis). Thus
w= 3,

We therefore assume

(3:x—-1P—-(32—-1)=6x - 2,

or ZI ~ 2724 Ox = Gr ~ 472,
and x= 24.
26 136

The part =, =.
€ parts ate 5, >

25. To divide a given number into three parts such that their
continued product gives a cube the side of which is equal to the
sum of the differences of the parts.

Given number 4.

Since the product is a cube, let it be 84% the side of
which is 2x.

Now (second part) — (first) + (third) — (second) + (third)
— (first) = twice difference between third and first.

Therefore difference between third and first = half sum of
differences = x.

Let the first be any multiple of x, say 2x; therefore the
third = 32,

Hence second = 8x%/62* = 4x; and, if the second had lain
between the first and third, the problem would have been
solved.

Now the second came from dividing 8 by 2. 3, and the
2 and 3 are not two numbers at random but con-
secutive numbers,

Therefore we have to find two consecutive numbers such
that, when 8 is divided by their product, the quotient
lies between the numbers.

Assume mm, m+ 1; therefore 8/(m?+ m) lies between
m and m + 1. :

 

Therefore +1>m+ 1,

m? + m
so that mE +m + 8 > md + 2m +m,
or 8 > md + nA
186 THE ARITHMETICA

I form a cube such that it has #3, #2 as terms, that is, the

cube (72 4%), which is greater than #2 +? and I put
(m+3y=3

therefore 7 + 4+ = 2, and m = 3.

Assume now for first number 4x; the third is $x, and
the second is 3».

Multiplying roto by 15, we take 25%, 272, 40x,
and the product of these numbers is a cube the
side of which is the sum of their differences.

The sum = 92x = 4, by hypothesis.

Therefore x = 5,

25 27 40 .
and & Bao 2) are the parts required.

[N.B. The condition 8/(m*+ m)<m+1 is ignored in
the work, and is zncidentally satisfied.]

26. To find two numbers such that their product added to
either gives a cube.
Let the first number be of the form #2, say 8x.
Second 22 — 1. Therefore one condition is satisfied, since

813 — 8x + Sr=a cube.

Also 81* —8xr+2*— 1 =a cube= (2x — 1)}, say.
Therefore 134% = 14%, and x = 1.

112
The numbers are 222, 27.

13 160

27. To find two numbers such that their product minus either
gives a cube.
Let the first be of the form 72x, say 8x, and the second
2+ 1 (since 82° + 8x — 8x =a cube).
Also 84°+8xr—22—1 must be a cube, “ which is impossible.”

! Diophantus means that, if we are to get rid of the third power and the absolute
term, we can only put the expression equal to (2x - 1)% which gives a negative and
therefore “impossible ” value for x. But the equation is not really impossible, for we can
get rid of the terms in 23 and x? by putting
1727
13752’

or we can make the term in x and the absolute term disappear by putting

 

3
83+ 8x—u?-1 = (2-2) , whence x=

3
8x3 48x — x21 =(%- 1) , whence 2=t1d
3 296

Diophantus has actually shown us how to do the former in IV. 25 just preceding.
BOOK IV 187

Accordingly we assume for the first number an expression
of the form mix + 1, say 8x + 1, and for the second
number #2 (since 82% + 2* —x* = a cube).

Also 82% +2? — 8x — 1 =a cube = (2x — 1) say.

Therefore x = 14,

13° 160"

23. To find two numbers such that their product * their
sum gives a cube.

Assume the first cube (product + sum) to be 64, and the
second (product — sum) to be 8.

Therefore twice sum of numbers =64 —8 = 56, and the
sum =28, while the product + the sum = 64; therefore
the product = 36.

Therefore we have to find two numbers such that their
sum is 28 and their product 36. If 14 +x, 14 —x are
the numbers, we have 106 — 2% = 36, or 2% = 160; and,
if 160 were a square, we should have a rational
solution.

Now 160 arises from 14-36, and 14=31.28=1.56
=} (difference of two cubes); also 36 =14 (sum of
the cubes).

Therefore we have to find two cubes such that

(1 of their difference)? — their sum = a square.

Let the sides of the cubes be (z + 1), (2 — 1);

therefore } of difference = 132? + 4, and the square of this
is 2} + 1422 + §;

3 the sum of the cubes is 2* + 32;

therefore 2} 24+ 112* + }— 2 — 32 = a square,

or 9z* + 65% + 1 — 42° — 122 =a square =(32* + 1 — 62)’, say;

whence 324 = 362% and z=}.

The sides of the cubes are therefore 1Z, {, and the cubes
1d st

Put now product of numbers + their sum = 4843, and pro-
duct — sum = ¢15.

Therefore their sum = 2436 and their product = 245%,

Now let the first number = x + half sum =x + 1228,

and the second = half sum — x = 1228 — x;

and the numbers are

15079 — 2457
therefore 15OTO84 _ 42 = 2480
and 2621442% = 250000.

TCL 1 ag
188 THE ARITHMETICA

= 500
Therefore x = 293,

and (Zz, = is a solution,

Otherwise thus.

If any square number is divided into two parts one of
which is its side, the product of the parts added to
their sum gives a cube.

[That is, x (x*— 2) + 22 — x + x =a cube.)

Let the square be x? and be divided into the parts x, 22—x.

Then, by the second condition of the problem,

2 — 28 — 2% = 1 — 222 =a cube (less than 2%) =(} x), say.

Therefore 84° — 1622 = 4% so that x= 1£,

and (%, a4) is a solution.
7° 49
29. To find four square numbers such that their sum added to
the sum of their sides makes a given number. :
Given number 12.
Now 2? +x + } =a square.
Therefore the sum of four squares + the sum of their sides
+ I =the sum of four other squares=13, by hypothesis.
Therefore we have to divide 13 into four squares; then, if
we subtract { from each of their sides, we shall have
the sides of the required squares.

! On this problem Bachet observes that Diophantus appears to assume, here
and in some problems of Book V., that any number not itself a square is the sum of
two or three or four squares. He adds that he has verified this statement for all
numbers up to 325, but would like to see a scientific proof of the theorem. These
remarks of Bachet’s are the occasion for another of Fermat’s famous notes: “I have
been the first to discover a most beautiful theorem of the greatest generality, namely this:
Every number is either a triangular number or the sum of two or three triangular
numbers ; every number is a square or the sum of two, three, or four squares; every
number is a pentagonal number or the sum of two, three, four or five pentagonal
numbers; and so on ad Znfinitum, for hexagons, heptagons and any polygons whatever,
the enunciation of this general and wonderful theorem being varied according to the
number of the angles. The proof of it, which depends on many various and abstruse
mysteries of numbers, I cannot give here; for I have decided to devote a separate and
complete work to this matter and thereby to advance arithmetic in this region of inquiry
to an extraordinary extent beyond its ancient and known limits.”

Unfortunately the promised separate work did not appear. The theorem so far as it
relates to squares was first proved by Lagrange (Noww. Mémoires de I’ Acad. de Berlin,
année 1770, Berlin 1772, pp. 123-133; Oeuvres, 111. pp. 189-201), who followed up
results obtained by Euler. Cf. also Legendre, Zaklentheorie, tr. Maser, 1. pp. 212 sqq.
Lagrange’s proof is set out as shortly as possible in Wertheim’s Diophantus, pp. 324-330.
The theorem of Fermat in all its generality was proved by Cauchy (Oeuwres, 11° série,
Vol. VI. pp. 320-353); cf. Legendre, Zaklentheorie, tr. Maser, II. pp. 332 sqq.
BOOK IV 189

Now 13=4 +0 = ($4 +38) + (4 +5),
and the sides of the required squares are 1}, 7%, 18, 13,

121 49 361 169

the squares themselves being I

30. To find four squares such that their sum 7znus the sum of
their sides is a given number.

Given number 4.

Now 2? —x + } = a square.

Therefore (the sum of four squares) — (sum of their sides)
+ 1 =the sum of four other squares=5, by hypothesis.

Divide 5 into four squares, as %, 3¢, $4, $¢.

The sides of these squares plus § in each case are the sides
of the required squares.

Therefore sides of required squares are 1, 13 21 17,

I 169 441 289

I2
and the squares themselves =—, 2 3 ;
100’ 100’ 100’ 100

31. To divide unity into two parts such that, if given numbers
are added to them respectively, the product of the two sums gives

a square.

Let 3, 5 be the numbers to be added; #, 1 —x the parts of I.
Therefore (x + 3)(6 — x) = 18 + 3x — 2? =a square = 44? say;
thus 18 + 32 = 54% which does not give a rational result.
Now 5 comes from a square + I; and, in order that the
equation may have a rational solution, we must sub-
stitute for the square taken (4) a square such that
(the square + 1). 18 + (3)? = a square.

Put (m* + 1)18 + 2} =a square,
or 72m? + 81 =a square = (8m + 9)? say,
and w= 18 mir=i324,

Hence we must put
*+3)(6—-2)=18 + 3r—2*= 3242"
Therefore! 325242 — 3x — 18 =0,
z=dh=d,"

and (= 2) is a solution.
25° 25

Otherwise thus.

The numbers to be added being 3, 5, assume the first of
the two parts to be x — 3; the second is then 4 — x.

Therefore x(9 — x) = a square = 44%, say,

and r=3.

But I cannot take 3 from £, and x must be > 3 and < 4.

! Observe the solution of a mixed quadratic equation.
190 THE ARITHMETICA

Now the value of x comes from g/(a square + 1), and, since
x > 3, this square + 1 should be< 3, so that the square
must be less than 2; but, since x <4, the square + 1
must be > §, so that the square must be > §.

Therefore I must find a square lying between £ and 2, or
between £9 and 128.

100 or 25 2 sokiaies the nition,

150
Put now x (9 —x)=232%
therefore x=
and (Z. 2) is a solution.
41° 41

32. To divide a given number into three parts such that
the product of the first and second + the third gives a square.
Given number 6.
Suppose third part =x, second = any number less than 6,
say 2; therefore first part =4— x.
The two remaining conditions require that 8 —2x +t r=a
square,
3—-» ;
x os are both squares. [Double-equation.]
This does not give a rational result (“is not rational”), since
the ratio of the coefficients of x is not a ratio of a square
lo a square.
But the coefficients of + are 2 — 1 and 2 + 1; therefore we
must find a number y to replace 2 such i
(7 + 1)/(¥ — 1) =ratio of Spe to A 4, say.
Therefore y+ 1 =4y - 4 and p=
Put now second part = §; Rs first=13 — x.

OF

Therefore 6p — 8x + x =a square.

That is, Ohm | are both squares,
65 — 24x

or Sk iy are both squares.
65 — 24x

The difference = 195 = 15. 13;
we put therefore } (15 —13)*=65 — 24x, and r=§.

Therefore the required parts are (2. 3 2)

1 Fermat observes: ‘‘The following is an easier method of solution. Divide the
number 6 into two in any manner, e.g. into 5 and 1. Divide their product less 1, that is

4, by 6, the given number: the result is 2 Subtracting this first from 5 and then from 1,
BOOK 1V 191

33. To find two numbers such that the first with a fraction
of the second is to the remainder of the second in a given ratio,
and also the second with the same fraction of the first is to the
remainder of the first in a given ratio.

Let the first with the fraction of the second = 3 times the
remainder of the second, and the second with the
same fraction of the first =§ times the remainder of
the first.

[The fraction may be either an aliquot part or not, To
avto pépos or Ta avra uépn as Diophantus says,
following the ordinary definition of those terms (“the
same. part” or “the same parts?): cf. Euclid VII.
Defl 3,4.]

Let the second =x + 1, and let the part of it received by
the first=1;

therefore the first= 3x — 1 (since 34 — 1 + I = 3x).

Since the second plus the fraction of the first = 5 times
the remainder of the first,

the second + the first = 6 times the remainder of the fiz t.

And first + second = 4x; therefore remainder of first
= 2% and hence the second receives from the first
3xr—1—3%x or fr—1.

We have therefore to secure that jxr—1 is the same
fraction of 34 — 1 that I is of x + I.

This requires that (jx —1)(x+1)=(3x—1).1;

therefore fa?+4xr—1=3xr—1,and r=4.

Accordingly the numbers are §, 42; and I is {5 of the
second.

 

; I 1 ;
we have as remainders 5 and 5 which are the first two parts of the number to be

divided; the third is therefore 3

That is, if £ », { be the required parts of the number #, Fermat divides a into two
parts x, a — x and then puts
x{a-x)-1_ a*+1
En

E=x- 3
aa lum y loom
PSR
2(ax—-x2-1) _2 {r(a-x)-1}
a

 

whence §=a—(E+9)=

The three general expressions in x satisfy the conditions, and x may be given any
value <a.
192 THE ARITHMETICA

Multiply by 7 and the numbers are 8, 12, and the fraction
is £5; but 8 is not divisible by 12: so multiply by 3,
and (24, 36) is a solution.

Lemma to the next problem.

To find two numbers indeterminately such that their product
together with their sum is a given number.

Given number 8.

Assume the first number to be z, he second 3.

Therefore 3x +x + 3 =given number =38; x=%, and the
numbers are £, 3.

Now 2 arises from (8 — 3)/(3 + 1), where 3 is the assumed
second number.

We may accordingly put for the second number (instead
of 3) any (undetermined) number whatever!; then,
substituting this for 3 in the above expression, we
have the corresponding first number.

For example, we may take x — 1 for the second number;

the first is then 9 — x divided by x, or 2 — IL

34. To find three numbers such that the product of any two
together with the sum of those two makes a given number?

1 The Greek phrase is éav dpa Tdfwper Tov B% °7 olovdimore (olovdfmore s in Lemma
to 1v. 36), “If we make the second’ (literally ‘‘ pus the second az”) ‘any s whatever.”
But the s is not here, as it is in the Lemma to 1v. 36, the actual x of the problem, for
Diophantus goes on to say ‘‘ E.g. let the second be x — 1.” In the Lemma to 1V. 34 the
corresponding expression is ‘‘any quantity whatever” (0covdjmore without s). The
present Lemma amounts to saying that, if xy +x +y=a, then x=(a-y)/ (y+ 1).

2 This determinate set of equations can of course be solved, with our notation, by
a simple substitution.

The equations yi+y+z=a
x+zt+x=0
a
(r+1) c+) =a+1, Rei,
are equivalent to (241) (x+1)=b+1, or {E=o0+1,
(x+1)(y+1)=c+1, p=c+1,
where E=x-h, g=yt1, =z 41.
The solution is E=at r= (Grr) arhle etc.
a+I

In order that the result may be rational, it is only necessary that (a+ 1) (641) (c+ 1)
should be a square ; it is not necessary that eack of the expressions a+1, é+1, c+1
should be a square, as Diophantus says.
BOOK IV 193

Necessary condition. Each number must be 1 less than some
square.

Let (product + sum) of first and second 8.
; 3 second and third = 18.
” third and first =74.
Ppthofnst evasion if we divide (8 —second)by (second +1),
we have the first number.
Let the second number be x — 1.

Therefore the first= =r ? 1.

-

Similarly the third number = 1,

The third equation remains, which gives

I
Hh I =24, and xr =12,
The numbers are 33, 7 © o
Zz) 3) 12’
: ; 165 84 340
or, when reduced to a common denominator, a Bo) aa

Lemma to the following problem.

To find two numbers indeterminately such that their product
minus their sum is a given number.
Given number 8.
First number x, second 3, suppose; therefore
(product) — (sum) = 3x —x—3=2vr—3=8, and r=5}.
The first number is therefore 51, the second 3.
But 54 comes from (8 + 3)/(3 — 1), and we may put for 3
any number whatever.
Eg put the second number =x + 1; the first is then +9

divided by z, or 1 +2.

35. To find three numbers such that the product of any two
minus the sum of those two is a given number?

Necessary condition. Each of the given numbers must be 1 less
than some square?

Let (product — sum) of first and second = 8.
’ IY) I) second and third = I5.
” 1 4 third and first = 24,

1 See last paragraph of preceding note.
2 The notes to 1V. 34 above apply, mutatis mutandis, to this problem as well.

H. D. 13
194 THE ARITHMETICA

By the first equation, if we divide (8 + second) by
(second — 1), we have the first number.

Assuming x + 1 for the second number, we have 1 +2

for the first.
a Bia : os
Similarly 2 is the third number, and two conditions
are satisfied.

The third gives the — 1=24, and r=12,

The numbers are 5 2 2
285 204 460
or, with a common denominator, 2, “=, ¢=.

Lemma to the following problem.

To find two numbers indeterminately such that their product
has to their sum a given ratio.

Let the given ratic be 3:1, the first aumber #, the
second 35.

Therefore sx =3(5 +2), x = 74; and the numbers are
7% 5.

But 74 arises from 15 divided by 2, while the 15 is the
second number multiplied by the given ratio, and
the 2 is the excess of the second number over the
ratio.

Putting therefore x (instead of 5) for the second number,
we have, for the first number, 3x divided by x — 3.

The numbers are therefore 3x/(x — 3), x.

36. To find three numbers such that the product of any two
bears to the sum of those two a given ratio.

Let product of first and second be 3 times their sum.

¥ 5 second and third be 4 times their sum.

: . third and first be 5 times their sum.
Let second number be x; the first is therefore 34/(x — 3),

by the Lemma, and similarly the third is 44/(x — 4).
Lastly gd =5 ( 37 J %
r—3'xr—4 r—3 x—4

or [2x2 = 352° — 120%,
Therefore x=120,

 

and the numbers are i 120, 480
51 23’ 28°
BOOK 1V 195

37. To find three numbers such that the product of any two
has to the sum of the three a given ratio?

Let product of first and second = 3 times sum of the three,

” » of second and third =4 . 5
% of third and first = 5 ”

First seek three numbers such that the product of any two
has to an arbitrary number (say 5) the given ratio.

Then product of first and second = 15; and, if x be the
second, the first is 15/z.

The product of second and third =20; therefore third
= 20/%.

It follows that 20. 15/22= 23.

And, if the ratio of 20. 15 to 25 were that of a square to a
square, the problem would be solved.

Now 15 =3.5, and 20 is 4. 5, the 3 and 4 being fixed by
the original hypothesis, but § being an arbitrary
number.

We must therefore find a number #2 (to replace 5) such
that 12/51 = ratio of a square to a square.

Thus 1272. 5m =601* = a square = goon, say; and m=135.

Let then the sum of the three numbers be 15.

Product of first and second is therefore 45, and first = 45/x.

Similarly third = 60/x.

Therefore 45 00/2 = 7%, and x =0,

Therefore the numbers are 73, 6, 10, and the sum of
these = 234.

Now, if this sum were 15 instead, the problem would be
solved.

! Loria (0p. cz¢. p. 130) quotes this problem as an instance of Diophantus’ ingenious
choice of unknowns. Here the equations are, with our notation,

yz=a(x+y+3),

sx=h(x+y+ 32),

ay=c(x+y+3),
and Diophantus chooses as his principal unknown the sum of the three numbers,
x+y +z=w, say.

; c
We may then write x=cw(y, z=awfy, so that szx=acw?®(y*=bw, and »=% Ww.

Putting w=% £2, we have

ac

2 a2 a ow
Zhwhe=a it, y= z=at, x=ck,

from which, by eliminating x, y, 2, we obtain &£ = (bc + ca + ab)/ac.
Hence x= (bc +ca+ab)la, y=(bc+ca+ab)lb, z=(bc+ca+ abd).

13—2
196 THE ARITHMETICA

Put therefore for the sum of the three numbers 152% and
for the numbers themselves 74x, 6x, 10x.
Therefore 2342 = 152% so that xr =4F,
and 3523 ey 479 is a solution.
20:%.30. 30
38. To find three numbers such that their sum multiplied
into the first gives a triangular number, their sum multiplied into
the second a square, and their sum multiplied into the third a cube.
Let the sum be 12, and let the numbers be 72/22, n/a? p/2*,
where 72, n, p are a triangular number, a square and
a cube respectively;
say first number = 6/2? second 4/4 third 8/22
But the sum is 22; therefore 18/2 = 2% or 18 = a*
Therefore we must replace 18 by some fourth power.
But 18 = sum of a triangular number, a square and a cube.
Let 2* be the required fourth power, which must therefore be
the sum of a triangular number, a square and a cube.
Let the square be 2* — 242 +1;
therefore the triangular number + the cube = 22% — 1.
Let the cube be 8; therefore the triangular number is
22: — 0.
But 8 times a triangular number + 1 = a square ; therefore
162% — 71 = a square = (4x — 1), say; thus z=g, the
triangular number is 153, the square 6400 and the
cube 8.
Assume then as the numbers 153/2?% 6400/2? 8/x%
Therefore 6561/2? = 22, or x*=6561, and x=.
Thus (E&, 80 =) is a solution.

1 The procedure may be shown more generally thus.
Let &, 7, ¢ be the required numbers; suppose
+n =a
a(att 2 ;
pas)

2x2 7? x2’

 

and t= 2
It follows that xi= gla +8245.

Suppose now that f=a?— 2% [Diophantus and Bachet assume z=r1].
Then etl i -t 4A

Eight times the left hand side plus 1 gives a square (by the property of triangular
numbers) ; that is,
(2a + 1)2=165%x — 82* — 8y3 + 1 =a square
= (42x — £)?, say,
BOOK :1V 197

39. To find three numbers such that the difference of the
greatest and the middle has to the difference of the middle and the
least a given ratio, and also the sum of any two is a square.

Ratio 3:1.) Since middle number + least = square, let the
square be 4.

Therefore middle > 2, let it be x + 2, so that least=2 x

Therefore difference of greatest and middle = 6x, whence
the greatest= 7x42.

Therefore bik gd are both Squares, [Double-equation.)

Take the difference 2x, split it into factors, say 7, 4, and
proceed by the rule; therefore x=112,

But I cannot take 112 from 2; therefore x must be found
to be < 2, so that 6x + 4< 16.

Thus there are to be three squares 8x44, 6x +4 and 4
(the 4 arising from 2.2), and the difference of the
greatest and middle is } of the difference of the
middle and least.

We have therefore to find three squares having this property
and such that the least = 4 and the middle < 16.

Let side of middle square be z+ 2; therefore excess of
middle over least = 2% + 42, whence excess of greatest
over middle = }#* I 142, and therefore the greatest

=1§2" + 555 +
This must be a square; therefore, le by 9, we have
125% + 485 + 36 = a square,

 

LSS
842

; ; 1
must be integral, and therefore a integral, so that S (4sr —4 — 1) must be

 

whence

But iat) 7)
2

8z? Ret By? zs 2
a

" Bachet assumes that it is necessary, with Diophantus, to take Z=1, observing that

trial will show that the problem can hardly be solved otherwise. On this Fermat remarks

that Bachet’s trial had not been carried far enough. We may, he says, put for 43 any

cube, for instance, with side of the form 322+ 1. Suppose, for example, we take 75.

Then [z being 1] we have to make

integral ; that is, must be integral.

2x? — 344 a triangle,

and therefore 1642-2751 a square, and we may take, if we please, 4x — 3 as the side of
this square [so that 4 is in this case 3].

By varying the cubes we may use an unlimited variety of odd numbers, besides 3,

as values for 2 which will satisfy the required condition.

Loria (0p. czt. p. 138) points out that the problem could have been more simply
solved by substituting x fcr x2 and z for 22 in the above assumptions. The ultimate
expression to be made a square would then have been 162x — 822 — 843 + 1, and we could
have equated this to A%, thus finding x.
198

THE ARITHMETICA

or 32+ 122 + 9 = a square = (ms — 3)? say.

It follows that z= (6772 + 12)/(77:2 — 3), which must be < 2.

Therefore 6m + 12 < 20 — 6, of 22 >6w +18.

“When we solve such an equation?, we multiply half the
coefficient of xinto itself—this gives 9—then multiply :
the coefficient of xr into the units—2. 18 =36—2add
this last number to the 9, making 45, and take the
side [square root] of 45, which is not less than 71
add half the coefficient of x—making a number not

                                             

of 22; the résult is not less than 5.”

[3+ 18.2=45, and }4/45.4+ 3 is not less than 3 + Z.]

We may therefore put m= $+1, or 5, and we thus have

3° +122 +9= (3 320%

Therefore z= 3}, and the side of the middle square is
43, the square itself being LEY.

Turning to the original problem, we put 6x + 4 = 18 and
x = 1385 which 4s less than 2.

 

The greatest of the required numbers = 7x + 2 = a
i = <2
the middle=x +2 = 226"
==
andtheleast = 2 — r= 6

The denominator not being a square, we can make it
a square by dividing out by 6; the result is
1834% 460% 144
eT
or again, to avoid the } in the numerators, we may

multiply numerators, and denominators, by 4; thus

733% 1873 58 ist
L220 0 22 is a solution?
484 ' 484° 484

! I have quoted Diophantus’ exact words here, with the few added by Tannery,
“making a number not less than ro...coefficient of 42,” in order to show the precise
rule by which Diophantus solved a complete quadratic. -
When he says A/45 is not less than 7, Diophantus is not seeking exact limits. Since
n/45 is between 6 and 7 we cannot take a smaller integral value than 7 in order to
satisfy the conditions of he problem (cf. p. 65 above).
2 A note in the Znventum Novum (Part 11, paragraph 26) remarks upon the prolix and
involved character of Diophantus’ solution and gives a shorter alternative. The problem

is to solve

E-n=m(n-§, (>> and m=3, say)
n+ $=’ i

$+E=2

E+ p=w2
BOOK IV 199

40. To find three numbers such that the difference of the
squares of the greatest and the middle numbers has to the differ-
ence of the middle and the least a given ratio, and also the sum of
each pair is a square,

Ratio 3 11.
Let greatest + middle number = the square 1642; therefore
greatest> 8x2: letitbe 32° 2: hence middle=322—3.
And, since greatest + middle > greatest + least,
162% > (greatest + least) > 84%;
let greatest + least = ga?, say; therefore least =12 — 2,
Now difference of squares of greatest and middle = 6442,
and difference of middle and least = 742
But 64 is not equal to. 3.7 or 21.
Now 64 comes from 32.2; therefore we must find a
number 72 (in place of 2) such that 32m = 21.
Therefore m = 4.
Assume now greatest number = 84% + 24, middle = 842 — 24,
least = 22 — 21.
[And difference of squares of greatest and middle
=212= 3. 74°]
The only condition left is: middle + least = square; that is,
9x* — 42 = a square = (3x — 6), say.
Therefore x= 3%,

(3009000 2633544 138681
331776 ° 331776 ° 331776

 

 

) is a solution.

Take an arbitrary square number, say 4, for the sum of #, ¢; suppose 2 +x =n, 2 —x =,
so that 5 -- {=2x; therefore £{ —n=3(n- {)=06x, whence {=2+ 7.x.
The last two conditions require that

a shall be squares.
+ .

Replace xv by syst 3. This will make 4+ 6x a square. It remains that

+8,41 ran square
4 > 3

zz it , say.
1 y.
5 16
Thus y(B-1=(2-5)

16 I, n

and Ams so that x=— y* += y=-—.

196 zo 00
121 121) 121

 

The numbers are therefore

~
200 THE ARITHMETICA

BOOK V

i. To find three numbers in geometrical progression such that
each of them muznus a given number gives a square.
Given number 12.
Find a square which exceeds 12 by a square. “This is
easy [IL 10]; 42} is such a number.”
Let the first number be 42%, the third 22; therefore the
middle number = 64x.
212
Therefore are both squares;
61x — 12
their difference = 2? — 63x = x (x — 6%); half the difference
of the factors multiplied into itself = 182; therefore,
putting 63x — 12 =182, we have x=381,

2346}% a) is a solution.
104 ’ 10816

 

and (421,

2. To find three numbers in geometrical progression such that
each of them when added to a given number gives a square.

Given number 20.

Take a square which when added to 20 gives a square,
say 16.

Put for one of the extremes 16, and for the other 22 so
that the middle term = 4x.

Therefore 2 ban

4% + 20

Their difference is > — 4x =x (x — 4), and the usual method
gives 4x +20=4, which is absurd, because the 4
ought to be some number greater than 20.

But the 4 =} (16), while the 16 is a square which when
added to 20 makes a square; therefore, to replace 16,
we must find some square greater than 4.20 and
such that when increased by 20 it makes a square.

Now 81 > 80; therefore, putting (72 + 9)? for the required
square, we have

| are both squares.

(m + 9)* + 20 = square = (m7 — 11)? say;
therefore 7 = 4, and the square = (94)? = go}.
BOOK V 201

Assume now for the numbers go}, 9}, 2% and we have
2% 4 20
9%x + 20
The difference =x (r —9}), and we put gir + 20=3§L.
Therefore # = 4%, and

389% : :
(01, Ee solution.

| both squares.

 

 

3. Given one number, to find three others such that any one of
them, or the product of any two of them, when added to the given
number, gives a square.

Given number 5.

“We have it in the Porisms that if, of two numbers, each,
as well as their product, when added to one and the
same given number, severally make squares, the
two numbers are obtained from the squares of con-
secutive numbers.”

Take then the squares (x + 3)% (x + 4)%, and, subtracting
the given number 5 from each, put for the first
number #*4+6x +4, and for the second +*+8x+11,
and let the third? be twice their sum 7zznus 1, or

4x + 28x + 29.

1 On this Porism, see pp. 99, 100 ante. :
2 The Porism states that, if a be the given number, the numbers x*-a, (x+1)® -a

satisfy the conditions.
In fact, their product + a={x (x +1)}2-a (24% + 2x +1) + a* +a
={x(@+1)}2-2ax (x+1)+a’={x (x +1) - a}?

Diophantus here adds, without explanation, that, if .X, ¥ denote the above two numbers,
we should assume for the third required number Z=2 (X + ¥) — 1. We want #47e¢ numbers
such that any two satisfy the same conditions as X, ¥. Diophantus takes for the third
Z=2(X+ Y)-1 because, as is easily seen, with this assumption two out of the three
additional conditions are thereby satisfied.

For Z=2(X+Y)-1=2(20*t20+1)—4a-1
=(2x+1)?- 1a;
therefore XZ+a=a%(2x+1)?—a{(2x+1)>+ 4a0?} + 4% +a

=a? (2x+1)2—a. 4x (200+ 1) + 40%
= {x (2x +1) — 2a}?
while YZ+a=(x+1)2(ex+1)2-a{(2x+1)2+4 (vr +1)%} +4a* +a
=(x+1)2(2x+1)2— a (84% + 12x + 4) + 4a°
={(x+1) (2x +1) — 2a}>
The only condition remaining is then
Z+a=a square,
or (2x + 1)% - 3a =a square = (2x — £)?, say,
and x is found.
Cf. pp. 100, 104 above.
202 THE ARUIHMETICA

Therefore 422+ 28x + 34 =a square = (2x — 6), say.

od 2861 7645 2) y Sil
Hence x =, and (=, 2, So) Isa solution.

4. Given one number, to find three others such that any one
of them, or the product of any two, minus the given number gives
a square.

Given number 6.

Take two consecutive squares 2%, x2 + 2x4 + I.

Adding 6 to each, we assume for the first number 22 + 6,
and for the second 22+ 2x + 7.

For the third? we take twice the sum of the first and
second minus 1, or 42° + 4x + 25.

Therefore third minus 6 = 42° + 4x + 19=square =(2x—6)?
say. ;

Therefore x = 1%,

and (0, 7, ay is a solution.

[The same Porism is assumed as in the preceding problem
but with a minus instead of a plus. Cf. p. 99 above.]

 

5. To find three squares such that the product of any two
added to the sum of those two, or to the remaining square, gives
a square.

“We have it in the Porisms” that, if the squares on any two
consecutive numbers be taken, and a third number
be also taken which exceeds twice the sum of the
squares by 2, we have three numbers such that the
product of any two added to those two or to the
remaining number gives a square?

! Diophantus having solved the problem of finding three numbers £, 7, { satisfying
the six equations
Eta=7 n¢+a=au?,
n+e=s2, (Et+a=7d,
+a=122, i +a=1w?
Fermat observes that we can deduce the solution of the problem

Zo find four numbers such that the product of any pair added to a given number
produces a square.

Taking three numbers, as found by Diophantus, satisfying the above six conditions,
we take x + 1 as the fourth number. We then have three conditions which remain to be
satisfied. These give a ‘‘ triple-equation ” to be solved by Fermat’s method.

? Diophantus makes this assumption for the same reason as in the last problem, v. 3.
The second note on p. 201 covers this case if we substitute — a for a Broughout.

3 On this Porism, see pp. 100-1 ante.
BOOK V 203

Assume as the first square 22+ 2x + 1, and as the second
22+ 42 + 4, so that third number =44% + 122 + 12.
Therefore 2° + 3x + 3 =a square = (x — 3), say, and x = }.
Therefore (%, 54) 196) is a solution.
9:9 9
6. To find three numbers such that each minus 2 gives a
square, and the product of any two minus the sum of those two,
or minus the remaining number, gives a square.
Add 2 to each of three numbers found as in the Porism
quoted in the preceding problem.
Let the numbers so obtained be 2242, 22+2x+43,
42° + 4x + 6.
All the conditions are now satisfied}, except one, which
gives
4x*+ 4x + 6 — 2 = a square.
Divide by 4, and #2 + x + I = a square = (x — 2)? say.
Therefore x =§,

59 l¢ 26) i ;
and (2. Tn is a solution.

Lemma I to the following problem.

To find two numbers such that their product added to the
squares of both gives a square.
Suppose first number x, second any number (72), say I.
Therefore x. 1 + 2+ 1 =2*+x + 1 = a square = (¥ — 2)’, say.
Thus x=42, and |

. 1) is a solution, or (3, 5).

Lenvina 11 to the following problem.
To find three right-angled triangles (Ze. three right-angled
triangles in rational numbers®) which have equal areas.
We must first find two numbers such that their product
+ the sum of their squares = a square, eg. 3, 5,as in
the preceding problem.

! The numbers are x2+2, (x+1)2+2, 2 {x?+ (x + 1)24+ 1} +2; and if X, ¥, Z denote
these numbers respectively, it is easily verified that
XY (X+ YV)= (22+ xt1)?, XYZ =(x2+x)3
XZ (X+2Z) =(ax2+x+2)?2 XZ-Y=(2x+tx+3),
and YZ ~(Y+2Z) =(@a2+3x+3), YZ-X=(2x2+30+4)2
2 All Diophantus’ right-angled triangles must be understood to be right-angled
triangles with sides expressible in rational numbers, In future I shall say ‘‘right-angled
triangle” simply, for brevity,
204 THE ARITHMETICA

Now form right-angled triangles from the pairs of
numbers!

(7,3), (7.5) (J, 5+)
[Z¢. the right-angled triangles (72+ 32, 72 — 3% 2.7. 3), etc).
The triangles are (40, 42, 53), (24, 70, 74); (15, 112, 113),
the area of each being 840.

1 Diophantus here tacitly assumes that, if ab+a2+42=c% and right-angled triangles
be formed from (c, a), (¢, 6) and (¢, a+) respectively, their areas are equal. The
areas are of course (c2—a?) ca, (2-62)ch and {(a+6)>—c*} (@+0)¢, and it is easy to
see that each =abc (a+b).

Nesselmann suggests that Diophantus discovered the property as follows. Let the
triangles formed from (7, mz), (g, 7), (», m2) have their areas equal ; therefore

1 (m2 — n2)=g (m2 — g2) =r (r2 — m2).

It follows, first, since mn — n3=m?g — ¢4,
that m2=(n3— ¢3)[ (12 — g) = 12+ ng +g
Again, given (¢, m2, 72), to find 7.
We have : g(mB- gd) =r (12 - m2),
and 72 — g2=72%+ ng, from above ;
therefore g (7+ ng)=r(r2- 12 — ng —¢?),
or q (122+ nr) +? (n+7)=7r (2 -n?).
Dividing by #»+ 72, we have gn+g*=r2-rn;
therefore (g+7)n=r*-g?
and r=g+n.

Fermat observes that, given any rational right-angled triangle, say z, é, ¢, where z
is the hypotenuse, it is possible to find an infinite number of other rational right-angled
triangles having the same area. Form a right-angled triangle from 2%, 24d; this gives
the triangle z*+ 462d% 2*— 40%@?%, 43%bd. Divide each of these sides by 2z(4*-d?),
b being >d ; and we have a triangle with the same area (Led as the original triangle.

Trying this method with Diophantus’ first triangle (40, 42, 58), we obtain as the new
triangle 1412881 1412880 1681

 

 

“180? “vi%p 7 189°
The method gives (LZ, ee. re as a right-angled triangle with area equal

to that of (3, 4, 5)

Another method of finding other rational right-angled triangles having the same area as
a given right-angled triangle is explained in the Znwentum Novum, Part 1, paragraph 38
(Euwres de Fermat, 111. p. 348).

Let the given triangle be 3, 4, 5, so that it is required to find a new rational right-
angled triangle with area 6.

Let 3, x+ 4 be the perpendicular sides ; therefore

the square of the hypotenuse =a%+ 8.x + 25=a square.

3
2

Again, the area is 2 x +6 ; and, as this is to be 6, it must be six times a certain square ;

that is, 3x46 divided by 6 must be a square, and this again multiplied by 36 must

be a square; therefore :
9x + 36 =a square.
Accordingly we have to solve the double-equation
x24 8x +25=u?
9x + 36= a ’
BOOK V 205

7. To find three numbers such that the square of any one +
the sum of the three gives a square.
Since, in a right-angled triangle,
(hypotenuse)? + twice product of perps. = a square,
we make the three required numbers hypotenuses and the
sum of the three four times the area.
Therefore we must find three right-angled triangles having
the same area, ¢.¢., as in the preceding problem,
(40, 42, 58), (24, 70, 74), (15, [12, 113).
Reverting to the substantive problem, we put for the
numbers 18%, 74x, 1132; their sum 2452 = four times
the area of any one of the triangles = 336042
Therefore x=,

and (2 518 791) is a solution.
g6’ 96° 9b
Lemma to the following problem.

Given three squares, it is possible to find three numbers such
that the products of the three pairs shall be respectively equal to
those squares.

 

 

This gives x= - Oyepave )
24050601
whence = yl .
3

The triangle is thus found to be

2896804 7776485
2405601’ 2405601
724201 851

A , the root of which is ——.
2405601 1551

 

9

 

The area is 6 times a certain square, namely

851
1551
4653 3404 77706485
851° 1331 1319901"

49

2d giving CR leads to

 

Dividing each of the above sides by , we obtain a triangle with area 6, namely

Another solution of the double-equation, x= —

Li 70 4 120} 1201

the same triangle (%. gi gs
The method of the Znwventum Novum has a feature in common with the procedure in
the ancient Greek problem reproduced and commented on by Heiberg and Zeuthen
(Bibliotheca Mathematica, Viiiz, 1907/8, p. 122), where it is required to find a rational
right-angled triangle, having given the area, 5 feet, and where the 5 is multiplied by a
square number containing 6 as a factor and such that the product * can form the area of
a right-angled triangle.” 36 is taken and the area becomes 180, which is the area of
(9, 40, 41). The sides of the latter triangle are then divided by 6, and we have the

required triangle (cf. p. 119, ante).

) as that obtained by Fermat’s rule (see above).
206 THE ARITHMETICA

Squares 4, 9, 16.

First number x, so that the others are 4/x,9/x; and 36/x*=16.

Therefore x=§, and the numbers are (14, 23, 6).

We observe that x=, where 6 is the product of 2 and 3,
and 4 is the side of 16.

Hence the following rule. Take the product of two sides
(2, 3), divide by the side of the third square 4 [the
result is the first number]; divide 4, 9 respectively
by the result, and we have the second and third
numbers.

8. To find three numbers such that the product of any two +
the sum of the three gives a square.

As in Lemma II to the 7th problem, we find three right-
angled triangles with equal areas; the squares of
their hypotenuses are 3364, 5476, 12769.

Now find, as in the last Lemma, three numbers such that
the products of the three pairs are equal to these
squares respectively, which we take because each
+ 4. (area) or 3360 gives a square ; the three numbers
then are

4292 y 321Ty [280182y Tannery],

4181 [618788x Tannery].

It remains that the sum of the three = ar

Therefore 32824806 1 [13120902247 Tannery] = 33602

Therefore x = JAE [1dYdsess6s OF tovssss 1annery],

781543 781543 781543
255380’ 109520° 67280

 

 

 

[and the numbers are

9. To divide unity into two parts such that, if the same given
number be added to either part, the result will be a square.

Necessary condition. The given number must not be odd and
the double of it + I must not be divisible by any prime number
which, when increased by 1, is divisible by 4 [Z.e. any prime number
of the form 47 — 1]

Given number 6. Therefore 13 must be divided into two
squares each of which >6. If then we divide 13 into
two squares the difference of which < 1, we solve the
problem.

1 For a discussion of the text of this condition see pp. 107-8, ante.
BOOK V 207

Take half of 13 or 6}, and we have to add to 6% a small
fraction which will make it a square,

or, multiplying by 4, we have to make = + 26 a square,

i.e. 26x + 1 = a square = (5x + 1)? say, whence x= 10.

That is, in order to make 26 a square, we must add 13, or,
to make 6% a square, we must add 35, and

bo + 64 = (33)

Therefore we must divide 13 into two squares such that their
sides may be as nearly as possible equal to 5%. [This
is the mapiaéryTos aywyn described above, pp. 95-8.]

Now 13=22+4 3% Therefore we seek two numbers such
that 3 minus the first = 31, so that the first = of, and
2 plus the second = 41, so that the second = 11.

We write accordingly (11x + 2)? (3 — 9x)? for the required
squares [substituting x for J].

The sum =2024*— 10x + 13 =13.

Therefore x= 331, and the sides are 251, 238,

Subtracting 6 from the squares of each, we have, as the
parts of unity,

4843 5358
10201’ 10201"

 

10. To divide unity into two parts such that, if we add different
given numbers to each, the results will be squares.
Let the numbers! be 2, 6 and let them and the unit be
represented ‘in the figure, where DA =a, AD =],
BE =6, and G is a point in AB so chosen that DG,
GE are both squares.

pilin

10

B E

 

Now DE =9. Therefore we have to divide 9 into two
squares such that one of them lies between 2 and 3.

Let the latter square be 2, so that the other is 9g —xa?
where 3 >>,

Take two squares, one >2, the other <3 [the former
being the smaller], say 28%, 361,

1 Loria (0p. cit. p. 15072.), as well as Nesselmann, observes that Diophantus omits to
state the necessary condition, namely that the sum of the two given numbers plies 1 must
be the sum of two squares. 3
208 THE ARITHMETICA

Therefore, if we can make 2? lie between these, we shall
solve the problem.
We must have > 1% and < 13.
Hence, in making 9 —2* a square, we must find
x>1% and < 13.
Put 9 — 4? = (3 — mx), say, whence x =6m[(m* + 1).
Therefore 1 « Gi a
12 m3 +1 12
The first inequality gives 72m > 17m*+ 17; and
36:—17.17 = 1007,
the square root of which! is not greater than 31;
: Ll enol.
17 17
Similarly from the inequality 197224 19> 72m we find?
m4 $8.
Let m= 34. Therefore 9 — 22=(3 — 34x), and x=§4.
Therefore x? = 2938,

 

 

therefore 72 $

1438 mam)

and the segments of 1 are (Le, 2809

11. To divide unity into three parts such that, if we add the
same number to each of the parts, the results are all squares.

Necessary condition®. The given number must not be 2 or any
multiple of 8 increased by 2.
Given number 3. Thus 10 is to be divided into three
squares such that each > 3.

Take } of 10, or 31, and find x so that a 33 may be a

square, or 304° + I =a square = (5x + 1), say.
Therefore x=2, 12=4, 1/2°=], and
35 + 33 = 1% = a square.

Therefore we have to divide 10 into three squares each of
which is as near as possible to 12L. [mapiaérnTos
ayy]

Now 10= 3+ 1*= the sum of the three squares 9, 4&, J.

Comparing the sides 3, ¢, 2 with 1%,
or (multiplying by 30) 9o, 24, 18 with 55, we must
make each side approach ;35.

1 Ze. the integral part of the root is +31. The limits taken in each case are a fortiori
limits as explained above, pp. 61-3.

2 See p. 61, ane,

3 See pp. 108-9, ante.
BOOK. V 209

[Since then $3=3—-38=4431=3431] we put for the
sides of the required numbers
3-357 31x+4, 374+4.
The sum of the squares = 3555242 — 1162 + 10 = 10.
Therefore x = #45;
and this solves the problem.

12. To divide unity into three parts such that, if three different
given numbers be added to the parts respectively, the results are
all squares.

Given numbers 2, 3, 4.. Then 1 have to divide 10 into
three squares such that the first > 2, the second > 3,
and the third > 4.

Let us add 4 of unity to each, and we have to find three
squares such that their sum is 10, while the first lies
between 2, 2}, the second between 3, 31, and the
third between 4, 43.

It is necessary, first, to divide 10 (the sum of two
squares) into two squares one of which lies between
2, 24; then, if we subtract 2 from the latter square,
we have one of the required parts of unity.

Next divide the other square into two squares, one of
which lies between 3, 3%;

subtracting 3 from the latter square, we have the second
of the required parts of unity.

Similarly we can find the third part’

_! Diophantus only thus briefly indicates the course of the solution. Wertheim solves
the problem in detail after Diophantus’ manner; and, as this is by no means too easy,
I think it well to reproduce his solution.

I. It is first necessary to divide 10 into two squares one of which lies between 2

and 3. We use the mapigéryros dywy.
The first square must be in the neighbourhood of 24%; and we seek a small fraction

: : I
2 which when added to 2% gives a square: in other words, we must make 4 (22+ 5)
or
: ; 2 ;
a square. This expression may be written 10+ (;) , and, to make this a square, we put

102+ 1=(3y +1)? say,
19

2
2) , which is an approximation

) I 361
whence y=6, y2=136, a%= I44, so that 2% + 25 Ti (

to the first of the two squares the sum of which is 10.

; a
The second of these squares approximates to 7%, and we seek a small fraction — such
ov
To 2\? I\>
that Tht 5 1s a square, or 30+ a =30 + pr , Say =a square.

H. D. 14
210 THE ARITHMETICA

13. To divide a given number into three parts such that the
sum of any two of the parts gives a square.
Given number 10.

 

Put 307° + 1=(57+1)% say;
olrt pun, Rey, Ale tele {iY
therefore y=12, y’=4, 2*=16, so that ni aT = (£2) = (3) ;
Now, since 1o=1%2+ 3% and aril, while Bley,
: 12 12 12 12
we put (1+7x)2+(3 - 3x) =10, [Cf. ¥. of
so that As,
Sh 29

14 43\?_ 1849

al ; A = J

(tgap= (r+ 34 - (4) 0
6 \2 81\2 06501

— 2 _— ={— = —,
(3-37) (5 2 &) 841

Therefore the two squares into which 10 is divided are 2 an nd - and the first of

these lies in fact between 2 and 23.

 

a 6561 : .
II. We have next to divide the square 2” into two squares, one of which, which

841
we will call x2 lies between 3 and 4. [The method of v. 10 is here applicable.]

Instead of 3, 4 take 2, Bs as the limits.

 

 

Therefore 2 <xi< i
or 7 =X = —-
4
And =— Bh — x2 must be a square = z i , say,
, ; 1624
which gives = 29 (1 +5 5
4 has now to be chosen such that :
(1) NE
29 (1 + 4° 37
from which it follows that k<o2 i :
1624 Ly
and (2) 5 oir 7) =
whence £223.10.
We may therefore put %= 225,
Therefore = 16v0 2 2024400 :
841 707281
and 6561 Al 2893401
841 707281

The three required squares into which ro is divided are therefore
1849 2624400 2893401
841° 707281 707281 °
And if we subtract 2 from the first, 3 from the second and 4 from the third, we obtain
as the required parts of unity
140447 592557 64277
707281’ 707281’ 707281"

 

 

 
BOOK V 214

Since the sum of each pair of parts is a square less than
10, while the sum of the three pairs is twice the
sum of the three parts or 20,

we have to divide 20 into three squares each of which
75 < 10,

But 20 is the sum of two squares, 16 and 4 ;

and, if we put 4 for one of the required squares, we
have to divide 16 into two squares, each of which is
< 10, or, in other words, into two squares, one of which
lies between 6 and 10. This we learnt how to do!
[v. 10].

We have, when this is done, three squares such that
each is < 10, while their sum is 20;

and by subtracting each of these squares from 10 we
obtain the parts of 10 required.

14. To divide a given number into four parts such that the
sum of any three gives a square.
Given number 10.
Three times the sum of the parts = the sum of four squares.
Therefore 30 has to be divided into four squares, each of
which is < 10.
(1) If we use the method of approximation (mraptaarys),
we have to make each square approximate to 7%;

1 Wertheim gives a solution in full, thus.
Let the squares be x? 16 — x? of which one, x? lies between 6 and ro.
: va TE
Put instead of 6 and 10 the limits —2 and g, so that
: : 4

Ben
Sash,

To make 16 — x2 a square, we put
16 — x2=(4 — Ax)?

 

T= A
whence ID

Now (1) 3

These tn give, as limits for 4, 2° ot and 2:21...

We may therefore e.g. put 2=2%.

Then r= % alte 16 —x%= 7016 :
29’ 841’ 841

6400 7056

S417 Su’

, and (2) a

 

The required three squares making up 20 are 4,

Subtracting these respectively from 10, we have the required parts of the given number
2010 1334
Bar’ Bur

 

10, namely )

14—2
212 THE ARITHMETICA

then, when the squares are found, we subtract each
from 10, and so find the required parts.

(2) Or, observing that 30=16+9 + 4 + I, we take 4,9 for
two of the squares, and then divide 17 into two squares,
each of which < 10.

If then we divide 17 into two squares, one of which lies
between 8} and 10, as we have learnt how to do!

~~ [cf. v. 10], the squares will satisfy the conditions.

We shall then have divided 30 into four squares, each of
which is less than 10, two of them being 4, 9 and the
other two the parts of 17 just found.

Subtracting each of the four squares from 10, we have the
required parts of 10, two of which are 1 and 6.

15. To find three numbers such that the cube of their sum
added to any one of them gives a cube.
Let the sum be » and the cubes 71%, 262%, 632°,
Therefore obi =x, or gba =1,
But 96 is not a square; we must therefore replace it by a
square in order to solve the problem.

! Wertheim gives a solution of this part of the problem.
2
As usual, we make 83+ » Or 34 + (2) , a square.

Loma IY
Putting z Ey we must make 34 + (2 a square.

Let 34y*+ 1=a square = (6y — 1)?
and we obtain 7=0, 2 =30, +i=144.
2\2
Thus 81 + = (3) ;
¥44 14g \12

and 2 is an approximation to the side of each of the required squares.

Next, si SED 3 rtm 13
ext, since 17=1%+ 4% and oe Fay #=ks
we put 17=(1+ 23x)? + (4 — 13x)2,
and we obtain Ze 2,
349
The squares are then (r+2 apr (L220 sony
4 = 349/ 12181 ’
2 8361
and = x= (200) = L2ge801
{4= 137) 349 121801

Subtracting each of these from ro, we have the third and fourth of the required parts
of 10, namely

185754 179649
121801’ 121801"

 

 
BOOK V 213

Now g6 is the sum of three numbers, each of which is 1 less -
than a cube;

therefore we have to find three numbers such that each
of them is a cube less 1, and the sum of the three
is a square.

Let the sides of the cubes! be 7 + 1, 2 — m, 2, whence the
numbers are wd + In + 3m, 7 — 12m + 6m — md, 7;
their sum = gm? — gm + 14 = a square = (372 — 4)? say ;

therefore m=

5
and the numbers are 338, 18257, 7,
Reverting to the original a we put x for the sum of
the numbers, and for the numbers respectively
103tat, AENe0 oo
whence 4374043 = 7,
that is (if we divide out by 15 and by x),
29162*= 225, and x= 12.
The numbers are therefore found.

16. To find three numbers such that the cube of their sum
minus any one of them gives a cube.

Let the sum be x, and the numbers 2% 384° £347

Therefore 43iT4° =x,

and, if 482% were the ratio of a square to a square, the
problem would be solved.

But 487% = 3 — (the sum of three cubes).

Therefore we must find three cubes, each of which < 1,
and such that (3 — their sum) = a square.

If, a fortiori, the sum of the three cubes is made < 1, the
square will be > 2.. Let?it be 21.

1 If g3, 83, 3 are the three cubes, so that a3+ 43+ ¢3 — 3 has to be a square, Diophantus
chooses ¢3 arbitrarily (8) and then makes such assumptions for the sides of @3, #, being
linear expressions in #2, that, in the expression to be turned into a square, the coefficient
of m3 vanishes, and that of 72 is a square. If a=, the condition is satisfied by
putting b= 34% —m, where 4 is any number.

2 Bachet, finding no way of hitting upon 2} as the particular square to be taken
in order that the difference between it and 3 may be separable into three cubes, and
observing that he could not solve the problem if he took another arbitrary square between
2 and 3, e.g. 2}, instead of 2}, concluded that Diophantus must have hit upon 24,
which does enable the problem to be solved, by accident.

Fermat would not admit this and considered that the method used by Diophantus for
finding 2} as the square to be taken should not be difficult to discover. Fermat accord-
ingly suggested a method as follows.

Let x — 1 be the side of the required square lying between 2 and 3. Then 3 — (x - 1)?
— 2 + 2x — x2, and this has to be separated into three cubes. Fermat assumes for the sides
214 THE ARITHMETICA

We have therefore to find three cubes the sum of which
= or 132;

that is, we have to divide 162 into three cubes.

But 162 =125+64 — 27;

and “we have it in the Porisms” that the difference of
two cubes can be transformed into the sum of two
cubes’.

Having thus found the three cubes? we start again, and
xr=2%}2% so that r=%.

The three numbers are thus determined.

 

of two of the required cubes two linear expressions in x such that, when the sum of their
cubes is subtracted from 2 + 2x — #2, the result only contains terms in x% and x3 or in x
and units.

The first alternative is secured if the sides of the first and second cubes are 1 — and
1+ x respectively ; for
. 8 26
24 2x —x%— (x - Lx) — (14+ 2)= — 43a? ——i3.
3 27
This latter expression has to be made a cube, for which purpose we put
26 m3x3

—43xi- == -—— sa

43 27 TE Ys
; ; : , 1

which gives a value for x. We have only to see that this value makes gr less than 1,

and we can easily choose #2 so as to fulfil this condition.

[Z.g. suppose 7 =5, and we find x= 3, so that
Del La TD
3 33 3. 33 33

and the side of the third cube is 5.

The square (x — =) , and in fact 3 — {ZY + (ZY = (21 = (2)
IT 33 33 33 I

We then have three cubes which make up the excess of 3 over a certain square ; but,
while the first of these cubes is <1, the second is >1 and the third is negative. Hence
we must, like Diophantus, proceed to transform the difference between the two latter
cubes into the sum of two other cubes.

It will, however, be seen by trial that even Fermat’s method is not quite general, for
it will not, as a matter of fact, give the particular solution obtained by Diophantus in
which the square is 23.

1 On the transformation of the difference of two cubes into the sum of two cubes, see
pp. 101-3, ante.

3 3
2 Vieta’s rule gives 43 —33= (32) + (£2) . It follows that

3. 10: /3N\S /ror)S 20\3

31-2) + (3%) = 273] ’
: 8 ;

and, since x3 = 0 the required numbers are

91 8 4998267 8 20338417 8

216° 27" 60283568 27° 20346417 27:

 

 

 
BOOK V 218

17. To find three numbers such that each of them minus the
cube of their sum gives a cube.

Let the sum be » and the numbers 22°, orf, 2318,

Therefore orf ="1;

and we must replace 39 by a square which is the sum
of three cubes + 3;

therefore we must find three cubes such that their sum
+ 3 is a square. :

Let their sides! be 72, 3 — 2, and any number, say I.

Therefore par + 31 — 27m =a square = (3m — 7 say, So
that 72 = §, and the sides of the cubes are §, 3, 1.

Starting again, we put x for the sum, and for the numbers

Hiv, 138°, 1884
whence 14452*=125, 2? =25%, and r=.
The required numbers are thus found.

18. To find three numbers such that their sum is a square and
the cube of their sum added to any one of them gives a square.

1 et the sum be 22 and the numbers 32% 34% 154%

It follows that 26x*= 1; and, if 26 were a fourth power, the
problem would be solved.

To replace it by a fourth power, we have to find three
numbers such that each increased by 1 gives a square,
while the sum of the three gives a fourth power.

Let these numbers be #* — 222%, 12* + 2mm, m* — 2m [the sum
being #24]; these are indeterminate numbers satisfying
the conditions.

Putting any number, say 3, for #2, we have as the required
auxiliary numbers 63, 15, 3.

Starting again, we put 2? for the sum and 34%, 152° 632° for
the required numbers,

and we have 814%=21, so that x={.

15-63

The numbers (4 2. 3).
are thus found Sn BE

19. To find three numbers such that their sum is a square and

the cube of their sum minus any one of them gives a square.
[There is obviously a lacuna in the text after this enunciation ;
for the next words are “ And we have again to divide 2 as before,”
1 Cf. note on Vv. 15. In this case, if one of the cubes is chosen arbitrarily and 7?
is another, we have only to put (34% - m2) for the side of the third cube in order that, in the

expression to be made a square, the term in 7% may vanish, and the term in 72 may be a
square. :
216 THE ARITHMETICA

whereas there is nothing in our text to which they can refer, and
the lines which follow are clearly no part of the solution of v. 19.

Bachet first noticed the probability that three problems inter-
vened between V. 19 and V. 20, and he gave solutions of them.
But he seems to have failed to observe that the eight lines or so in
the text between the enunciation of v. 19 and the enunciation of
V. 20 belonged to the solution of the last of the three missing
problems. The first of the missing problems is connected with
v. 18 and 19, making a natural trio with them, while the second and
third similarly make with v. 20 a set of three. The enunciations
were doubtless somewhat as follows.

19a. To find three numbers such that their sum is a square
and any one of them minus the cube of their sum gives a square.

190. To find three numbers such that their sum is a given
number and the cube of their sum plus any one of them gives a
square.

19c. To find three numbers such that their sum is a given
number and the cube of their sum minus any one of them gives a
square,

The words then in the text after the enunciation of Vv. 19
evidently belong to this last problem.]

The given sum is 2, the cube of which is 8.

We have to subtract each of the numbers from 8 and
thereby make a square.

Therefore we have to divide 22 into three squares, each
of which is greater than 6;

after which, by subtracting each of the squares from 8, we
find the required numbers.

But we have already shown [cf. V. 11] how to divide 22
into three squares, each of which is greater than 6—
and less than 8, Diophantus should have added.

[The above is explained by the fact that, by addition,
three times the cube of the sum minus the sum itself
is the sum of three squares, and three times the
cube of the sum minus the sum = 3.8 —2=22.]!

1 Wertheim adds a solution in Diophantus’ manner. We have to find what small

) 66 :
fraction of the form 4 we have to add to > or = and therefore to 66, in order to

make a square. In order that 66+ may be a square, we put

66x% + 1 =square = (1 + 8x)?, say,
which gives x=8 and 2%=04.
BOOK V 217

20. To divide a given fraction into three parts such that any
one of them sinus the cube of their sum gives a square.

Given fraction #.

Therefore each part = 4 + a square.

Therefore the sum of the three =] = the sum of three
squares + 4.

Hence we have to divide {2 into three squares, “ which is
easyL”

21. To find three squares such that their continued product
added to any one of them gives a square.
Let the “ solid content” = a7.
We want now three squares, each of which increased by 1
gives a square.
They can be got from right-angled triangles? by dividing
the square of one of the sides about the right angle
by the square of the other.

Let the squares then be

164% TH Tost
The continued TE = AAA00- 15 = 22 by hypothesis.
Therefore 123242 = 1; and, if 139 were a square, the problem

would be a

 

We have therefore to increase 66 by 7 , and therefore 73 by or , in order to make a

 

square. And in fact 73+ 6= =(2Y.

Next, since 22 =3%+ 32+ 22, and 65 — 48=17, while 72 — 65=7, we put
: =(3- 74) + (3 — 72) + (2 + 17%)
16
and X= ps s
1049 1049 1046
3% 3%
1100401 1100401 1094116

149769 * 149769 ' 149769 ’

and the required parts of 2 are SE

149769" 149769" 149769
1 As Wertheim observes, HEL + on + 9. and the required fractions into which
64 O04" 2g joO
1. a 250 8 61
3 is divided are eon Tee Tene

Therefore the sides of the squares are

the squares themselves

 

 

 

 

 

2 If a, 6 be the perpendiculars, ¢ the hypotenuse in a right-angled triangle,

7 1 L=7,=a Square.

Diophantus uses the triangles (3, 4, 5), (5, 12, 13), (8, 15, 17).
218 THE ARITHMETICA

As it is not, we must find three right-angled triangles such
that, if 4’s are their bases, and p’s are their perpen-
diculars, p, p, ps 6,0.0, = a square;

and, if we assume one triangle arbitrarily (3, 4, 35), we
have to make 122, £,6,6, a square, or 3p,6,/p.0, a square.

“This is easy!” and the three triangles are (3, 4, 3),
(9, 40, 41), (8, 15, 17) or similar to them.

1 Diophantus does not give the work here, but only the result. Bachet obtains it
in this way. Suppose it required to find three rational right-angled triangles (%y, #;, 61),
(A2y po, 69) and (43, ps, bs) such that pypeps/b1bsbs is the ratio of a square to a square.
One triangle (%;, #,, 6,) being chosen arbitrarily, form two others by putting
ha=hi+p?, pa=MlP-piP=0%, by=2Mp1,
h3=hP+b%, ps=ht-02=p1%, b3=2/41by,

15223 1'Y?
ans = (2) =a square.

If now %y=3, p1=4, b1=3, the triangles (42, pa, 62) and (43, ps3, b3) are (41, 9, 40)
‘and (34, 16, 30) respectively. Dividing the sides of the latter throughout by 2 (which
does not alter the ratio), we have Diophantus’ second and third triangles (9, 40, 41) and
{3, 13, 17).

Fermat, in his note on the problem, gives the following general rule for finding two
right-angled triangles the areas of which are in the ratio nz: 2 (n> n).

and we have

Form (1) the greater triangle from 222+ 7, 7 — 72, and the lesser from mz +22, m — 1,
“or (2) the greater from 2m —n, m+n, the lesser from 2722z—m, m+n,
or (3) the greater from 6m, 2m —n, the lesser from ym +2, 4m — 21,
or (4) the greater from m+ 47, 2m — 4n, the lesser from 612, m—2n.

The alternative (2) gives Diophantus’ solution if we put m=3, 2=1 and substitute
m—a2n for 21 —m. :

Fermat continues as follows: We can deduce a method of finding #4ree right-angled
triangles the areas of which are in the ratio of three given numbers, provided that the
sum of two of these numbers is equal to four times the third. Suppose e.g. that 7, 7, ¢
are three numbers such that m+ ¢g=4n (m>g¢). Now form the following triangles :

(1) from m+ 4m, 2m — 4n,
(2) from 61, m— an,
(3) from 47n+gq, 47-24.

[If Ay, A, A3 be the areas, we have, as a matter of fact,

Am = Ag|n=As]qg= — 6m3 + 36m2n + 144mn® — 38425.)

We can derive, says Fermat, a method of finding three right-angled triangles the areas
of which themselves form a right-angled triangle. For we have only (o find a triangle
such that the sum of the base and hypotenuse is four times the perpendicular. This is
easy, and the triangle will be similar to (17, 15, 8); the three triangles will then be formed

(1) from 1744.8, 2.17—-4.8 or 49, 2,
(2) from 6.8, 17-2.8 or 48,1,
(3) from 4.8415, 4.8-2.15 or 47, 2.

[The areas of the three right-angled triangles are in fact 234906, 110544 and 207270,
and these numbers form the sides of a right-angled triangle.]

Hence also we can derive a method of finding three right-angled triangles the areas
of which are in the ratio of three given squares such that the sum of two of them is equal to
BOOK V 219

Starting again, we put for the squares

Ter? Zhao (Bloat
Equating the product of these to +? we find x to be
a 16 100
rational [x =18, and the squares are i 1.

22. To find three squares such that their continued product
minus any one of them gives a square.
Let the solid content be 2? and let the numbers be
obtained from right-angled triangles, being
$847 1554 IE |

Therefore the continued product Es + Ap

 

5:13.17
2 600
or (55:2 Yom 2000 23 4 =
2.13.17 1221025
If then —23°2° were a fourth power, ze. if 4.0.2 were
1221025 2.13.17

a square, the problem would be solved.

We have therefore to find three right-angled triangles
with hypotenuses /,, /,, /; respectively, and with
D1 2 15 28 one of the perpendiculars in each re-
spectively, such that

hei, py po ps = a square.

Assuming one of the triangles to be (3, 4, 5), so that eg.

0: = 5. 4=20, we must have
5/u pie, po = a square.

This is satished if £, 2, = 54. 2,.

With a view to this we have first (cf. the last proposition)
to find two right-angled triangles such that, if x, 7,
are the two perpendiculars in one and z,, 9, the two
perpendiculars in the other, 1,9, =54,5,. From such
a pair of triangles we can form two more right-
angled triangles such that the product of the
hypotenuse and one perpendicular in one is five times

the product of the Zypotenuse and one perpendicular in
the other. .

 

Sour times the third, and we can also in the same way find three right-angled triangles of
the same area; we can also construct, in an infinite number of ways, fwo right-angled
triangles the areas of which are in a given ratio, by multiplying one of the terms of the
ratio or the two terms by given squares, etc. ;

1 Diophantus’ procedure is only obscurely indicated in the Greek text. It was
explained by Schulz in his edition (cf. Tannery in Oeuvres de Fermat, 1. p. 323, note).
220 THE ARITHMETICA

Since the triangles found satisfying the relation x, y,=5x, 9,
are (35, 12, 13) and (3, 4, 3) respectively’, we have in
fact to find two new right-angled triangles from them,
namely the triangles (4, p,, 6,) and (4, p,, 8), such that

Aip=30and 4,0,=6
the numbers 30 and 6 being the areas of the two
triangles mentioned.

These triangles are (64, 42, [W?)) and (24, 22, [FH] re-
spectively.

Starting again, we take for the numbers

182 B04 Ht

[42 divided by 2} gives £4, and & 0 divided by 64 gives 128.]

The product =22;

therefore, taking the square root, we have

4-24.120
5.235.160
so that x=$3, and the required squares are found.

De
SE )

23. To find three squares such that each mznus the product of
the three gives a square.

 

Having given a rational right-angled triangle (z, x, y), Diophantus knows how to find a

rational right-angled triangle (4, p, 4) such that hp== xy. We have in fact to put

h=lz, p= , whence 62=/2-p2= LE) - (22)
5 4

22 2z
Thus, having found two triangles (3, 12, 13) and (3, 4, 5) with areas in the ratio of 3
to 1 (see next paragraph of text with note thereon), Diophantus takes

h=>.13=6} == 50; and similarly y=. 5=01, pp=34 = 12

3 5 5

Cossali (after Bachet) gives a formula for three right-angled triangles such that the
solid content of the three hypotenuses has to the solid content of three perpendiculars
(one in each triangle) the ratio of a square to a square ; his triangles are

i HB 4p B

(1) 7, b, p [{=ipotenusal], (2). HH u 4 3 2 ; 2 ur

02+ 4ip  b. 4bp +p (402-89) i 46p—06 (42-09) _

) ag oy STE Th ETT =42,
b b 0
£02 + 42) (0 + 4ip?)
I

 

2 (H2 2)2
ip. 47. pf PF : 4 0%02
If i=5, b=4, p=3, we can get from this triangle the triangles (13, 5, 12) and
16 22
3.65

1 These triangles can be hia putting 7z=35, z=1 in Fermat’s fourth formula
(note on last proposition). By that formula the triangles are formed from (9, 6) and
(6, 3) respectively ; and, dividing out by 3, we form the triangles from (3, 2) and (2, I)
respectively.

and, in fact,

(65, 63, 16), and our equation is ur =T.
BOOK V 221

Let the “solid content” be #2 and let the squares be
formed from right-angled triangles, as before.

If we take the same triangles as those found in the last
problem and put for the three squares

2342 82072 28501 +3

 

6% 218%, 1336s
each of these minus the continued product (x?) gives a
square. a

It remains that their product =22;
this gives x=48 and the problem is solved.

24. To find three squares such that the product of any two
increased by I gives a square.

Product of first and second + 1 =a square, and the third

is a square; therefore “solid content” 4 each=a
square.
The problem therefore reduces to v. 21 above.

! De Billy in the Zuwentum Novum, Part 11. paragraph 28 (Oeuvres de Fermat, 111.
PP. 3734), extends this problem, showing how to find fox numbers, three of which (only)
are squares, having the given property, 7.c. to solve the equations

axl t1=72 afayti=dl,
xxl +1=s2, alayti=22
2a? 1=02, a?ayt1=wd
First seek three square numbers satisfying the conditions of Diophantus’ problem
9, 25 256

V, 24, SAY (Z 2% ) the solution of v. 21 given in Bachet’s edition. We have then

to find a fourth number (x, say) such that
9 \
7 X+1 |
2 v+1) areall squares.
256

Baws

: 6 :

Substitute Ri 4 Ty for a, so as to make the first expression a square. We have

then to solve the double-equation
100 200
2 =
=== YZ: y+i=u
9 9 Sey,

8192

4096 2 44 —Z= y+ 1=22
729 729
which can be solved by the ordinary method.

 

,

De Billy does not give the solution, but it may be easily supplied thus.

; 10 %) (2 4) 5
The difference sl ag 2d Nl Sa 242
e differ & FE (3242)

_154 (26 2.26
ts (Ds 27 ):

 
222 THE ARITHMETICA

25. To find three squares such that the product of any two
minus 1 gives a square.
This reduces, similarly, to v. 22 above.

26. To find three squares such that, if we subtract the product
of any two of them from unity, the result is a square.
This again reduces to an earlier problem, Vv. 23.

27. Given a number, to find three squares such that the sum of
any two added to the given number makes a square.

Given number 15.

Let one of the required squares be 9;

I have then to find two other squares such that each
+ 24 = a square, and their sum + 15 = a square.

To find two squares, each of which + 24 = a square, take
two pairs of numbers which have 24 for their pro-
duct?

Let one pair of factors be 4/x, 6x, and let the side of one

square be half their difference or : — 3x.

Let the other pair of factors be 3/#, 8x, and let the
side of the other square be half their difference or
1}
x

Therefore each of the squares + 24 gives a square.

It remains that their sum + 15 =a square;

therefore (2 47)

9 a

/ 5 (C- 3) + 15 = a square,

 

Equating the square of half the sum of the factors to the larger expression, we have

(2 BY oe 00, T,

 

 

 

3 27 9
whence y=- Ts and y2+2y= — a .
Therefore »etl (P2+2y)=— onan which satisfies the equations. In fact
fr (B0. 2s ROY ni ene LY.
But even here, as the value of x which we have found is negative, we ought, strictly
speaking, to deduce a further value by substituting y — Ee for x in the equations

and solving again, which would of course lead to very large numbers.

1 The text adds the words “and [let us take] sides about the right angle in a right-
angled triangle.” I think these words must be a careless interpolation : they are not
wanted and give no sense; nor do they occur in the corresponding place in the next
problem.
BOOK V 223
6%
7
Therefore x= 3, and the problem is solved.

or + 25x%— g = a square = 257% say.

28. Given a number, to find three squares such that the sum of
any two minus the given number makes a square.

Given number 13.

Let one of the squares be 25;

I have then to find two other squares such that each
+ 12 =a square, and (sum of both) — 13 = a square.

Divide 12 into factors in two ways, and let the factors be
(3%, 4/x) and (4x, 3/%).

Take as the sides of the squares half the differences of the
factors, ze. let the squares be

(rar-2), (2x - HY,

Each of these + 12 gives a square.
It remains that the sum of the squares — 13 =a square,

6} 6}

2 a: ard
= t0l — 25 =asquare = 7, say.

Therefore x = 2, and the problem is solved?

or

1 Diophantus has found values of £, #, { satisfying the equations
72+ (24 a= 2?
2 E+a= “| .
E+ pt a=w?
Fermat shows how to find fox numbers (not squares) satisfying the corresponding
conditions, namely that the sum of any two added to a shall give a square. Suppose a=15.
Take three numbers satisfying the conditions of Diophantus’ problem, say g, = a s
Assume x2 — 15 as the first of the four required numbers; and let the second be 6x +9
(because g is one of the square numbers taken and 6 is twice its side); for the same
2
i
Three of the conditions are now fulfilled since each of the last three numbers added to
the first (x2 15) plus 15 gives a square. The three remaining conditions give the triple-
equation

: I I
reason let the third number be — x + Se and the fourth g x
5

2
ir+ 9+ —+15= ota + (42) = 22,
\ 2
5 5
2 2
3 pil 8% ie 20h (2) =u?
15 100 225% 15 6

2 Fermat observes that four numbers (not squares) with the property indicated can
be found by the same procedure as that shown in the note to the preceding problem.
If a is the given number, put x2+a for the first of the four required numbers.
224 THE ARITHMETICA

29. To find three squares such that the sum of their squares is
a square.
Let the squares be x? 4, 9 respectively.
Therefore x* + 97 = a square = (x? — 10)? say;
whence x? = 5. :
If the ratio of 3 to 20 were the ratio of a square to a
square, the problem would be solved ; but it is not.
Therefore 7 have to find two squares (p* ¢°, say) and a
number (m, say) such that m?—p—q* has to 2m the
ratio of a square to a square.
let A=20=4 and m=r*+4.
Therefore m2 — pt —g* = (22+ 4) — 2* — 16 = 822
Hence 82%/(22% + 8), or 42/(s* + 4), must be the ratio of a
square to a square.
Put 2+ 4= (241) say;
therefore z= 1}, and the squares are f= 2%, ¢*=4, while
wm =06};
of, if we take 4 times each, f° =0, "= 16, m = 25.
Starting again, we put for the squares x? 9, 16;
then the sum of the squares=2x*+4 337 =(2*— 25), and
z=
The required squares are i, g, 16.

.

“ho

30. [The enunciation of this problem is in the form of an
epigram, the meaning of which is as follows.]

A man buys a certain number of measures (oes) of wine, some

at 8 drachmas, some at 5 drachmas each. He pays for them a

square number of drachmas; and if we add 60 to this number, the

result is a square, the side of which is equal to the whole number
of measures. Find how many he bought at each price.

let x= the whole number of measures; therefore 2? — 60

‘was the price paid, which is a square = (x — m)? say.

 

If now 42, /2, m? represent three numbers satisfying the conditions of the present
problem of Diophantus, put for the second of the required numbers 24a + 42, for the third
2/x +172, and for the fourth 27x + m2. These satisfy three conditions, since each of the
last three numbers added to the first (42+ a) less the number a gives a square. The
remaining three conditions give a triple-equation.

1 « Why,” says Fermat, ‘‘does not Diophantus seek zwo fourth powers such that
their sum is a square? This problem is in fact impossible, as by my method I am in
a position to prove with all rigour.” It is probable that Diophantus knew the fact
without being able to prove it generally. That neither the sum nor the difference of
two fourth powers can be a square was proved by Euler (Commentationes arithmeticae, 1.
PP- 245qq., and Algebra, Part II. c. XIII).
BOOK V : 225

Now 1 of the price of the five-drachma measures + { of
the price of the eight-drachma measures =x;

so that a?— 60, the total price, has to be divided into
two parts such that { of one +} of the other = x.

We cannot have a real solution of this unless

x>3(@*—60) and < 1 (x%— 60).

Therefore sr < x? —060« 82

{1) Since 22> tv 4-60,

x%= 5x + a number greater than 60,

whence x is! not less than 11.

(2) rt < 8x +60

or x?= 8x + some number less than 60,
whence x is! not greater than 12.

Therefore [I< Z<12

Now (from above) x = (722 + 60)/2m;
therefore 22m < m* + 60 < 24m.
Thus (1) 22m =»? + (some number less than 60),
and therefore 7: is® not less than 19.
(2) 24m =m*+ (some number greater than 60),
and therefore wm is? less than 21.
Hence we put # = 20, and
x%— 60 = (x — 20),
so'that z= 113, 22= 132}, and 2° — 60 = 721,
Thus we have to divide 72} into two parts such that }
of one part plus } of the other = 114.
Let the first part be 52.
Therefore § (second part) = 11} — 3,
or second part = 92 — 87;
therefore 52 + 92 — 82 = 724,

and z= 72
12
Therefore the number of five-drachma ydes = 3
» ” » , eight-drachma ,, =,
12

1 For an explanation of these limits see p. 60, ante.
? See p. 62, ante.

15
220 THE ARITHMETICA

BOOK VI

1. To find a (rational) right-angled triangle such that the
hypotenuse minus each of the sides gives a cube.

Let the required triangle be formed from zx, 3.

Therefore hy pottause = = 2? +9, perpendicular =6x, base
=2—Q,

Thus 22 + 9 — (2? — 9) = 18 should be a cube, but it is not.

Now 18 = 2.32; therefore we must replace 3 by 72, where
2.77 1s a cube; and m=12.

We form, Tn a right-angled triangle Pos YZ
namely (#244, 44, 2*—4); and one condition is
satisfied.

The other gives 2° — 4x + 4 =a cube;

therefore (x — 2)? is a cube, or x — 2 is a cube = 3, say.

Thus x= 10, :

and the triangle is (40, 96, 104).

2. To find a right-angled triangle such that the hypotenuse
added to each side gives a cube.

Form a triangle, as before, from two numbers; and, as
before, one of them must be such that twice its
square is a cube, Ze, must be. 2.

We form a triangle from x, 2, namely #2 4-4, 4%, 4 — 2%;
therefore 2° + 4r + 4 must be a cube, while 2* must
be less than 4, or z< 2. :

Thus x + 2 =a cube which must be <4 and >2 = 2%, say.

Therefore x= 11, :

and the triangle is i 5%, m,
or, if we multiply by os common denominator, (135,
352, 377)

3. To find a right-angled triangle such that its area added to
a given number makes a square.
Yet 5 be the given number, (32, 47, 35) the ns
triangle.
1 Diophantus’ expressions are 6 év 73 vmorewoboy, ‘‘the (number) in (or represent-
ing) the hypotenuse,” 6 év éxarépg Tv 8pfdv, “the (number) in (or representing) each
of the perpendicular sides,” 6 év 7¢ éuBad, ‘the (number) in (vr representing) the area,”

etc. It will be convenient to say ‘‘the hypotenuse,” etc. simply. It will be observed
that, as between the numbers representing sides and area, all idea of dimension is ignored.
BOOK VI 227

Therefore 64% + 5 =a square = 91? say,

or 3at=35.

But 3 should have to 5 the ratio of a square to a square.

Therefore we must find a right-angled triangle and a
number such that the difference between the square
of the number and the area of the triangle has to 5 the
ratio of a square to a square, ze. = 1 of a square.

J 3 { :
Form a right-angled triangle from (om, >)

9

, I
thus the area is 7.2 — —.
7m

2.7

Let the number be mz + el so that we must have

101

wm’

therefore ' 4.25 + te

or 100m:®+ 505 = a square = (10m + 5)? say,

and m= 24,

The auxiliary triangle must therefore be formed from 2t,
+, and the auxiliary number sought is 413.

Put now for the original triangle (/x, px, bx), where (4, p, 0)
is the right-angled triangle formed from 2%, 2; ;

this gives p01 + 5 =17080942

and we have the solution.

[The perpendicular sides of the right-angled triangle are

 

4.5 +—=1} of a square;

 

= a square,

24% 52
o@ 2 331151 -
& a= dubly and 2x,
’ 331 2 = 170569 ,2
whence BBLALLY + § = 11086942
rez 2d
53)

and the triangle is
331151

Bd 43, HE
4. To find a right-angled triangle such that its area minus a
given number makes a square.
Given number 6, triangle (3x, 4, 52), say.
Therefore 62% — 6 = square = 44? say.
Thus, in this case, we must find a right-angled triangle
and a number such that
(area of triangle) — (number)?=} of a square.

- I
Form a triangle from 7, i

15—2
228 THE ARITHMETICA

i 1
., and let the number be 72 — §. nk

I
Its area'is 72 — ——
77

Therefore 6 — — = 1 (a square),

or 3677 — LI a square = (672 — 2)? say.

Therefore 7: =§, and the auxiliary triangle is formed from
(8, 3), the auxiliary number being 3%

We start again, substituting for 3, 4, § in the original
hypothesis the sides of the auxiliary triangle just
found, and putting (32)*+* in place of 44%; and the
solution is obvious.

[The auxiliary triangle is (4848, 2, 4177), whence
41ot— 6=(31)22* and x = §,

so that the required triangle is (4215, 16 41173]
5. To find a right-angled triangle such that, if its area be
subtracted from a given number, the remainder is a square.
Given number 10, triangle (3, 4x, 5x), say.
Thus 10—62?=a square; and we have to find a right-
angled triangle and a number such that
(area of triangle) : (number)? = J; of a square.

I
Form a triangle from 72, — , the area being m?* — hs
m’

and let the number be — + 572.

Therefore 26° + 10 =; of a square,

or 260m2® + 100 =a square,
or again 65m + 25 = a square = (8m + §)?, say,
whence 7#= 30,

The rest is obvious.
The required triangle is 4889849 2. 40960001
6. To find a right-angled triangle such that the area added
to one of the perpendiculars makes a given number.

Given number 7, triangle (3x, 47, 52).

Therefore 6222 =7,

In order that this might be solved, it would be necessary that
(half coefficient of x): + product of coefficient of x* and
absolute term should be a square ;

but (14)*+ 6.7 is not a square.

Hence we must find, to replace (3, 4, 5), a right-angled
triangle such that

(3 one perpendicular)’ + 7 times area = a square.

Let one perpendicular be 7, the other 1.
BOOK VI 229

Therefore 34 +4 =a square, or 147 + 1 = a square.
Also, since the triangle is rational, 7? + 1 =a
The difference #72 — 14m =m (mn — 14);

and putting, as usual, 72 = 14m + 1,

we have m= 2%,

The auxiliary triangle is therefore (24, 1, 2%) or (24, 7, 25).
Starting afresh, we take as the triangle (24x, 7x, 257).
Therefore 8422 4-72=17,

and r=1.

We have then (6 7 : 5) as the solution.
7. To find a right-angled triangle such that its area minus one
of the perpendiculars is a given number.
Given number 7.
As before, we have to find a right-angled triangle such that
(4 one perpendicular)? + 7 times area=a square ;
this triangle is (7, 24, 253).
Let then the triangle of the problem be (7.x, 24, 25%).
Therefore 842° — 72 =7,
=,
and the problem is solved?

1 Fermat observes that this problem and the next can be solved by another method.
“Form in this case,” he says, ‘“‘a triangle from the given number and 1, and divide
the sides by the sum of the given number and 1; the quotients will give the required
triangle.”

In fact, if we take as the sides of the required triangle

(@+1) x, (@2-1)x, 20x,
where a is the given number, we have
(a2 ~-1) ax? + 2ax=a,
1 a I

one root of which is w= ooo
a@-1 at-1 a+1’

 

and the sides of the required triangle are therefore
a?+1 a’-1 2a
491 ati 2%i

The solution is really the same as that of Diophantus.

2 Similarly in this case we may, with Fermat, form the triangle from the given number
and 1, and divide the sides by the difference between the given number and 1, and we
shall have the required triangle.

In v1. 6, 7, Diophantus has found triangles ¢, £ 7 (¢ being the hypotenuse), such that

(1) lenti=a,

 

 

I
and (2) Ty =g=a,
Fermat enunciates the third case
I
3) t-=uy=a
230 THE ARITHMETICA

8. To find a right-angled triangle such that the area added to
the sum of the perpendiculars makes a given number.
Given number 6.
Again I have to find a right-angled triangle such that
(3 sum of perpendiculars)? + 6 times area = a square.
Let m2, 1 be the perpendicular sides of this triangle ;
therefore } (m+ 1): + 3m = bm*+ 33m + } = a square,
while 722 + 1 must also be a square.
Therefore ee Ip !
7
The difference is 272.7, and we put
wi—ym+ 12k =m +1,
whence m = §3, :
and the auxiliary triangle is (42, 1, $2), or (45, 23, 53).
Assume now for the triangle of the problem
(45%, 28x, 53%).
Therefore 630% + 732 =6;
x is rational [= {], and the solution follows.

| are both squares.

‘9. To find a right-angled triangle such that the area minus the
sum of the perpendiculars is a given number.
Given number 6.
As before, we find a subsidiary right-angled triangle such
that ({sum of perpendiculars)’+ 6 times area=a square.
This is found to be (28, 45, 53) as before.
Taking (28x, 452, 53%) for the required triangle,
flor: —- 73x =06;
x =, and the problem is solved
10. To find a right-angled triangle such that the sum of its
area, the hypotenuse, and one of the perpendiculars is a given
number.

observing that Diophantus and Bachet appear not to have known the solution, but that
it can be solved ‘‘by our method.” He does not actually give the solution ; but we may
compare his solutions of similar problems in the Znwentum Novum, e.g. those given in
the notes to vi. 11 and VI. 135 below and in the Supplement. The essence of the method
is that, if the first value of x found in the ordinary course is such as to give a negative
value for one of the sides, we can derive from it a fresh value which will make all the
sides positive. ;
1 Here likewise, Diophantus having solved the problem

I
Lin (Et =a,
Fermat enunciates, as to be solved by his method, the corresponding problem

F+n-—in=a
n rE »
BOOK VI 231

Given number 4.
[f we assumed as the triangle (Lv, pr, bx), we should have
pbx + fix + bx = 4;
and, in order that the solution may be rational, we must
find a right-angled triangle such that
1 (hyp. + one perp.)?+ 4 times area =a square.
Form a right-angled triangle from 1, 72 + 1.
Then 1 (hyp. + one perp.) = 1 (m* + 2m + 2 + m* + 2m)?
=r + 4m + 6 + 4m + 1,
and 4 times area =4 (me + 1) (12> + 2m)
= 418° + 1218 + 8m.
Therefore ;
mt + Sued + 18m + 1272 + 1 = a square = (6m + 1 — mm?) say,
whence 72 = 4, and the auxiliary triangle is formed from
(1, 9) or (5,-9).; This triangle is (56, oo, 106) or
(28, 45, 53).
We assume therefore 28x, 45x, 53x for the original triangle,
and we have. 630vi+4 Sir=4.
Therefore x = $5, and the problem is solved.

11. To find a right-angled triangle such that its area minus
the sum of the hypotenuse and one of the perpendiculars is a given
number.

Given number 4.
We have then to find an auxiliary triangle with the same
property as in the last problem ;
therefore (28, 45, 53) will serve the purpose.
We put for the triangle of the problem (28x, 45%, 53x), and
we have 6301*— Sir =4;
z=}, and the problem is solved’

1 Diophantus has in VI. 10, 11 shown us how to find a rational right-angled triangle

¢, & 7 (¢ being the hypotenuse) such that

(1) én CHE =a

(2) En-(+i=a.

Fermat, in the Znwzentum Novum, Part 111. paragraph 33 (Oeuvres de Fermat, 111.
p- 389), propounds and solves the corresponding problem

(3) §+E-2tn=a.

In the particular case taken by Fermat a=4. He proceeds thus:
First find a rational right-angled triangle in which (since a= 4)

Lira).

$n =a square.

w=
232 THE ARITHMETICA

Lemma I to the following problem.
To find a right-angled triangle such that the difference of the
perpendiculars is a square, the greater alone is a square, and further
the area added to the lesser perpendicular gives a square.
Let the triangle be formed from two numbers, the greater
perpendicular being twice their product. !
Hence I must find two numbers such that (1) twice their
product is a square and (2) twice their product exceeds
the difference of their squares by a square.

This is true of any two numbers the greater of which
= twice the lesser.

Form then the triangle from x, 2x, and two conditions are
satisfied.

The third gives 6x + 34%=a square, or 62+ 3 = a square.

I have therefore to find a number such that 6 times its
square + 3 =a square;

one such number is 1, and there are an infinite number of
others,

If x= 1, the triangle is formed from 1, 2.

 

Suppose it formed from x+ 1, xv; the sides then are
$=2x%+2x +1, E=2x+1, B=242+42a.
1 2 I
Thus i’ (¢ + a} =a. n= 4B + 0a? + qx +1 — 4 (223+ 32% +0)
=at— y3— 6a + 1

=a square
=(a%—2x+1)% say.
Therefore ~ 6x?=06x2— 4x, w=, and x41 =! y
The triangle formed from z ; is (2. i, >) . Thus we may take as the auxiliary

triangle (17, 15, 8).
Take now 17x, 15%, 8x for the sides of the triangle originally required to be found.
We have then

RAR 2
§+E--En=32x-6ox’=4;
whence =t, and the required triangle is (= 3 2, 5) :
; 3
[The auxiliary right-angled triangle was of course necessary to be found in order to
make the final quadratic give a rational result.]
Bachet adds after v1. 11 a solution of the problem represented by
I,
- &N— {=a,

to which Fermat adds the enunciation of the corresponding problem
1
f= 1 fn=a.

! Though there are an infinite number of values of x for which 6x2 + 3 becomes a square,
the resulting triangles are all similar. For, if x be any one of the values, the triangle is
BOOK ‘V1 233

Lemma I1 to the following problem.

Given two numbers the sum of which is a square, an infinite
number of squares can be found such that, when the square is multi-
plied by one of the given numbers and the product is added to
the other, the result is a square.

Given numbers 3, 6.

Let 2+ 2x4 1 be the required square which, say, when
multiplied by 3 and then increased by 6, gives a square.

We have 32% + 6x + 9g =a square;

and, since the absolute term is a square, an infinite number
of solutions can be found.

Suppose, eg. 322+ 6x + 9=(3 — 3r),

and E=q,

The side of the required square is 5, and an infinite
number of other solutions can be found.

12. To find a right-angled triangle such that the area added
to either of the perpendiculars gives a square.

Let the triangle be (52, 122, 137)

Therefore (1) 302+ 122 =a squate = 3617, say,

and x =2,

But (2) we must also have

30x? 4 5x =a square.

Tis is however not a square when x= 2.

Therefore I must find a square 722 to replace 362% such
that 12/(s — 30), the value of x obtained from the
first equation, is real and satisfies the condition

304% + 5x = a square,

This gives, by substitution,

(60m2* + 2520)/(1m* — 60m + OO) = a square,

or 60m?* + 2520 = a square.

This could be solved [by the preceding Lemma II] zf
60 + 2520 were equal to a square.

Now 60 arises from §.12, Ze. from the product of the
perpendicular sides of (5, 12, 13);

2520s 30.12.(12—35), 7.e. the continued product of the
area, the greater perpendicular, and the difference
between the perpendiculars.

 

formed from x, 2x, and its sides are therefore 3x2, 4x%, 52%; that is, the triangles are all
similar to (3, 4, 5). Fermat shows in his note on the following problem, VI. 12, how to
find any number of triangles satisfying the conditions of this Lemma and #o/ similar to
{3, 1, 5). See p. 233, note.
234 THE ARITHMETICA

Hence we must find an auxiliary triangle such that
(product of perps.) + (continued product of area,
greater perp. and difference of perps.) =a square.

Or, if we make the greater perpendicular a square and
divide out by it, we must have
(lesser perp.) + (product of area and diff. of perps.)

=a square.

Then, assuming that we have found two numbers, (1) the
product of the area and the difference of the perpen-
diculars and (2) the lesser perpendicular, satisfying
these conditions, we have to find a square (72?) such
that the product of this square into the second of
the numbers, when added § to the first number, gives
a square’.

! The text of this sentence is unsatisfactory. Bachet altered the reading of the MSS.
So did Tannery, but more by way of filling out. The version above follows Tannery’s text,
which is as follows: drdyerac els 76 dvo dpifuovs elpévras [for dvras of MSS.]<7dv Te mo >
700 éuBadol kal Tis Umepoxfs T@v dpfdv, <kal Tov év Tf éNdogove TGV 6pfdv >, alls [for
avrfis of MSS.] {prety gov Twa, bs moAamhaciaclels émi &va Tov dobévTa, <kal mpoohafy
Tov éTepov >, mwoiel TeTpdywrov. :

The argument would then be this. If (4, p, 4) be the triangle (46> p), we have to make

bp+ : bp (b- p) b a square,

or, if 4 is a square, p+ bp (6— p) must be a square.
The ultimate equation to be solved (corresponding to 6on:?+ 2520=a square) is

bpm? +2 bp (0 — p) b=a square,

a a1

or, if 4 is a square, p+ = bp (6 - p)=a square ;

and therefore, according to Tannery’s text, “the problem is reduced to this: Having found
two numbers 20 (b—p) and p [satisfying the conditions, namely that their sum is a

square, while & is also a square], to find after that a square such that the product of it
and the latter number added to the former number gives a square.”
The difficulty is that, with the above readings, there is nothing to correspond exactly to

the phraseology of the enunciation of Lemma I, which speaks, not of making p+ > bp (6-p)
a square when 4 is a square, but of making é — p, é and p++ bp all simultaneously squares.

But the particular solution of the Lemma is really equivalent to making 4 and p+ : bp (6-2)
simultaneously squares. For the triangle is formed from a, 2a; oe method of maling

b a square (= 44?) incidentally makes 6 — p a square (=a?), and A 0 becomes ate 6at,

while p+2bp (6—p) becomes 3a%+6a8. Since the solution actunlly need is a=1, the

effect is the same whichever way the problem is stated. And in any case, whether the
expression to be made a square is 3a%72+ 6a* or 3a*m>+ 64%, the problem equally reduces
to that of making 32246 a square.
BOOK: V1 235

How to solve these problents is shown in the Lemmas.
The auxiliary triangle is (3, 4, 5). [Lemma I.]
Accordingly, putting for the original triangle (3x, 4x, 5x),

 

we have eh both squares.

62" + 32
Let r=_t G be the solution of the first equation ;
then 2 4

Tw — 12024 36°
The second equation therefore gives

os 2 _a square
mt — 12m + 36 © mt —6 1
whence 12m + 24 = a square,

and we have therefore to find a square (2?) such that
twelve times it + 24 is a square; this is possible, since
12 + 24 is a square [Lemma II].

A solution is mm? = 23,

whence z=

19
and [Z. 2) is the required triangle.
19) 19’ 19

13. To find a right-angled triangle such that its area minus
either perpendicular gives a square.
We have to find an auxiliary triangle exactly as in the
last problem ;

 

Bachet’s reading is dwdyerar els 70 dVo aplfudy dofévTwy Tol Te éuPadod, kal Ths
éNdogovos TQv wep Tv Opfiv, avTols {nTelv TeTpdywvév Twa, bs moANamAaciaclels éml
&va Tv dobévTwy, kal mpoahafiw TOV éTepoy, mo TETPAyWYOV.

1 Fermat observes that Diophantus gives only one species of triangle satisfying the
condition, namely triangles similar to (3, 4, 5), but that by his (Fermat’s) method an infinite
number of triangles of different species can be found to satisfy the conditions, the first
being derived from Diophantus’ triangle, the second from the new triangle, and so on.

Suppose that the triangle (3, 4, 5) has been found satisfying the condition that

X,
tien), ~ én =a square,
where £, 7 are the perpendicular sides and £> 17.

To derive a second such triangle from the first (3, 4, 3), assume the greater of the two
perpendicular sides to be 4 and the lesser 3 +x.

Then fn+E(E-m) © n=36-120-8x’=a square.
Also P= +9*=125+ 6x + x? =a square.
We have therefore simply to solve the double-equation
: aa
25+ Got 2=22f’
which is a matter of no difficulty. As a matter of fact, the usual method gives

_ 20667 . 2 of 200667 el
I and the triangle is ( oe 49 prep li

 
236 THE ARITAMETICA

this triangle is (3, 4, 5), and accordingly we assume for
the triangle of the problem (3x, 4x, 5x).
One condition then gives 64*— 4x = a square = 7222? say

 

(#2: < 6),
: 4
and X= pm .
The second condition gives 64% — 3x = a square; and, by
substitution,
5 pi si square
nt — 12m + 36 6 — mt :
or 24 + 127° = a square.

This is satisfied by m= 1,

whence r=4#, and the required triangle is (= : =, 4) :

Or, if we do not wish to use the value 1 for ,
let 72 = z+ 1, and (dividing by 4) we have
3m: +6 = 32° + 62 + 9 = a square;
z must be found to be not greater than 13 (in order that
m* may be less than 6), and #2 will not be greater than
22, The solution is then rational.
14. To find aright-angled triangle such that its area minus the
hypotenuse or minus one of the perpendiculars gives a square.
1 ct the triangle be (37, 4%, 52).
Ste = are both squares.
64% 3x
Making the latter a square (= 72%), we have

a3 22
IR (m* <6).

Therefore

 

! Diophantus having solved the problem of finding a right-angled triangle §, », &
(¢ being the hypotenuse) such that
I (= Y
5 Ek
are both squares,
I
Sn
Fermat enunciates, as susceptible of solution by his method, but otherwise very difficult,
the corresponding problem of making
1
fe 3 én
both squares.
I
n-Lin
; This problem was solved by Euler (Novi Commentarii Acad. Petropol. 1 749, 11. (1751)
PP: 49 qq. = Commentationes arithmeticae, 1. pp. 62-72). :

3
BOOK VI 237

The first equation then gives
54 “ar 95
mt — 12m* +36 6 — nt
or 157% — 36 = a square.
This equation we cannot solve because 15 is not the sum of
two squares. Therefore we must change the assumed

= a square,

 

triangle.

Now (with reference to the triangle 3, 4, 5) 15722 = the
continued product of a square less than the area, the
hypotenuse, and one perpendicular ;

while 36 = the continued product of the area, the perpen-
dicular, and the difference between the hypotenuse
and the perpendicular.

Therefore we have to find a right-angled triangle (%, p, 4,
say) and a square (72%) less than 6 such that

mp — Spb. p(f—p) is a square.

If we form the triangle from two numbers X,, X, and
suppose that p=3X,X,, and if we then divide
throughout by (X,— X,)* which is equal to Zz —p, we
must find a square 2*[= 72/(X, — X,)*] such that

hp —L pb. pis a square.

The problem can be solved if X,, X, are “similar plane
numbers®” :

Form the auxiliary triangle from similar plane numbers
accordingly, say 4, I. [The conditions are then
satisfied? ]

[The equation for 72 then becomes

8.17m% —4.15.8.9=a square,

or 136m® — 4320 = a square. |
Let* m*= 36. [This satisfies the equation, and 36 < area
of triangle.]

1 See p. 70 above.
* Diophantus states this without proof. [A ¢‘ plane number” being of the form a . 4,
6 i mm m?
a plane number similar to it is of the form Se b or - ab.)
The fact stated may be verified thus. We have
2? (XP + Xp?) 2X) Xp — XX, (X12 — X9?) 2X7 Xp =a square.
The condition is satisfied if 22= JX} Xj, for the expression then reduces to 4.2.12. X52.
In that case X;X; is a square, or X;/X; is a square.
3 Since X1=4, Xa=1, we have h=17, p=8, b=13, 22=X1 Xp=4, and

1
Php-_pb-p=4. 17.8-4.15.8=12. 32=04, a square.

4 The reason for this assumption is that, by hypothesis, z>=m2/(X] ~ Y4)2, or
+=m2[32, and m2=30.
238 THE ARITHMETICA

The triangle formed from 4, 1 being (8, 15, 17), we assume
8x, 15x, 172 for the original triangle.

We now put 6612 — 32 = 360°,

and xr=1.

The required triangle is therefore E. 5; 3.

Lemma to the following problem.

Given two numbers, if, when some square is multiplied into
one of the numbers and the other number is subtracted from
the product, the result is a square, another square larger than
the aforesaid square can always be found which has the same
property.

Given numbers 3, 11, side of square 5, say, so that

3.25 —11=064, a square.
Let the required square be (x + 5)
Therefore

3(x+ 5) — 11 =322+ 304 + 64 = a square
= (8 — 2x)? say,

and L=03,
The side of the new square is 67, and the square itself

4489.

15. To find a right-angled triangle such that the area added
to either the hypotenuse or one of the perpendiculars gives a
square.

In order to guide us to a proper assumption for the
required triangle, we have, in this case, to seek a
triangle (4, p, 0, say) and a square (#®) such that
m? > % pb, the area, and

wip —% pb. p (fe —p) is a square.

Let the triangle be formed from 4, 1, the square (7%)
being 36, as before ;

but, the triangle being (8, 15, 17), the square is nol
greater than the area.

We must therefore, as in the preceding Lemma, replace
36 by a greater square.

Now /%p=136,and }pb.p (f—p)=60.8.9=4320,

so that 13672 — 4320 = a square,

which is satisfied by 72 = 36 ; and we have to find a larger
square (2%) such that

136% — 4320 = a square.
BOOK VI 239

Put z=m + 6, and we have a
(m*+ 12m + 36) 136 —4320=a square,

or 136m* + 1632m + 576 = a square = (km — 24) say.
This equation has any number of solutions; eg., putting

£ = 10, we have :

m= 20, 2=20, and 22= 0676,

We therefore put (8x, 15x, 17x) for the original triangle,

and then assume

6042 + 8x = 67612

whence xr =.,%, and the problem is solved.

! In VI. 14, 15 Diophantus has shown how to find a rational right-angled triangle
$, m, £ (where { is the hypotenuse) such that

he (1) Zyl

!

2
are both squares,
1 i
Ege
2 fn=2
T
or: .. {2} “tte
7 are both squares.
I
= fy +2

In the Znventum Novum, Part 1. paragraphs 26, 40 (Oeuvres de Fermat, 111. pp. 341
-2, 349-50) is given Fermat's solution of a third case in which"
= : én
are both squares.
SF
t E én
This depends on the Lemma: Tq find a rational right-angled triangle in which
I: I
G (e+ on) - Sip=n square.
Form a right-angled triangle from x +1, 1; the sides are then
ator ta, aitaov, ayia,
. We must therefore have ll
(2+ 22 +2) (a+ 3x +1) — (x +1) (22 + 2x) =a square,
or a+ 423+ 622+ 6x + 2 =a square
= (22+ 2x +1)? say.

I : = . :
Therefore x= = and the triangle has one of its sides x?+ 2x negative. Instead

. J I :
therefore of forming the triangle from ror from 1, 2, we form it from x +1, 2 and

repeat the operation. The sides are then
x2t+2x+5, x*+2x-3, 4x+4,
and we have
(224+ 2x +5) (22+ 4x — 1) — (22+ 3) (22+ 224 — 3) =a square,
or xt 4 4x3 + 6x2 + 20x + 1 =a square

= (1 + 10x — x2)? say,
240 THE ARITHMETICA

16. To find a right-angled triangle such that the number
representing the (portion intercepted within the triangle of the)
bisector of an acute angle is rational’.

A

4x
C 3—3z DZ B
Suppose the bisector AD = 5x, and one segment of the
base (DB) = 3x; therefore the perpendicular = 4x.
Let the whole base CB be some multiple of 3,say 3; then
(Dm=3 32,

But, since 4D bisects the angle CAB,

dC: CP=4B: BD:
therefore the hypotenuse AC = $(3 — 3%) = 4 — 4x.

2 ; is ; . 2
whence x = 22, and the required auxiliary triangle is formed from 2. 2 or from 29, 12,

the sides being accordingly 983, 697, 696.
(Fermat observes that the same result is obtained by putting y= for x in the’
expression at + 423+ 642+ 6x +2; for we must have :
Arapedped ye =a square (1+ 37-7) say,

: 2 HEE | . . 2
whence =, so that pegs 1 and the triangle is formed from 2, 1 or from
2 a7 12 12

29, 12, as before.)
We now return to the original problem of solving

Lei?
£~ hi=w

£-fy=o

We assume for the required triangle (985x, 697x, 696x) and we have = fn =2425561%

so that
o83w- a must both be squares.
697x — 242556x2
Assume that 6972 — 24255027 = (697x)2,
and we have x —34822=069742,

281 by Ry
1045’ 1045’ 1045)"
[The ¢85x — 24255647 is a square by virtue of the sides 985, 697, 696 satisfying the
985 1 697.696
1045 1° (1045)

whence gmat , and the required triangle is (
. 42

conditions of the Lemma; for 98sx — 242556x%= , which is a square

if 985.1045 ~=.697 . 696 is a square, and 1045=697 + .696.]

1 Why did not Diophantus propound the analogous problem ‘To find a right-angled
triangle such that the sides are rational and the bisector of the right angle is also rational ’?
Evidently because he knew it to be impossible, as is clear when (a, ¢ being the perpen-

diculars) the bisector is expressed as =v (Loria, op. cit. p. 148 7.)
BOOK VI 241

Therefore [by Eucl. 1. 47]
164° — 32x + 16 = 162° + 9,
and x=.

If we multiply throughout by 32, the perpendicular = 28,
the base = g6, the hypotenuse = 100, and the bisector
=25

17. To find a right-angled triangle such that the area added
to the hypotenuse gives a square, while the perimeter is a cube.

Let the area be x» and the hypotenuse some square
minus x, say 16 — x.

The product of the perpendiculars = 2x;

therefore, if one of them be 2, the other is », and the
perimeter = 18, which is not a cube.

Therefore we must find some square which, when 2 is
added to it, becomes a cubel

! “Did Diophantus know that the equation 2?+2=2* only admits of oe solution
#=35, v=3? Probably not” (Loria, 0p. cit. p. 155). The fact was noted by Fermat
(on the present proposition) and proved by Euler.

Euler’s proof (Algebra, Part 11. Arts. 188, 193) is, I think, not too long to be given
here. Art. 188 shows how to find x, y such that ax®+¢y*> may be a cube. Separate
ax? +cy? into its factors a n/a+y J (—¢), xafa—ya/(-¢), and assume

x fatyJ(-d={pJatgJ(-c)}%
xafa-yJ(-)={pa-g/(-a)}
the product (ap? + cg?)? being a cube and equal to ax? + cy

To find values for x and y, we write out the expansions of the cubes in full, and

xna+yN(-c)=ap’Ja+3ap*gN (=) - 3p Na cg*J(-¢)
xna-yN(-d)=ap*Ja-3ap°7 J (=o) = 3cpg* Na+ eg’ (=),
whence x=ap® — 3cpg’,
k y=3ap’q —cq®.
For example, suppose it is required to make x*+4 3? a cube. Here a=1 and ¢=1,
so that x=p%— 39%,
y=3% 4%
while 22+ p2=(p?+¢*3. If now p=2 and ¢=1, we find x=2 and y=11, whence
x24 yt=125=35"%

Now (Art. 193) let it be required to find, if possible, in integral numbers, other squares
besides 25 which, when added to 2, give cubes.

Since 42+ 2 has to be made a cube, and 2 is double of a square, let us first determine
the cases in which x%+ 232 becomes a cube. Here a=1, c=2, so that

x=p3-6pg% y=3p%-124;
therefore, since y= + 1, we must have
38% - 2° or ¢ (347-2?) =*1;
consequently ¢ must be a divisor of 1.

Let, then, ¢=1, and we shall have 3p2—2= *1.

With the upper sign we have 322=3 and, taking p= — 1, we find x= 5; with the lower
sign we get an irrational value of which is of no use.

H. D. : 16
242 THE ARITHMETICA

Let the side of the square be 72 + 1, and that of the cube
m— 1.

Therefore m2 — 3m + 3m — 1 =m? + 2m + 3,
from which 7 is found! to be 4.

Hence the side of the square = 5, and that of the cube = 3.

Assuming now x for the area of the original triangle,
25 — x for its hypotenuse, and 2, x for the perpen-
diculars, we find that the perimeter is a cube.

But (hypotenuse) = sum of squares of perpendiculars;
therefore 2°—s02+0625=2"4 4;

x=2821 and the problem is solved.

18. To find a right-angled triangle such that the area added
to the hypotenuse gives a cube, while the perimeter is a square.

Area x, hypotenuse some cube 7zzn2us x, perpendiculars z, 2.

Therefore we have to find a cube which, when 2 is added
to it, becomes a square.

Let the side of the cube be m2 ~1.

Therefore #2 — 3m* + 3m + 1 = a square = (1{m + 1), say.

Thus 7 = 21, and the cube = (41)? = 4313,

Put now x for the area, x, 2 for the perpendiculars, and
Ae for the hypotenuse ;

 

and x is found from the equation (4 3 EEA,
24121 a 4153953
[x= 28121 and the triangle is (2, 25121185 a 8).]

19. To find a right-angled triangle such that its area added to
one of the perpendiculars gives a square, while the perimeter is
a cube.

Make a right-angled triangle from some indeterminate odd
number? say 2x + 1;

then the altitude = 22+ 1, the base = 222+ 2x, and the
hypotenuse = 22% + 2x + 1.

 

It follows that there is no square except 25 which has the required property.

Fermat says (‘‘Relation des nouvelles découvertes en la science des nombres,”
Oeuvres, 11. pp. 433—4) that it was by a special application of his method of descente,
such as that by which he proved that a cube cannot be the sum of two cubes; that he proved
(1) that there is only one integral square which when increased by 2 gives a cube, and
(2) that there are only two squares in integers which, when added to 4, give cubes. The
latter squares are 4, 121 (as proved by Euler, 4/gebra, Part 11. Art. 192).

1 See pp. 66, 67 above.

? This is the method of formation of right-angled triangles attributed to Pythagoras.
If m is any odd number, the sides of the right-angled triangle formed therefrom are nz,

2 2
= (m= 7) ~ (m+ 1), for m2 + i (722 of = it (72 + of . Cf. Proclus, Comment.
on Eucl. 1. (ed. Friedlein), p. 428, 7sqq., etc. etc.
BOOK VI ; 243

Since the perimeter =a cube,
42+ 6x +2= (47+ 2) (r+ 1)=a cube;
and, if we divide all the sides by x + 1, we have to make
4x + 2 a cube.
Again, the area + one perpendicular = a square.
283-32 y ov 1
(r+ 1) t xr+1
28 4 5a + 4x + 1
Bor +1
But 42 + 2 =1a cube;
therefore we must find a cube which is double of a
‘square ; this is of course 8.
Therefore 4x + 2=38, and z= 13.

The required triangle is CE. z 7) :

Therefore = 9 Square;

that is, = 2x + I = a square.

20. To find a right-angled triangle such that the sum of its
area and one perpendicular is a cube, while its perimeter is a
square.

Proceeding as in the last problem, we have to make
4x + 2 a square
2r+1 a cube b
We have therefore to seek a square which is double of a
cube ; this is 16, which is double of 8.
Therefore 4x + 2 = 16, and x= 3d.
The triangle is (%, 2 %) .
9: 9 9
21. To find a right-angled triangle such that its perimeter is
a square, while its perimeter added to its area gives a cube.

Form a right-angled triangle from x, I.

The perpendiculars are then 2x, 22 — 1, and the hypotenuse
bo

Hence 22? + 2x should be a square,
and 2* + 22° +x a cube.

It is easy to. make 2x*+ 2x a square; let 22% 4 2x = mr?
therefore x=2/m’=2)

By the second condition,

3 8 2
Ente a toon must be a cube,
; 27
2.0, Zr = 8 cube
(m2® — 2)

16—2
244 THE ARITHMETICA

Therefore 2m*=a cube, or 2m =a cube = 3, say.

Thus m=4, x=%=1%, **=4.

But one of the perpendiculars of the triangle is x*— 1, and
we cannot subtract 1 from J.

Therefore we must find another value for x greater than 1 ;
hence 2 id

And we have therefore to find a cube such that 1 of the
square of it is greater than 2, but less than 4.

If 22 be this cube,

2<i®< 4,

or S3< 28 < 16

This is satisfied by 2% = 222, or #2 =2T.

Therefore m= 2%, m*=122, and xr=§12, the square of
which is > 1.

Thus the triangle is known [1224 215055 309233]

22. To find a right-angled triangle such that its perimeter is
a cube, while the perimeter added to the area gives a square.

(1) We must first see how, given two numbers, a triangle
may be formed such that its perimeter = one of
the numbers and its area = the other.

Let 13, 7 be the numbers, 12 being the perimeter, 7 the

area.

Therefore the product of the two perpendiculars

I
= 14 — . 14x.
If then =, 14x are the perpendiculars,
hypotenuse = perimeter — sum of perps. = 12 - — 14%.
Therefore [by Eucl. 1. 47]

! 2 24 A 2.
BY 19642 + 172 3302 =_; + 19622;

that is, 172= 3362+ 4,
or 172% = 3362 + 24.

This equation gives no rational solution, because 86*— 24 . 336
is not a square. :
Now 172 = (perimeter)® + 4 times area,
24.336 = 8 times area multiplied by (perimeter)?
(2) Let now the area =m, and the perimeter = any
number which is both a square and a cube, say 64.
BOOK VI 245

Therefore { (04* + 4m)}* — 8.64%. m must be a square,

or 4m* — 24576m + 4104304 = a square.

Therefore m2 — 6144m + 1048576 = a square

Also m+ 64 =a iE :

To solve this double-equation, multiply the second by
such a number as will make the absolute term the
same as the absolute term in the first.

Then, if we take the difference and the factors as usual,
the equations are solved.

[After the second equation is multiplied by 16384, the
double-equation becomes

mm? — 6144m + 1048576 = a square)
16384 + 1048576 = a iL ?

The difference is m?*— 22528m.

If 1m, 1 — 22528 are taken as the factors, we find 7 = 7680,
which is an impossible value for the area of a right-
angled triangle of perimeter 64.

We therefore take as the factors 11m, {4m — 2048 ; then,
when the square of half the difference is equated to
the smaller of the two expressions to be made squares,
we have

(89m + 1024) = 16384m + 1048576,

1
= 39424
and m = 33424,

: Joe I
Returning now to the original problem, we put oo 2x

for the perpendicular sides of the required triangle,
and we have

(64 LF 2m) =i + 4m’x?
x 25 ?
which leads, when the value of ms is substituted, to
the equation
788484% — 8432x + 225 = 0.
The solution of this equation is rational, namely
2h 23 2% or
9856 448 176
Diophantus would of course use the first value, which
would give (48 116 5368) as the required right-
angled triangle. The second value of x clearly gives
the same triangle.]

 
246 THE ARITHMETICA

23. To find a right-angled triangle such that the square of its
hypotenuse is also the sum of a different square and the side of
that square, while the quotient obtained by dividing the square
of the hypotenuse by one of the perpendiculars of the triangle is
the sum of a cube and the side of the cube.

Let one of the perpendiculars be x, the other 22
Therefore (hypotenuse)*=the sum of a square and its

. Bt 2
side ; also

 

= 2% + 2+ = the sum of a cube and its

side.
It remains that 2* 4+ 2* must be a square.
Therefore 2* +1 = a square = (¥ — 2}, say.
Therefore x = 4, and the triangle is found [£, i%, 13].

24. To find a right-angled triangle such that one perpendicular
is a cube, the other is the difference between a cube and its side,
and the hypotenuse is the sum of a cube and its side.

let the hypotenuse be 2*+x, and one perpendicular
r—r ;

Therefore the other perpendicular = 24%?=a cube = 23, say.

Thus z= 2, and the triangle is (6, 8, 10).

It is on Bachet’s note to v1. 22 that Fermat explains his method of solving
triple-equations, as to which see the Supplement, Section v.

[No. 20 of the problems on right-angled triangles which Bachet
appended to Book vi. (“To find a right-angled triangle such that its area
is equal to a given number”) is the occasion of Fermat's remarkable note
upon the theorem discovered by him to the effect that Z%e area of a right-
angled triangle the sides of which are rational numbers cannot be a square
number.

This note will be given in full, with other information on the same
subject, in the Supplement. ]
ON POLYGONAL NUMBERS

All numbers from 3 upwards in order are polygonal, containing
as many angles as they have units, eg. 3, 4, 3, etc.

“ As with regard to squares it is obvious that they are such
because they arise from the multiplication of a number into
itself, so it was found that any polygonal multiplied into a
certain number depending on the number of its angles, with
the addition to the product of a certain square also depending
on the number of the angles, turned out to be a square. This
I shall prove, first showing how any assigned polygonal
number may be found from a given side, and the side from
a given polygonal number. I shall begin by proving the pre-
liminary propositions which are required for the purpose.”

1. If there are three numbers with a common difference, then
8 times the product of the greatest and middle + the square of the
least = a square, the side of which is the sum of the greatest and
twice the middle number. :
Let the numbers be AB, BC, BD in the figure, and we
have to prove 848 .BC + BD*=(AB+ 2BC).
5 AD B

 

By hypothesis 4C= CD, AB=BC + CD, BD=BC—- CD.
Now 8A48.8C=4A48.8C+WQBC+48C.CD).
Therefore 845.BC+ BD?
=4AB.BC+4BC*+(4BC. CD + BD?
=4A8.BC+4BC*+ AB", [Eucl. 11. 8]
and we have to see how AB? + 4AB. BC+ 4BC* can
.be made a square.
[Diophantus does this by producing BA to FE, so that
AE = BC, and then proving that
AB + 44AB . BC+ 4BC*=(BE+ EAY]
It is indeed obvious that
AB: + 4AB. BC+ 4BC*=(AB +2BCr

2. If there are any numbers, as many as we please, in A.P,,
the difference between the greatest and the least is equal to the
common difference multiplied by the number of terms less one.
248 ON POLYGONAL NUMBERS

[That is, if in an AP. the frst term is a, the common
difference 6 and the greatest term /, 7 being the
number of terms, then

[—a=(n—1)0.]
Let AB, BC, BD, BE have a common difference.

A ewe B
Now AC, CD, PF are all equal.
Therefore EA = AC x (number of terms AC, CD, DE)
= A C x (number of terms in series — 1).

3. If there are as many numbers as we please in AP, then
(greatest + least) x number of terms = double the sum of the
terms.

[That is, with the usual notation, 2s = (/+ a).]
(1) Let the numbersbe A. 5, CD, E, FZ, the number of
them being ever.

A... B.C Di iE e,.D
G L M K y : H

 

Let GH contain as many units as there are numbers,
and let GF, being even, be bisected at X. Divide
GK intounits at Z, #7.

Since F=D=C—4,

Fr A=0+D0,
But Fd =(F+4 A). GCL;
therefore C+D=(F+4).LM.
Similarly E+ B=(F+4). MK.

Therefore, by addition,
A+B+C+D+E+F=(F+4). GK.
Therefore 2(Ad +B +... )=2(F+4).CK

 

= (FAY. CH,
(2) Let the number of terms be odd, the terms being
1, BCE.
A. 8 = DF
F H L K G

Let there be as many units in FG as there are terms,
so that there is an odd number of units.
Let FH be one of them ; bisect ZG at K, and divide #K

into units, at ZL.
ON POLYGONAIL NUMBERS 249

Since E=C=0C- A,
Zt Ad=cl=2C.L8,
Similarly Bt+D=2C LAH.
Therefore A+ EB4+ Ba D=2C, HE
= C. HG.
Also Cel .HE;

therefore, by addition,
A+ B+C+D+ E=C.FG;
and, since 2C = 4 + Z,
2(A+B+C+D+E)y=(A+E). FG.

4. If there are as many numbers as we please beginning with
1 and increasing by a common difference, then the sum of all
x 8 times the common difference + the square of (common
difference — 2) = a square, the side of which diminished by 2
= the common difference multiplied by a number which when
increased by 1 is double of the number of terms.

[The A.P. being 1, 1+ 4,... 1 +(7 — 1), and s the sum,

we have to prove that

5.80 +(b—2r=1{b(2n—1)+2}}

2.0, 8bs = 40% — 4 (b— 2) nb,

or 2s=bn*—(b—2)n

=n{2+(n—1)0}

The proof being cumbrous, I shall add the generalised
algebraic equivalent in a column parallel to the
text] :

Let AB, (CD, EF be the terms in 146, 14+20,1+ 30,....
A.P. after 1.

 

 

 

G
P A B o’
+ }emeememmmmsmmammsmmememmeesece;—————— M
K N
Cr—t D
- 0
x
H
Let GH contain as many units 7

as there are terms including 1.
Difference between £/ and 1 :
= (diff. between AB and 1) x (GH —1). I—1=mn—1)6
[Prop. 2]
250 ON
Put AX, EL, GY each equal
to unity.
. Therefore LE=KB M/A.
Make KN =2, and we have to
inquire whether
(sum of terms) x 8X5 + N5*
={2+ KB(GH+ HM)"
Now sum of terms
=4 (FE+ELY.GH
= (LF +2EL).GH
=3(XDB WH. CH+2GH),
since LF = KB. MHA [above]
Bisecting MH at O, we have
(sum of terms)
sKE CH .HO+ CH.
have therefore to

(Prop. 3]

We
whether
(KB.CH.HOLGH)Y.3KB + NB*
is a square.

Now AB.GH.HO.83KEB

=8CH. HO KB?
s ACH. Hil. KDB

Is then :
4GH. HAM .KB'+3K8.CH + NE®
a square ?

Now S8GH.KB

=AGCM. KB + 4(GH+ HM) KZ,

Also ACH. KB=2NE. XB;
and, adding NVA? the right-hand side
becomes K5%+ KN= {Euel. 11. 7]

Isthen AGH. HM. .KB"

+4(GH+ HM) KB + KB*+ KN?
a square?

Again, KB*+ 4GH.HM . KB?
= RB + 4GH. HA. KB
=(CH + HAY. KB. [Eucl 11.8]

Isthen (CH + AMY. KDB?

+4(GH+ HM)KB + KN?
a square?
Make the number NO’ equal to
(CGH + BMY. KB;

inquire

 

POLYGONAL NUMBERS

Call the expression on
the left-hand side X.

s=3+{+1)n
=f({/—1+2)n
=4{(n—1)bn+ 2n}

72 =

Rtn Sr B04 Bin
+ (fF

= 4gn(n—1)+8bn+(b—2)*

=4n(n—1)08
+4n+(n—1)0+0+ 2°

=(n+n—1250
+4{n+n—1)}bo+ 2°
ON POLYGONAL NUMBERS 251

thus (CH + HMY KB2= N03, as
will be proved later’

Also i 4gNO = 2 . NO . NX, since
VK =2,

Is then NO? + NK?*+ 2NO'. NK
a square ?

Yes; it is the square on KO.

And Therefore

OK-2=NO=KB(GH+ HM), | X={n+n—1)6+2}*

while GH + HM + 1 = (twice number ={(2n—1) b+ 2}~
of terms).

Thus the proposition is proved.

 

“The above being premised, I say that,

[5] If there be as many terms as we please in A.P. beginning
from 1, the sum of the terms is polygonal; for it has as many
angles as the common difference increased by 2 contains units, and
its side is the number of the terms set out including 1.”

The numbers being as set out in the figure of Prop. 4, we
have, by that proposition,
(sum of terms). 8K B+ NB: = KO~>
Taking another unit AP, we have XP = 2, while KV =2;
therefore 2B, BK, BN are in A.P., so that
8PB.BK + NB '=(PB+2KBy%; {Prop. 1]
and PB+a3KB-2=PB+2KB~PK =13KD,
while 3+ 1=2.2, or 3 is one less than the double of 2.
Now, since the sum of the terms of the progression

1 Deferred lemma.
To prove that (GH + HM)?2. KB?
={(GH+ HM). KB}2.
Let a= GH + HM,
B=K2, D a B F
v=\GH+ HN). KB.
Place DZ (equal to a) and ZZ (equal to 8) in a
straight line. iY

Describe squares DH, EL on DE, EF and com-
plete the figure.

 

 

 

 

 

 

Then DE Ef=DH: Hr, G H M
and HE: ER=FF ZL.

Therefore /Z7is a mean proportional between the two squares,
that is DE. FR=1F?3,

or a. B2= (ap)?
252

ON POLYGONAL NUMBERS

including unity satisfies the same formula® [literally
“does the same problem ”] as £25 does,

while PAB is any number and is also always a polygonal,
the first after unity (for A2 is a unit and 428 is the
term next after it), and has 2 for its side,

it follows that the sum of all the terms of the progression
is a polygonal with the same number of angles as £5,
the number of its angles being the same as the
number of units in the number which is greater by 2,
or PK, than the common difference KB, and that its
side is GH which is equal to the number of terms
including I.

And thus is demonstrated what is stated by Hypsicles in
his definition, namely, that,

“If there are as many numbers as we please beginning
from I and increasing by the same common difference,
then, when the common difference is 1, the sum of all
the terms is a triangular number ; when 2, a square;
when 3, a pentagonal number [and so on]. And the
number of the angles is called after the number
exceeding the common difference by 2, and the side
after the number of terms including 1.”

[In other words, if there be an arithmetical progression

I, 1+6, 1+20,...14+(2—1)0,
the sum of the # terms, or 47 {2 + (2 —1) 4}, is the
nth polygonal number which has (& + 2) angles.]

Hence, since we have triangles when the common dif-
ference is 1, the sides of the triangles will be the
greatest term in each case, and the product of the
greatest term and the greatest term increased by I
is double the triangle.

1 Nesselmann (pp. 475-6), exhibits this result thus.

Take the A.P. I, 1406, 1426, ... 1+ (re—1)o.
If s is the sum, 8sbo+ (b—2)2=14b (212-1) +2}
If now we take the three terms é— 2, 4, 4+ 2, also in A.P.,

8b(b+2)+(6—2)2={(b+2)+20}2
=(38+2)%

Now 6&2 is the sum of the first two terms of the first series, and corresponds there-
fore to s when #2=2; and 3=2. 2 — 1, so that 3 corresponds to 27 — 1.

Hence 5s and 4 + 2 are subject to the same law ; and therefore, as 6+2 is a polygonal
number with 4 + 2 angles, s is also a polygonal number (the 72th) with é+ 2 angles.
ON POLYGONAL NUMBERS 253

And, since PB is a polygonal with as many angles as
there are units in it,

and 8PFB.(PB —2)+ (PB —4) = a square (from above,
BK being equal to PS — 2, and NB to PB — 4),

the definition of polygonal numbers will be as follows :

Every polygonal multiplied by 8 times (number of angles
— 2) + square of (number of angles — 4) =a square’.

The Hypsiclean definition and the new one being thus
simultaneously proved, it remains to show how, when
the side is given, the prescribed polygonal is found.

For, having given the side G/ and the number of angles,
we know KB.

Therefore (GH + HM) KB, which is equal to VO’, is also
given ; therefore KO (=NO'+ NK or NO'+ 2) is given.

Therefore £0 is given ;

and, subtracting from it the given square on NAB, we
obtain the remaining term which is equal to the
required polygonal multiplied by 8X5. Thus the
required polygonal can be found.

Similarly, given the polygonal number, we can find its

side CH, Q.E.D,

Rules for practical use.

(1) Zo find the number from the side.

Take the side, double it, subtract 1, and multiply the
remainder by (number of angles — 2). Add 2 to the
product; and from the square of the sum subtract
the square of (number of angles — 4). Dividing the
remainder by 8 times (number of angles — 2), we
have the required number.

! Hultsch points out (art. Diophantos in Pauly-Wissowa’s Real-Encyclopidie der
classischen Altertumswissenschaften) that this formula
82 (a-2)+ (a— 4)*=a square
shows that Diophantus intended it to be applied not only to cases where a is greater than
4 but also where a=4 or less. For 36, as Diophantus must have known, besides being
the second 36-gon, is also a triangle, a square, and a 13-gon, inasmuch as
8.36(3~-2)+(3~4F= 280=172,
8.36(+—2)+(4-4)2= 576=124%
8.36 (13-2) + (13-4)? =3249=357%
And indeed it is evident from Def. 9 of the Arithmetica that (3 — 4)2=1, while it is
equally obvious that (4 — 4)2=o0.
254 ON POLYGONAL NUMBERS

[If P be the nth a-gonal number,
P.8(a—2)+(a—4r={2+(2n—-1)(a- 2),
(24 (en— 1) (a= 2) = (aay
or P= =) 3

(2) Zo find the side from the number.

Multiply the number by 8 times (number of angles — 2);
add to the product the square of (number of angles — 4).
We thus get a square. Subtract 2 from the side of
this square and divide the remainder by (number of
angles — 2). Add 1 to the quotient, and half the
result gives the side required?

[emi (AS Rem leah m2, 0).|
2

 

 

a—2
Given a number, to find in how many ways it can be polygonal.

Let AB be the given number, BC | [Algebraical equivalent.]
the number of angles, and in BC take | Number AB = P.

 

 

 

CD=32,CF =4 Number of angles BC= a.
H ue L
A ! B y 7 GC Y K
G N M
F a

Since the polygonal AB has BC

angles,

(1) 84B8.BD + BE*=asquare= FG? | 8P(a—2)+(a—4)

say. ={2+(2n-1)(a—2)}
Cut off 4H equal to 1; = X? say.

therefore SA B.8D But 8P(a— 2)

=4AH .BD+4(AB+BH)BD. | =4(a-2)+4(2P—1)(a—2)
Make DX equal to 4 (AB + BH), | =2(a—2).2+4(2P-1)(a—2).
and for 4442.80 put 28D. DE. DE=4(CP—-1)

 

1 Fermat has the following note. ‘“ A very beautiful and wonderful proposition which
I have discovered shall be set down here without proof. 77, in the series of natural
numbers beginning with 1, any number n be multiplied into the next following, n+ 1,
the product is twice the nth triangular number; if n be multiplied into the (n+ 1)th
triangular number, the product is three times the nth tetrahedral number; if n be
multiplied into the (1+ 1)th tetrahedral number, the product is four times the nth triangulo-
triangular number [ figured number of sth order]; and so on, ad infinitum. 1 do not
“think there can be, in the theory of numbers, any theorem more beautiful or more
general. The margin is too small, and T am not at liberty, to give the proof.” (Cf.
Letter to Roberval of 4 November 1636, Oeuvres de Fermat, 11. pp. 84, 85.) For a proof,
see Wertheim’s Diophantus, pp. 318-20.
ON POLYGONAL NUMBERS

Therefore
(2) FC=KD. DB +280.DF + BE?

(3) = KD.DB+ BD DE?
[Bucl. 11. 7}
(4) = KB.BD+ DE? [Eucl. 11. 1]
But, since DK =4 (AB + BH),
DE > 44H > 4,
and DC=hal{40r2;
therefore CK > 0D:

Therefore, if DK be bisected at
ZL, Z falls between C and KX:
And, since DX is bisected at Z,
KB.BOALD=105"
whence XB .0D=1B'~101%
Therefore, by (4) above,
(5) FC:=Bl -LD:+DF5
or FG 4pL'=RBL+ DES,
6) or LD DE = LB FG

Again, since ED =D, and DC
is produced to Z,
EL.LCALLi= DL:
therefore EL .LC=DL*— D(C?

=I DE"
7) sl FC
Put. FM =BL {or .BL>FGC,
since FG'+DL'sBL*+ ED?
while DL > EDA,

Therefore FI —-FG2=F£L. LC.

Now, DK being bisected at LZ and
being equal to 4 (AB + BH),
DL =2(A8+ BH)

And DC=240,
Therefore (1 = 454,
or BH=1CL.
But AB {=1)=1EC;
therefore AB=3E1,
while BH =1CL,

Therefore 48. BH = FEL . LC,
or EL .LC=1648.8H.

 

253

X2=402P—-1)(a—2)
+2(a—2).2+(a—4)}
= 4{2F~- 1}(a- 2)
+ (a — 2) + 2?
={4(2P—1)+a—2}(a 2)
+ 22

DIL =202L~1)
[BL=2(2P —-1)+a—-2]

Xt={202P—-1)+a-2}
—-{2(2P—1))* + 2°

{2(2P-1))- 2°

={202P—1)+a—2}—-X"?

[EL=2(2P 1) +2,
CL=20P—1)~12]

(22P-1)+2}{2(2P—1)-2}
= {3(aP~ )+a—2 17

Fl =20P~-1)+a=2

(CL =4(P— 1)

ZL=AP
256 ON POLYGONAL NUMBERS

(8) Therefore 16° (P —1)
1648. BH = MF*— FG? ={2(02P—-1)+a—-2}- X"
(9) = GCM: +2FG.GM. | ={2(GP-1)+a-2-X)
+2X{2(2P—-1)+a—2-X}
Therefore GM is even. [={2(z2P-2)-2(n-1)(a-2)}?
Yet GA be bisected at &. ......... +242 + (or —i1) (a ~ 2)

CE ORS A x{2(2P-2)-2(n-1)(a—2)}]

[Here the fragment ends, and the question of course arises whether
Diophantus ever actually solved the problem of finding in how many
different ways a given number can be a polygonal. Tannery went so far
as to call the whole of the fragment, from and including the enunciation
of the problem, the “vain attempt of a commentator” to solve itl.
Wertheim? has however shown grounds for thinking that Diophantus did
solve the problem and that the fragment is a genuine part of his argument
leading to that result. The equation

8P(a—-2)+(a—4)32={2+(2n-1)(a—2)}
easily reduces (by algebra) to
8P(a—2)=4n(a—2){z2 + (n-1)(a—2)},
or 2P=ni{2+ (n—1)(a—2)}

Wertheim has shown how this result can be obtained by a continuation
of the work, from the point where the fragment leaves off, in the same
geometrical form which is used up to that point®, and how, when the

1 Dioph. 1. pp. 476-7, notes.

2 Zeitschrift fiir Math. u. Physik, hist. litt. Abtheilung, 1897, pp. 121-6.

3 The only thing, so far as I can see, tending to raise doubt as to the correctness of

this restoration is the fact that, supposing it to be required to prove geometrically, from
the geometrical equivalent of
8P(a—2)+(a—4)2={2+ (22-1) (a—2)}%,

that 2P=n{z2+(n-1)(a-12)},
it can be done much more easily than it is in Diophantus’ proposition as extended by
Wertheim.

For let FG=2+(212—1)(a—2). Cut off ZR equal to 2, and produce R/£ to S so that
RS=a-19.

 

 

2n(a—2)
rT Ey
ee : =
hs 7 8
We have now 8P.SR=FG2-SF?
=(SG-SF)-SF?
=85G2-125G. SF.
Bisect SG at 7), and divide out by 4;
therefore 22. SR=S72-ST7.SF
= ST(ST~ SF)
=ST. TT

=ST. (FR+ RT).
Now S7=n.SR, and FR=2, while R7=(n-1). SR=(n-1)(a-2).
It follows that 2P=n{2+ (2-1) (a-2)}.
ON POLYGONAL NUMBERS 25%

formula is thus obtained, it can be used for the purpose of finding the
number of ways in which 2 can be a polygonal number. The portion of
the geometrical argument which has to be supplied is, it is true, somewhat
long, and its length and difficulty may, as Wertheim suggests, account for
the copyist having failed, as it were, to see his way through it and having

stopped through discouragement when he had lost his bearings.
I shall now reproduce Wertheim’s suggested restoration of the rest of
the problem. The figure requires some extension, and I accordingly give

a new one after Wertheim.

 

 

A Ha |

B Bop S L XK.
D

F $ & G N M

The last step in the above fragment is

(9) 2FG.GM + GM*=16A4B. BH.

Bisect GAM in WV,
so that GN = NA7

Therefore, if we divide by 4,
(10) FG. GN + GN*= 448. BH,

or

(11) FN.NG=4A4B. BH.

Put now ZR =248, and BS= GHA,
so that G.S= RN, and we have
FS=FR—-RS=248—- RS,

FN=FR+RN=248 + RA,
GN=RS=24B- FS.
Substituting in (11), we have
(12) (24 B+ RN)(2AB-FS)=4AB. BH,

 

2{2+ (27-1) (a-12)}
{2(2P-2)—2(n—1)(a—2)}
+12(2P-2)—2(n—1)(a-2)?

=16P(P~1)

GN = NVM
=2(P-1)-(n-1)(a— 2)

{2+ (22-1) (a—2)}
{2(P-1)—(n—1)(a—2)}
+ {2 (P= 1)=(r-1)(a 2)
=4P(P-1)

{2P+n(a-2)}

{2(P—1)—(n-1)(a-2)}
= APP 1)

ZS= 27
—{2(P-1)— (n~1)(a—2)}
=2+(n—1)(a—2)
FN=2P+n(a-2), from above
GN=2(P-1)-(n-1)(a-2)
RN=FNV—-24AB=n(a-2)
{2P+n(a—2)}
{2(P—1)-(n—-1)(a-2)}
=P (P~1)
17
258 ON POLYGONAL NUMBERS

 

“or
(13) 4452 24B(FS—- RN) 4.
— RN. FS=4AB*— 4AB. AH. —2P{(n—1)(a—2)+2—n(a-2)}
—n(a-2){(n-1)(a—2)+2}
=A —4P
Therefore
(14) 24B(fS—RN)Y+ RN. FS 2P {2 — (a — 2)}
=aA4B. AH, +n(a—-2){(n-1)(a—2)+2}
=A
or
(15) 2AB (AH + RN-FES)=RN.FS. | 2P(a—2)
=n(a—2){(n-1)a-2)+2}

Now RN = FN - FR=FM—-NM—-FR=FM-LGM- FR
=BL—-YGM—-24B=BD + }DK—-LGM— 248
RD 1 eAB rR EN SAB
=BD+2BH—- LGM,

and HS=FR— RS
=248-3GAM.
Therefore RN —-FS=BD+ 2BH—248
oe =BD— 24H,
and RN —FS+2AH = BD.

Again, we have
RN=BD+2BH-GM=BD +:2BH-}BL+3L1FG

=BD+2BH—-1BD - LDL +31 FG
=31BD + 2BH-1DL +1 FG
=4BD + 2BH— (AB + BH) +3 FG
=3BD+BH— AB +L FG
=3BD—- AH +1 FG
=3(BD+FG-24H).

But, from the rule just preceding this proposition,

FG=BD(2n—1)+2;

therefore BD + FG =2n. BD + 2,
or BD + FG ~ 24 H=2n. BD;
therefore BN=n. Bl),
Accordingly the equation (15) above becomes
(10) 24B.8BD=n.8D. FS, 2P(a— 2)
-or =n(a—2){(n—1)(a—2) + 2}
(17) 24B8=1. FS 2P=n{(n—1)(a—2)+2}

Thus the double of any polygonal number must be divisible by its
side, and the quotient is the number arrived at by adding 2 to the product
of (side — r) and (number of angles — 2).

For a triangular number the quotient is z + 1, and is therefore greater
ON POLYGONAL NUMBERS 259

than the side; and, as the quotient increases by = — 1 for every increase
of 1 in the number of angles (a), it is always greater than the side.

We can therefore use the above formula (17) to find the number of .
ways in which a given number 2 can be a polygonal number. Separate 2 2
into two factors in all possible ways, excluding 1.2/2. Take the smaller
factor as the side (7). Then take the other factor, subtract 2 from it,
and divide the remainder by (2 - 1). If (2-1) divides it without a
remainder, the particular factors taken answer the purpose, and the quotient
increased by 2 gives the number of angles (a). If the second factor
diminished by 2 is not divisible by (z—1) without a remainder, the
particular division into factors is useless for the purpose. The number of
ways In which 2 can be a polygonal is the number of pairs of factors
which answer the purpose. There is always one pair of factors which will
serve, namely 2 and 2 itself.

The process of finding pairs of factors is shortened by the flowing

considerations.
2P=n{{n-1)(a— es +2};

therefore 2Pln=q+an—a—2n,.
Dl
and enh a
nin-1)]
therefore not only 2/2/~ but also 1s must be a whole number and,
as a is not less than 3,
2(P—n)
> or =1,
n(n—1)

and consequently
oe + JC +87)

Thus in choosing values for the factor » we need not go beyond thiat
shown in the right-hand expression.

Example 1. In what ways is 325 a polygonal number?

Here — 1 + ./(1 + 8P)=~-1 + (2601) = = go. ‘Therefore # cannot be
greater than 25. Now 2.325=12.5.5.13, and the only possible values
for »n are therefore 2, 3, 10, 13, 25. The corresponding values for a are
shown in the following table.

 

7 2 5 Io | 13 | 25

e325 ufo l6]s

 

 

 

Esample 2. P=120.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

17—:2
ed ed ee
i ay

or
or

—

—

Beef SEC bm peg pen

CONSPECTIUS OF THE ARITHMETICA

Equations of the first degree with one unknown.
7. x—a=m(x-20).
8. xta=mix+ od)
9. a—x=m(b—x)

10. X+b=m{a—x)

«IL K+ o=wln ~a)
: 30: (os )istisecile iin)

(a+b) x+(b+x)a=2(a+x)b,
(a+b) x+(a+x)b=2(b+ x) ot
Determinate systems of equations of the first degree.

(a> 0).

I... xty=a a-y=4,

2. Xty=a, v=).

4. x—y=a, =m,

3 xty=a x=myrl
Poy I

5. XH y=—a, Ee

I I
6. xt p=ua, Sing r=t

212. mt mam nhs, BSW Y=, (02 5%, HER)
L {3. MtomshhV,=n 0m

(%1> Xay > 02) 50> 22)
X= MYs, Y= 13a, 5 = a

15. X+a=m(y—a), y+b=n(x—05).

16. yrosa at-a=d, Rty=,

17. yYtztw=a stwta=) Wwtaty=4 a tyta=d.
18. ytz-x=a stw—y=b, ty —a=e

10, yi+ztw—x=a, 3tWte-y=b Wwtaty—s=s

x+y +z-w=d.

220, wt ytas=g xty=me yhr=ay.

I
7

I I
. 2l p= == g= qi = 1 X>yY>2)
PEt, yma, a Lope)

wal Lie)=p-[ y )+ (2 )
«13% x (n=re)+ (G+e)=s (r+ & eva)

=2=(52+) + (3540)

Determinate systems of equations reducible to the first degree

x+ytz=a.

. 20. ax=o} fx=0.
J20. xryv=a alo gR=4,

* Probably spurious.
CONSPECTUS: OF ARITHMETICA 261

S30. a=, lr Paley)
32. x=my, 2+ F=n(@=y)
33 x=my A-y=alet rp)
234. =a, Pima lx—y),
2340. Cor. 1. x=my, xy =n (x+y)
Cor. 2. "=m, qp= nim =1
133" x=, Rs,
+ 306. x=my, =a,
“37. x=, Y=alney.
+ 38. a= Peulz-)
. 38. Cor. x=my, a2=ny.
x= ayy als a,
x = gy, = (xt)
= =u, = ple —1)
W. 6% y-y=a F-Psa-pid
wv. 30. r=wm (y+), x=nln+ a), y=p(xr+y)
Determinate systems reducible to equations of second degree.
Livia. x+y=a, x)= 5.
se 3 kT y=aixy=10.
1. 28 a vy=aaiv=s,
v. 1. Yr yP=a aty=d
re 2. X-Y=a x-y=4
W.15 (rx=a (f+)y=6, EL) z=¢
Iv. 34. y+ (p+0)=af-1, 2x + (a+ 0)=F—1, y+ (x +P) =2F-1.
i 35. yp—-(+D=—1, 2x-(Z+0)=F~—1, xy—(x+y)=—1.
v.37. y=mxty+o), sx=nE+y+7), gp=ple+y+2)
Lemmatov. 8. yi=d zsx=l3 ay=e

Pe bd eed pt

Pht bd bet ed

Systems of equations apparently indeterminate but really reduced, by
arbitrary assumptions, to determinate equations of the first degree.
1. 14. xy=m (x+y) [value of y arbitrarily assumed].
wit w=merNoadar=mix-y)y
irk (hl) Be yv=micty)

{mo* (ef 1 34) L—yV=min-—y) [x assumed = 2y].
1. 4% (cf 1.32.) B3v=nlx-3)
iL 5% (cl 1.33) A-F=mlxiy)
IL 7%.  x*—)*=m (x —y) +a [Diophantus assumes x —y = 2].
I I I I 1 :
1 22. Rs Ut impels EAE Et [value of y assumed].
roa I I
4 Lizz Be msgs es yh

m q
I

2

 

I T I
loi Yara ris ;
ar ty z [value of y assumed]

* Probably spurious.
262 CONSPECTUS OF ARITHMETICA

1 1 %, 1
I. 24. a al LR
[value of y + z assumed].

I I
1.328, ses lpreemi=yi-(zew ra)

1 Tr,
—g+-(W+ax+p)=w+-(x+y +2
i 7) Slums)
[value of y + z + 7 assumed].

iw. 17% {cl 1.2.) w= (La+a) + {3 0)

y= (Ly+o)+ (Lera)=s-(G+) (Gr+0)

[ratio of x to y assumed].
Iv Hal =m 5) Laon (a-ta)
33 whoymmlpecy), bmn 2
| Diophantus assumes sy EL).

Indeterminate equations of the first degree. :
Lemma to 1v. 34. ay +(x +3) =a [Solutions év dopiore.
vw, »IV.35 ap-lery= | y practically found in
i» IV. 30. my=min ty) terms of «.]
Indeterminate analysis of the second degree.
%. 3 Zr i=4.
H. 0 +tPr=0+4
I. 10. F—-P=a.
go 17. xta=l c+ i=7~
n.12. a-x=0, i—x=7"
la 1} k—a=v, c-b=0"
IL I4=11.21. ®+y=a, ot8=i yptal=0’
IL. 15=1L 20. Xty=a L-x=9) £=y=7.
1m, 10, s=my, dts, CLy=1
I. 10, B—-Vi=w OF — 2)
IL. 20. Bty=d Vix=2"
{ i. 21. FE =yz1l P-w=il
L222. BrlxtN=0 rites =o
{ IL. 23. (wip =uf Vs y)=0-
11-24. Et ra=05 (tyY +y=9~
{ 0.25. (x+y —n= (B+ fF —-y=7~
i. 26, xytx=, ap ty=7, u+o=a
{ WW. 27. sy—x=17, xy—y=7, uty=a
I: 28. r+ =0 B+ P=00
{ H..20. ¥-—ai=d aly ~-P=l
H.30. xy+{xty=u, gy (x+1)=0"
¥ Probably spurious.
111.

III.

Iv.

|
1V.

"13,
TA
, 16,
1s

19.

20,

21.
22.

23,

CONSPECTUS OF ARITHMETICA 263

ay + (x+y)=0t ay— (x+y) =7% x+y=1w

Pred frtao=d drys

Pom P-n=th y=’

+(x +y+2)=u V+ (x+y+2)=0% 22+ (x+y +2)=w
= (x+y+3)=u} Y—(x+y+2)=0% FF— (x+y +3) =w
(x+y+2)-P= (x+y+2)-y=0% (x+y +2) -F =u’
(x+y +l +x=0F (x+y +2 +y=70, (x+y+2)+z=0"
(x+y+2)—x=0 (X+y+2)—y=1% (x+y +2)? -2=0"
x—{(x+y+2f= y-(@+ty+af=0’, s-(x+ty+ =u"
Zry+z=0 v4e—K=0 s+x—y=10, xty—z=a
Z+y+ asl Y4tusdh ata=0, x+y=wl

R—-y=y—a, ytd, z t= ot y=

x+y+art+a=0 ytita=1 1+x+ta=0, x+y +a=w’
x+y+z—a=0 y+r—a=u, 2+x—-a=0,, x+y-a=w
yz a=1 wmta=tl, Ly +ta=ul

y2—-a=u m—-a=0% xy—a=uw’

Fri x=0t, a+ y=t, Xp te=at

Vi—-x=1f m—-y=tf, Xy—z=ul

r=, wt Y=, xy + 22 = wh

y+ (y+) =u x +(z+x)=0% ay+ (x+y) =n’
yi—(y+3) =u, sx—(2+X)=0" ay—(x+y)=u’

x {xt =0f xy +x=00, xy +y=ut

xy —(x+P)=0’ xy—x=0, xy—y=uwt

72
(oy +x, + x +24 y= I ;
2

(+ Xp +25 +2) + =A)

ia
(Xp + Xp + Ty +X) x x 1 2s

*

70°
{B+ tata ns ut

Bry=u, a+ y=
Bty=y, xty=t

xZ+1=f y+-1=20 atyt r= x-pL1=0s
Bye = (Pp) + (Fo) + (=F) (x55).
x+y+e=l, Sty=, ¥+z=0, 2+x=0
x+y+ta=l B—y=if, Y-z=1f 2-x=0’
PE+ I=, xt I=, Ayr sel :

Ml = Mm = mits 2,

X04 mal wn wy =e a, =k

x=), x=y=if, x-1=0f, y-z=wf (x>y>2)
E+ x=U, Xyp+y=1%, Xy5+s=ut

yr —-R=dh Xyn—y=1t, Xy5- z= ur,

* Probably spurious.
264

-
IV.

FV.

29.

CONSPECTUS OF ARITHMETICA

+++ w+ x+y tz +tw=a.
+++ t— (x+y +s+w) =a.

x+y=1, (x+a)(y+0)=u’

xtytz=a, xp+i=u, xy—-s=0"

x—y=m(y—3), y+ta=0’ z+x=0, x+y=w’
—P=m(y—2), y+ti=u 1+x=0, X+y=w’

v= x—asl, y—a=0, 1—a=w"

y=, Atami, yta=7, 2ta=0"

fta= grass ata=l,

yr+a=u, sx+a=1%, xy+a=w’

x—a=7, ya=s 2—a=/

yE—a=2 m—a=v, ¥y-a=u’ :
rrr img St B= r=,

YL red=0f int f ri =f, Yr af,
E=2=72 y—2=f rnd :
yE—-y—2=0, Ix —2~X=70, Ay—x-y=uh

Vi=g= ut ay? ay —a=wl

Yemmartow 7. xp+ ais i=aS

Vv.

——_ ——_——— —— =
<<

=

9.
IL
IO.
12,
13
14.

21.

22.

w27
. 24.
- 25.

26.

\ 2s
23.

» 30.

uw z
x (w+yen) =|}, Pirlx+y+a)= i

sr@eyrg={).

2 *

por(eryen={2}, mi leey+a= 5),
2
watery ra={o).
w
{ef 11.11.) tym), ata yra=0f
xLptr=1, tamil, ptasl’ sta=w’
XtyRy, xta=0 ytb=2"
atytasl, atasw, yhé=h steal
Xty+tz=a, presi, star, x+y=0’
X+y+s+w=a,
Xty+a=5 prot w=0 nt wta=0, wtxty=2~

nye? al =, EE +P =1707 ERE po = af,
a?yig?— x= 7 al 2a? — 2 — 7? atytz?— 22 = 20>.
a2 alia P= 22 =f, B—atP=a
Yri=uf Zatni=v Fit a=al
YE-1=0 Zxl_n=9 FPl-1=7"
I-Yei=w 1-2 =0 1-3 =u’
Pr ta=, 2+a2+a=0, +P: ta=al
Y+22—a=u F+at—-a=7" FP+yi-a=ut
mx +ny=1, w+a=(x+y)>

Lemna 2 to vi, 12. ex’ +é=ut (where a+b=7°).
Lemma to v1. 15.  ax*—d=u* (where ad®*—4=¢is known).
CONSPECTUS OF ARITHMETICA 263

[n. 18]. ay +x ty=dd wt1=2 (y+ 0).

2
(tr. 16]. —(Ee)=th x—1=2 (y-1).

[rv. 32} wt = 1):

[v.21].

Brr=0 Pri=0h BA tr=al

Indeterminate analysis of the third degree.

po
1v.

Vv.

[1v.

IV.
s 15.
: 10.
17.
13.

—r
<i oo

Vv. 19a.
v.19 0.
V. Ige
V.

[1v. 8].

PD .go =u ove

I0.

12.
18.
24.
235i
206.
2%.
28.
28].

38.

19.

20.

By=u, y=
BLy=gp 2 Pm
Bryl=nl 24 Pup,
x r= wt+y=u
2 + P= wty=u
PL ty
oh a
Pry =yr+x

Really reducible to second
degree.
} the same problem

Bry=f YLasil
x+y=a, y=u'—n
x+y+z=a, ays={{x—p) + (x—2)+(y—2)F (x>y>32).
yt x=ul, pt y=d,
xy—x=u, xy—y=0"
xy +(x+p)=1, xy— (x+y) =7"
RE r 1p) =i
(x+y+2)x=tu(m+1), (x+y+2)y=0% (x+y+2)z=0’.
(x+y+2f+a=u8 (x+y +2 +y=0% (X+y+2P+z=00".
(x+y+af—x=0 (x+y +2f—y=70% (x+y+2P-z=0’.
x—(x+y+2Pf=u’ y—(x+y+2=0 s— (x+y +2 =u’
x+y+z=0 (x+y+3Pf+a=u? (x+y+2P+y=107
(x+y+2l+z=0
x+y+z=0 (x+y+2)—x=u (x+y+2P—y=127
(x+y+2P—z=0
xty+a=0 x—(x+y+aP=1f y—(x+y+2¥=107
z— (x+y +2)P°=w
x+y+tai=a, (X+y+2P+x=0 (X+y+2f+y="7,
(x+y+2l+a=ut
x+y+z=a, (x+y+3f—x=u} (x+y+2P—y=107
(x+y +2 —z=1"
pris, x—(x+y+al=u y—(x+y+3@=0"
z— (x+y +2 =’
od

foal
[1v. 9; 10} (22+ yf = (x +p).

jv. 11]. Py = (x—p).
266 CONSPECTUS OF ARITHMETICA

v.15) Arie f-3=4~
jv. 16] s(t +F + D=4
[v.17 2+ Fa + 3=04
Indeterminate analysis of the fourth degree.
v.20. typist
[v.18]. S++ g=v

Problems of constructing right-angled triangles with sides in rational
numbers and satisfying various other conditions.

[N.B. I shall use «x, y for the perpendicular sides and z for the
hypotenuse in all cases, so that the condition x*+ »*=2* must be under-
stood to apply in every case in addition to the other conditions specified. ]:

Lemma to V. 7. Xy=% =; Ys.

VI. 1. z=x=45 09s.
0 L 2. zha=el rhs
YI. 3. ayta=ut
fo 4 Fxy—a=it
VI. 5. a—jay=
VI. 6. dmye =a,
& 5. lxp-x=a
vi. 8. lay+(x+y)=a.
VI. 3 lay —(x+y)=a
VI. lay +(x +2)=a.
VI. lay —(x +2) =a.
lt Yo Vl. 12. x=0f a—y=1 lap y=nt
VI. 12. Lay+a=et lay y=9t
I 1. 13. 2ay-x=0f day-yp=17
IL 14. lay—a=u:, Jay—x=0"

bo 15. ay tz=d, ay+ta=7~

nerd

1.16. ftu=z, y=
VI. 17: dp +a=2, x+y +a=o.
V1. 13. luoy+a=4, xtyte=7~
V1.:10. Yay ta=0f, x+y te=
VI. 20. lay ta=ut, xtryptz=if
VI. 21. +y+z=0% Lay +(x+y+2)=10~
Vi. 22. X+y+a=u day (uty + a)=0~
L.23. Z=seita Plast
Vi. 24. z=ilty x=0-9 p=
(v1.6,77 QaY+imup=n’
[vi. 8,9] {}(x+2)12+ dmxy=0u
[vi 10, 11] : (z+ he F may = ut.
[v.12]. yilx=-7).Yav=2 x=0" (x>3).
[Vi. 14, 15]. vmx —day.x(z—x)=2* (#F<or> fay).
SUPPLEMENT

ADDITIONAL NOTES, THEOREMS AND PROBLEMS BY FERMAT,
TO WHICH ARE ADDED SOME SOLUTIONS BY EULER

I HAVE generally referred to the notes of Fermat, and allied propositions
of his, on the particular problems of Diophantus which were the occasion
of such notes, illustrations or extensions; but there are some cases where
the notes would have been of disproportionate length to give in the places
where they occur. Again, some further explanations and additional
theorems and problems given by Fermat are not in the notes to Diophantus
but elsewhere, namely in his correspondence or in the Doctrinae Analyticae
Inventum Novum of Jacques de Billy “based on various letters sent to
him from time to time by Pierre de Fermat” and originally included at the
beginning of the 2nd (1670) edition of Bachet’s Diophantus (the Znventum
Novum is also published, in a free French translation by Tannery, in
Oeuvres de Fermat, Vol. uL pp. 323-398). Some of these theorems and
problems are not so closely connected with particular problems in Dio-
phantus as to suggest that they should be given as notes in one place
rather than another. In these circumstances it seemed best to collect the
additional matter at the end of the book by way of Supplement.

In the chapter on the Porisms and other assumptions in Diophantus
(pp. 106-110 above) I quoted some famous propositions of Fermat on the
subject of numbers which are the sums of two, three or four square numbers
respectively. The first section of this Supplement shall be devoted to
completing, so far as possible, the story of Fermat's connexion with these
theorems.

SECTION. 1.

ON NUMBERS SEPARABLE INTO INTEGRAL SQUARES.

As already noted, Fermat enunciated, on Diophantus 1v. 29, a very
general theorem of which one part states that Zwvery number is either a
square or the sum of two, three or four squares. We shall return to this
later, and shall begin with the case of numbers which are the sum of
two squares.
268 SUPPLEMENT

1. On numbers which are the sum of two squares.

I may repeat the beginning of the note on 11. 19 already quoted (p. 106).

«A prime number of the form 4» + 1 is the hypotenuse of a right-angled
triangle in one way only, its square is so in two ways, its cube in three, its
biquadrate in four ways, and so on ad infinitum.

“The same prime number 47 + 1 and its square are the sum of two
squares in one way only, its cube and its biquadrate in two ways, its fifth
and sixth powers in three ways, and so on ad infinitum.

“If a prime number which is the sum of two squares be multiplied into
another prime number which is also the sum of two squares, the product
will be the sum of two squares in two ways ; if the first prime be multiplied
into the square of the second prime, the product will be the sum of two
squares in three ways; if the first prime be multiplied into the cube of the
second, the product will be the sum of two squares in four ways, and so on
ad infinitum.”

Before proceeding further with this remarkable note, it is natural to
ask how Fermat could possibly have proved the general proposition that
(a) Every prime number of the form an + 1 is the sum of two square
numbers, which was actually proved by Euler'. Fortunately we have
in this case a clear statement by Fermat himself of the line which his
argument took. In his ¢ Relation des nouvelles découvertes en la science
des nombres” sent by Fermat to Carcavi and shortly after (14 August,
1659) communicated by the latter to Huygens, Fermat begins by a refer-
ence to his method of proof by indefinite diminution (descente infinie or
indéfinte) and proceeds? thus: “I was a long time before I was able to
apply my method to affirmative questions because the way and manner
of getting at them is much more difficult than that which I employ with
negative theorems. So much so that, when I had to prove that every
prime number of the form an + 1 is made up of two squares, I found myself
in a pretty fix. But at last a certain reflection many times repeated gave
me the necessary light, and affirmative questions yielded to my method,
with the aid of some new principles by which sheer necessity compelled me
to supplement it. This development of my argument in the case of these
affirmative questions takes the following line: if a prime number of the
form 47 + 1 selected at random is not made up of two squares, there will
exist another prime number of the same sort but less than the given
number, and again a third still smaller and so on, descending ad infinitum,
until you arrive at the number 5 which is the smallest of all numbers of

U Novi Commentarii Academiae Petropolitanae 1752 and 1753, Vol. 1v. (1758),
PP. 3-40, 1754 and 17353, Vol. Vv. (1760), pp. 3-58= Commentationes arithmeticae
collectae, 1849, 1. pp. 155-173 and pp. 210-233.

2 Oeuvres de Fermat, 11. p. 432.
THEOREMS AND PROBLEMS BY FERMAT 269

the kind in question and which the argument would require 7o# to be made
up of two squares, although, in fact, it is so made up. From which we
are obliged to infer, by reductio ad absurdum, that all numbers of the kind
in question are in consequence made up of two squares.”

The rest of the note to Diophantus 111. 19 is as follows.

“From this consideration it is easy to deduce a solution of the problem

“70 find in how many ways a given number can be the hypotenuse of
a right-angled triangle.

“Take all the primes of the form 47 + 1, e.g. 5, 13, 17, which measure
the given number.

“If powers of these primes measure the given number, set out the
exponents of the powers; eg let the given number be measured by the
cube of 5, the square of 13, and by 17 itself but no other power of 17;
and set out the exponents in order, as 3, 2, 1.

“Take now the product of the first of these and twice the second, and
add to the product the sum of the first and second : this gives 17. Multiply
this by twice the third exponent and add to the product the sum of 17 and
the third exponent: this gives 52, which is ke number of the different right-
angled triangles which have the given number for hypotenuse. [If a, b, ¢ be
the exponents, the number of the triangles is 4adc + 2 (bc + ca + ab) +a + b + ¢.]
We proceed similarly whatever the number of divisors and exponents.

“Other prime factors which are not of the form 4+ 1, and their
powers, do not increase or diminish the number of the right-angled triangles
which have the given hypotenuse.

“PROBLEM 1. 70 find a number which is a hypotenuse in any assigned
number of ways.

“Let the given number of times be 7. We double 7: this gives 14.
Add 1, which makes 15. Then seek all the prime numbers which measure
it, ze. 3 and 5. Next subtract 1 from each and bisect the remainders.
This gives 1 and 2. [In explanation of the process it is only necessary to
observe that, for example, 2 {4abc+ 2 (bc+ca + ab) +a+b+c}+ 1 1s equal
to (2a+1)(26+1)(2¢+1), and so on.] Now choose as many prime
numbers of the form 47 + 1 as there are numbers in the result just arrived
at, 7.2. in this case two. Give to these primes the exponents 1, 2 re
spectively and multiply the results, ze. take one of the primes and multiply
it into the square of the other.

“It is clear from this that it is easy to find the smallest number which
is the hypotenuse of a right-angled triangle in a given number of ways.”

[Fermat illustrates the above further in a letter of 25 December 1640
to Mersenne'. 7

70 find a number which is the hypotenuse of 367 different right-angled
triangles and no more.

1 Oeuvres de Fermat, 11. pp. 214 sq.
270 SUPPLEMENT

Double the number and add 1; this gives 735. Take all the divisors
which are prime numbers: these are 3, 5, 7, 7. Subtract 1 from each and
then divide by 2; this gives 1, 2, 3, 3. We have then to take four prime
numbers of the form 47 + 1 and give them 1, 2, 3, 3 respectively as ex-
ponents. The product of these powers is the number required.

To find the least such number, we must take the four least primes of the
form 47 + 1, i.e. 5, 13, 17, 29, and we must give the smallest of them,
in order, the largest exponent ; ie. we must take 5° 13% 17” and 29 in this
case, and the product of these four numbers is the least number which is
the hypotenuse of 367 right-angled triangles and no more. -

If the double of the given number + 1 is a prime number, then there is
only one possible divisor. Suppose the given number is 20; the double
of it plus 1 is 41. Subtracting unity and bisecting, we have zo, so that the
number to be taken is some prime number of the form 47 + 1 to the power
of 20.]

“PROBLEM 2. 70 find a number which shall be the sum of two squares
in any assigned number of ways.

“Let the given number be 10. Its double is 20, which, when separated
into its prime factors, is 2.2.5. Subtract 1 from each, leaving 1, 1, 4.
Take three different prime numbers of the form 47 + 1, say 5, 13, 17, and
multiply the biquadrate of one (the exponent being 4) by the product
of the other two. The result is the required number.

“By means of this it is easy to find the smallest number which is the
sum of two squares in a given number of ways.

“In order to solve the converse problem,

“Tov find in how many ways a given number is the sum of two squares,
“let the given number be 325. The prime factors of the form 47 + 1
contained in this number are 5, 13, the latter being so contained once only,
the former to the second power. Set out the exponents 2, 1. Multiply
them and add to the product the sum of the two: this gives 5. Add 1,
making 6, and take the half of this, namely 3. This is the number of ways
in which 325 is the sum of two squares.

“If there were three exponents, as 2, 2, 1, we should proceed thus.
Take the product of the first two and add it to their sum: this gives 8.
Multiply 8 into the third and add the product to the sum of 8 and the
third: this gives 17. Add 1, making 18, and take half of this or 9. This
is the number of ways in which the number taken in this second case is
the sum of two squares. [If a, 4, ¢ be the three exponents, the number
of ways is §{abc+ (bc+ca+ab)+ (a+b +c)+ 1} provided that the number
represented by this expression is an integer. |

“If the last number which has to be bisected should be odd, we
must subtract 1 and take half the remainder.
THEOREMS AND PROBLEMS BY FERMAT 271

“But suppose we are next given the foilowing problem to solve:

“70 find a whole number which, when a given number is added fo if,
becomes a square, and which is the hypotenuse of any assigned number of
right-angled triangles.

“This is difficult. Suppose eg. that a number has to be found which
is a hypotenuse in two ways and which, when 2 is added to it, becomes
a square. J ;

“The required number will be 2023, and there are an infinite number
of others with the same property, as 3362 etc.”

2. On numbers which cannot be the sum of two squares.

In his note on Diophantus v. g Fermat took up a remark of Bachet’s
to the effect that he believes it to be impossible to divide 21 into two
squares because ‘it is neither a square nor by its nature made up of two
squares.” Fermat's note was: “The number 21 cannot be divided into
two squares (even) in fractions. That I can easily prove. And generally
a number divisible by 3 which is not also divisible by g cannot be divided
Into two squares either integral or fractional.”

He discusses the matter more generally in a letter of August 1640
to Roberval’.

“I have made a discovery a propos of the 12th [9th] proposition of
the fifth Book of Diophantus (that on which I have supplied what Bachet
confesses that he did not know and at the same time restored the corrupted
text, a story too long to develop here). I need only enunciate to you my
theorem, while reminding you that I proved some time ago that

“A number of the form an — 1 is neither a square nor the sum of two
squares, either in integers or fractions.”

[This proposition was sent by Mersenne to Descartes, on 22 March
1638, as having been proved by Fermat.]

“For the time I rested there, although there are many numbers of the
form 472 + 1 which are not squares or the sums of squares either, e.g. 21,
33, 77, etc., a fact which made Bachet say on the proposed division of 21
into two squares ‘It is, I believe, impossible since 21 is neither a square
nor by its nature made up of two squares,” where the word 7eor (I think)
clearly shows that he was not aware of the proof of the impossibility.
This 1 have at last discovered and comprehended in the following general
proposition. :

“If a given number is divided by the greatest square which measures it,
and the quotient is measured by a prime number of the form an — 1, the given
number is neither a square nor the sum of two squares cither integral or
Jractional.

L Oeuvres de Fermat, 11. pp. 203—4.
272 SUPPLEMENT

“ ExaMPLE. Let the given number be 84. The greatest square which
measures it is 4, and the quotient is 21 which is measured by 3 or by
7, both 3 and 7 being of the form 42 — 1. I say that 84 is neither a square
nor the sum of two squares either integral or fractional.

“Let the given number be 77. The greatest square which measures it
is 1, and the quotient is 77 which is here the same as the given number
and is measured by 11 or by 7, each of these numbers being of the form
4n—1. I say that 77 is neither a square nor the sum of two squares,
either in integers or fractions.

“I confess to you frankly that I have found nothing in the theory of
numbers which has pleased me so much as the proof of this proposition,
and I shall be glad if you will try to discover it, if only for the purpose
of showing me whether I think more of my discovery than it deserves.

‘““ Following on this I have proved the following proposition, which
is of assistance in the finding of prime numbers.

“If a number is the sum of two squares prime to one another, I say

that it cannot be divided by any prime number of the form an — 1.
“For example, add 1, if you will, to an even square, say the square
10 000 000 000, making 10 000 000 oor. I say that 10000 005001 cannot
be divided by any prime number of the form 42 — 1, and accordingly,
when you would try whether it is a prime number, you need not divide by
3. J, LI ete.”

(The theorem that Numbers which are the sum of two squares prime to
one another have no divisors except such as are likewise the sum of two squares
was proved by Euler'.)

3. Numbers (1) which are always, (2) which can never be, the sum
of three squares.

(x) Zhe number which is double of any prime number of the form
8n— 1 is the sum of three squares (Letter to Kenelm Digby of June 1658)

L.g. the numbers 7, 23, 31, 47 etc. are primes of the form 82—1; the
doubles are 14, 46, 62, 94 etc.; and the latter numbers are the sums of
three squares.

Fermat adds “I assert that this proposition is true, though I do so in
the manner of Conon, an Archimedes not having yet arisen to assert it
or prove it.”

Lagrange® remarks that he has not yet been able to prove the pro-
position completely. The form 8% — 1 reduces to one or other of the three

V Novi Commentarii Acad. Petropol. 1752 and 1753, Vol. 1v. (1758), pp. 3-40=
Commentationes arithmeticae, 1. pp. 155-173.

2 Oeuvres de Fermat, 11. pp. 402 sqq.

8 «“Recherches d’Arithmétique” in Berlin Mémoires 1773 and 1775= Oeuvres de
Lagrange, 111. p. 795.
THEOREMS AND PROBLEMS BY FERMAT 473

forms 2472 — 1, 247 + 7, 247 + 135, of which the first two only are primes.
Lagrange had previously proved that every prime number of the form
24n + 7 1s of the form x? + 6y% The double of this is 24% + 123% and

22° + 1297 = (x + 29)’ + (x — 2)* + (27),
that is, 24° + 12)? is the sum of three squares.

The theorem was thus proved for prime numbers of the form 8x — 1,
wherever 7 is not a multiple of 3, but not for prime numbers of the form
247 — 1.

Legendre, however, has the theorem that Every number which is the
double of an odd number is the sum of three squares.

(2) No number of the form 24n +7 or 4™(24n +7) can be the sum
of three squares.

This theorem is substantially stated in Fermat’s note on Dioph. v. 11.
We may, as a matter of fact, substitute for the forms which he gives the
forms 87 + 7 and 4™ (8x + 7) respectively.

Legendre® proved that numbers of the form 8x + 47 are the only odd
numbers which are not the sum of three squares.

4. Every number is either a square or the sum of two, three or
Jour squares.

This theorem is also mentioned in the “Relation des nouvelles dé-
couvertes en la science des nombres” already quoted, as a case to which
Fermat ultimately found himself able to apply the method of proof by
descente. He says® that there are some other problems which require new
principles in order to enable the method of descente to be applied, and the
discovery of such new principles is sometimes so difficult that they cannot
be arrived at except after very great trouble.

“Such is the following question which Bachet on Diophantus admits
that he could never prove, and as to which Descartes in one of his letters
makes the same statement, going so far as to admit that he regards it as
so difficult that he does not see any means of solving it.

« Every number is a square or the sum of two, three or four squares.

“I have at last brought this under my method, and I prove that, if
a given number were not of this nature, there would exist a number smaller
than it which would not be so either, and again a third number smaller
than the second, etc. ad infinitum ; whence we infer that all numbers are
of the nature indicated.”

In another place (letter to Pascal of 25 September, 1654)*, after quoting
the more general proposition, including the above, that every number 1s

1 Legendre, Zaklentheorie, tr. Maser, 1. p. 387.
2 bid. p. 386.

3 Qeuvres de Fermat, 11. p. 433.

3 Lid. p. 313:
274 SUPPLEMENT

made up (1) of one, two, or three triangles, (2) of one, two, three or four
squares, (3) of one, two, three, four or five pentagons, and so on ad infinitum,
Fermat adds that “to arrive at this it is necessary—

(1) To prove that every prime number of the form 4n+ 1 is the sum
of fwo squares, e.g. 5, 13, 17, 29, 37. €¢. ;

(2) Given a prime number of the form 47 + 1, as 53, to find, by a
general rule, the two squares of which it is the sum.

(3) Every prime number of the form 3n+ 1 is of the form x* + 3),
eg 7 13y 10, 3% 37, Cc

(4) Every prime number of the form 8n + 1 or 8n + 3 is of the form
Atay? a0, 11,175, 10, 41, 43, 0

(5) Zhere is no rational right-angled triangle in whole numbers the
area of which is a square.

“This will lead to the discovery of many propositions which Bachet
admits to have been unknown to him and which are wanting in Diophantus.

“I am persuaded that, when you have become acquainted with my
method of proof in this kind of proposition, you will think it beautiful, and
it will enable you to make many new discoveries, for it is necessary, as you
know, that multi pertranseant ut augeatur scientia [Bacon].”

Propositions (3) and (4) will be mentioned again, and a full account
will be given in Section 111. of this Supplement of Fermat's method, or
methods, of proving (5).

The main theorem now in question that every integral number is the
sum of four or fewer squares was attacked by Euler in the paper! (1754—
1755) in which he finally proved the proposition (1) above about primes
of the form 47 + 1; but, though he obtained important results, he did not
then succeed in completing the proof. Lagrange followed up Euler's
results and finally established the proposition in 1770% Euler returned
to the subject in 1772; he found Lagrange’s proof long and difficult, and
set himself to simplify it?

(The rest of the more general theorem of Fermat quoted above, the
portion of it, that is, which relates to numbers as the sum of # or fewer
n-gonal numbers, was proved by Cauchy.)

! Novi Commentarii Acad. Petropol. for 1754-5, Vol. Vv. (1760), pp. 3-58= Com-
mentationes arithmeticae collectae, 1849, 1. pp. 210-233.

2 Nouveaux Mémoires de I’ Acad. Roy. des Sciences de Berlin, année 1770, Berlin 1772,
pp. 123-133 = Oeuvres de Lagrange, 111. pp. 187-201: cf. Wertheim’s account\in his
Diophantus, pp. 324-330.

# “Novae demonstrationes circa resolutionem numerorum in quadrata,” Acta Erudit.
Lips. 1773, p- 193; Acta Petrop. 1. 11. 1775, p. 48; Comment. arithm. 1. pp. 538-548.

# Cauchy, “ Démonstration du théoréme général de Fermat sur les nombres polygones,”
Oeuvies, 11¢ Série, Vol. VI. pp. 320-333. See also Legendre, Zahlentheorie, tr. Maser,
11. Pp. 332-343:
THEOREMS AND PROBLEMS BY FERMAT 275

Under this heading may be added the further proposition that

“Any number whatever of the form 8n— 1 can only be represented as the
sum of four squares, not only in integers (as others may have seen) but in
Jractions also, as 1 promise that I will prove.” ;

5. On numbers of the forms x*+ 2) «+ 3y% x* + 53° respectively.

(1) Zwery prime number of the form 8n+1 or 8n+3 is of the form
22+ of

This is one of the theorems enunciated in the letter of 25 Sept., 1654,
to Pascal? and also in the letter of June, 1658, to Kenelm Digby?

[In a paper of 1754 Euler says that he does not yet see his way to
prove either part of the theorem® In 1759 he says® he can prove the
truth of the theorem for a prime number of the form 8x + 1, but not for
a prime of the form 8x +3. Later, however, he proved it for prime
numbers of both forms®. Lagrange” also proved it for primes of the form
87+ 3.]

(2) Every prime number of the form 3n +1 is of the form x°+ 32

The theorem is stated in the same two letters to Pascal and Digby
respectively.

Lagrange naturally quotes it as ‘ All prime numbers of the form 67 + 1
are of the form x? + 33%” for of course 37 + 1 is not a prime number unless
7 1s even.

The proposition was proved by Euler®. Lagrange proved® (a) that all
prime numbers of the form 122 — 35 are of the form x®+ 33? (4) that all
prime numbers of the form 122-1 are of the form 3x*-3% and (c) that
all prime numbers of the form 12x + 1 are of do#/ the forms a* + 3y* and
at — 392,

(3) No number of the form 3n—1 can be of the form x*+ 3)°

In the “Relation des nouvelles découvertes en la science des nombres”
Fermat says that this was one of the negative propositions which he proved
by his method of descente.

1 Letter to Mersenne of Sept. or Oct. 1636, Oeuvres de Fermat, 11. p. 66.

2 Oeuvres de Fermat, 11. p. 313.

® 2bdd. 11. p. 403.

4 «Specimen de usu observationum in mathesi pura (De numeris formae 2aa +44)” in
Novi Commentarii Acad. Petrop. 1756-7, Vol. vi. (1761), pp. 185-230= Comment.
arithm. 1. pp. 174-192.

5 Novi Commentarii Acad. Petrop. 1760-1, Vol. vil. (1763), pp. 126-8 = Comment.
arithm. 1. p. 296.

8 Commentationes arithmeticae, 11. p. 607.

7 ¢ Recherches d’Arithmétique ” in Oeuvres de Lagrange, 111. pp. 776, 784. ;

8 « Supplementum quorundam theorematum arithmeticorum, quae in nonnullis de-
monstrationibus supponuntur (De numeris formae aa+ 368)” in Novi Comment. Acad.
Petrop. 1760-1, Vol. VIIL (1763), pp. 105-128 = Comment. arithm. 1. pp. 287-296.

¥ Op. cit., Oeuvres de Lagrange, 111. pp. 784, 791.

Y Qcuvres de Fermat, 11. p. 431.

18—2
276 SUPPLEMENT

(4) If two prime numbers ending in either 3 or 7 which are also of the
Jorm an + 3 are multiplied together, the product is of the form af 4 gt

This theorem also is enunciated in the letter of June, 1658, to Kenelm
Digby. Fermat instances 3, 7, 23, 43, 47, 67 etc. as numbers of the kind
indicated. Take, he says, two of these, e.¢. 7 and 23. The product 161
will be the sum of a square and 5 times another square, namely 81 + 5. 16.

_ He admits, however, that he has not yet proved the theorem generally:

«J assert that this theorem is true generally, and I am only waiting for
a proof of it. Moreover the square of each of the said numbers is the sum of
a square and 5 times another square : this, too, I should like to see proved.”

Lagrange proved this theorem also’. He observes that the numbers
described are either of the form 207+ 3 or of the form 207+ 7, and he
proves that all prime numbers of these forms are necessarily of the form
2x” + 2x9 + 35°. He has then only to prove that the product of two
numbers of the latter form is of the form «?+ 5)”

This 1s easy, for

(22* + 2xy + 397) (2x2 + 22y + 39") = (2x + xy + 3&5 + 397) + 5 (ay — ya’

6. Numbers of the forms x*— 29° and 2x2—)°

Fermat’s way of expressing the fact that a number is of one of these
forms 1s to say that it is the sum of, or the difference between, the two smaller
sides, 7.e. the perpendicular sides, of a right-angled triangle. Like Diophantus,
he “forms” a rational right-angled triangle from two numbers x, y, taking
as the three sides the numbers &® + 3? a? —j?% 2xy respectively. The sum
therefore of the perpendicular sides is x? + 2xy —3* or (x +3)? — 23% and
their difference is either a?—2xy—3* or 2xy—(x?—3?), that is, either
{2=3)=21 or 2 (x7)

The main theorem on the subject of numbers of these forms is, as
a matter of fact, contained, not in a letter of Fermat’s, but in two letters
of Frénicle to Fermat dated 2nd August and 6th Sept., 1641, respectively?
It is, however, clear (cf. the letter in which Fermat had on 15th June, 1641,
propounded to Frénicle a problem on such numbers) that the theorem was
at any rate common property between the two.

Frénicle’s two statements of the theorem are as follows :

“ Every prime number of the form 8x + 1 is the sum of the two smaller
sides of a (right-angled) triangle, and every number which is the sum of the
two smaller sides of a (right-angled) triangle with sides prime to one another
1s of the form 87 + 1.” 5

“Every prime number of the form 8 + 1, or which is the product
of such prime numbers exclusively, is the difference between the two
smaller sides of an infinite number of primitive right-angled triangles.”

1 Op. cit., Oeuvres de Lagrange, 11. pp. 784, 788-9.
2 Qeuvres de Fermat, 11. pp. 231, 235.
THEOREMS AND PROBLEMS BY FERMAT 277

Lagrange! quotes the theorem in the form

All prime numbers of the form 8n+ 1 are-of the form y* — 2£>

Lagrange himself proves? that all prime numbers of the form 8% — 1 are
of bot the forms x? — 29” and 2a? — 3”, and observes? that this theorem is more
general than that of Fermat so far as prime numbers of the form 82 — 1 are
concerned. This, however, seems scarcely correct if the further explanations
given by Frénicle are taken into account. For Frénicle shows clearly,
in the second of the two letters referred to*, that he was fully alive to
the fact that numbers which are of the form a*— 2)? are also of the form
2x —?; and indeed it is obvious that he was aware that

82-29% =2(x +p): — (x + 2y)%

Lagrange proved in addition? that

Lvery prime number of the form 8n + 1 is at the same time of the three
Jforms a+ 29°, A2—2y} extn

This is, I think, really included in Frénicle’s statements when combined
with Fermat's theorem (1) above to the effect that every prime number
of the form 8x + 1 1s of the form &* + 2)~

The problem propounded by Fermat to Frénicle in connexion with the
numbers now under consideration was: —-

Given a number, to find in how many ways it can be the sum of the two
smaller sides of a right-angled triangle.

Frénicle replied that this involved also the problem of finding a number
which will be the sum of the two smaller sides of a right-angled triangle in
an assigned number of ways and no more, and tried, but unsuccessfully’,
to bring these problems under a rule corresponding to that by which
Fermat found the number of ways in which a prime number of the form
4n + 1 can be the hypotenuse of a right-angled triangle (see p. 269 above),
but with a prime number of the form 8x + 1 substituted for the prime
number of the form 42 + 1. I cannot find that Fermat ever communicated
his own solution, at all events in the correspondence which we possess.

SECTION 11

EQUATION x%— Ay*=1.

History of the equation up to Fermat's time.

Fermat was not the first to propound, or even to discover a general
method of solving, the problem of finding any number of integral values of
x, y satisfying the above equation, wherein 4 is any integral number not

at. ying q y 8
a square. But Fermat rediscovered the problem and was perhaps the first

1 Op. cit., Oeuvres de Lagrange, 111. p. 775. 2 7bid. p. 784. 2 Zid. p. 788.

4 Qeuvres de Fermat, 11. pp. 235-240.

5 0p. cit., Oeuvres de Lagrange, 111. p. 790.

5 See Oeuvres de Fermat, 11. pp. 231, 238 sqq.
278 SUPPLEMENT

to assert that the general solution is always possible whatever be the
(non-square) value of 4. The equation has a history of over 2000 years,
and that history, even in outline, requires, as it has now obtained, a book
to itself’. This note will therefore be confined, practically, to recalling, in
the briefest possible way, the recorded stages anterior to Fermat, and then
to setting out somewhat fully the passages in Fermat's writings which throw
the most light on his connexion with the subject.

‘The Pythagoreans.

We have seen (p. 117 above) that the Pythagoreans had already
discovered a general solution of a particular equation of this type, namely

28 —yi=+1, ;
by which all the successive values of x, y satisfying the equation were

ascertained. If x =p, y = ¢ satisfies the equation 2a? — »* = + 1, they proved
that the equation 2x? — y*=¥ 1 is satisfied by

HLH=P+9 G=20+g,
the equation 24? —)?=+ 1 again by
Lh ron BY tn
and so on. As p=1, ¢=1 satisfies 2a?—»?=+ 1, we have all the suc-
cessive solutions of 2x? — 3? = + 1 by forming ( ,, ¢;), (#1, #2) etc. in accord-
ance with the law.

Archimedes.

The solution of the above equation by the Pythagoreans was evidently
used in order to obtain successive approximations to ,/z.

Consequently, when we find Archimedes giving, without explanation, the
fractions $25 and 42S: as being approximately equal to ,/3, the hypothesis of
Zeuthen and Tannery that he arrived at these approximations by obtaining
successive solutions of equations of a similar form, but with 3 substituted
for 2, is one of the most natural that have been suggested The equations
are in this case

rd
23 ==02.

Tannery shows how the law for forming successive solutions of such
simple cases as these can easily be found when we have found by trial
(which is not difficult) the three simplest solutions. If we take the more
general equation

2—ay=7,

1 H. Konen, Geschichte der Gleichung t* — Di? =1, Leipzig (S. Hirzel), 1gor.

? Zeuthen, ‘“ Nogle hypotheser om Arkhimedes kvadratrodsberegning,” Zidsskrift for
Mathematik, V1. Raekke, 3. Aargang, pp. 1505qq.; P. Tannery, “Sur la mesure du cercle
d’Archimede ” in Mémoires de la soc. des sciences phys. et nat. de Bordeaux, 11 Sér. 1v.,
1882, p. 303; see Giinther, ‘‘ Die quadratischen Irrationalititen der Alten und deren
Entwickelungsmethoden” in Abkandlungen zur Gesch. der Mathematik, Heft 1v. 1882,
pp- 87-91; Konen, op. cit. p. 15.
THEOREMS AND PROBLEMS BY FERMAT 279

of which x = 4, y = ¢ is a known solution, and put
P=ap+ Bg, qi=1p +3,

it is sufficient to know the three simplest solutions in order to find a, 8, ,
8; for, substituting the values of (2, ¢), (1, ¢,) and (py, ¢,) where (2, gs)
are formed from (p,, ¢,) by the same law as (2,, ¢,) are formed from (p, ¢),
we have four simultaneous equations in four unknown quantities. Taking
the particular equation :
BR=3l=1,

we easily find the first three solutions, namely (p=1, g=0), (1 =2, /,=1)
and (p,= 7, ¢,= 4), whence

2=uq0, T=)

j=g0+pB, 4=2y+38,
and e=3, S=3,y=1,8=2 sothat

Li=03 80, i= 20,
But there is evidence that Archimedes dealt with much more difficult
equations of the type, for (as stated above, p. 123) the Cattle-Problem
attributed to him requires us to solve in positive integers the equation

X°— 4729494) = 1.

There is this difference between this equation and the simpler ones
above that, while the first solutions of the latter can be found by trial,
the simplest solution of this equation cannot, so that some general method,
e.g. that of continued fractions, is necessary to find even the least solution
in integers. Whether Archimedes was actually able to solve this particular
equation is a question on which there is difference of opinion; Tannery
thought it not impossible, but, as the smallest values of x, y satisfying the
equation have 46 and 41 digits respectively, we may, with Glinther, feel
doubt on the subject’. There is, however, nothing impossible in the
supposition that Archimedes was in possession of a general method of
solving such equations where the numbers involved were not too great for
manipulation in the Greek numeral notation.

Diophantus.
Tannery? was of opinion that Diophantus dealt with the equation
R—dfi=y
somewhere in the lost Books of the A7ithmetica. Diophantus does indeed
say (Lemma to VI. 15) that, if a, 4 are any numbers and aa®— 4 is a square
when «x is given a certain value p, then other values of x greater than p can
also be found which have the same property ; and Tannery points out that

1 Giinther, 0p. czt., pp. 92-93 note. Cf. Konen, op. cit., p. 14.
2 Tannery, ‘‘L’Arithmétique des Grecs dans Pappus” in Mémoires de la soc. des
sciences phys. et nat. de Bordeaux, 11° Sér. 111., 1880, pp. 370 sq.
280 SUPPLEMENT

we can, by making suppositions of the same kind as Diophantus makes,
deduce a more general solution of the equation

when one solution (2, ¢) is known.
Put pH=mx—p, 1=%X+g,
and suppose
PE Adgi=nit—ampx + Pp — Ax — 2 dgx—- Ag’ = 1;

therefore (since 2? — 4g = 1)
mp + Ag
a

and, by substitution in the expressions for ,, ¢,, we have -
(m*+ A) p + 24mg amp + (m* + A) g
LESEET ar
and in fact p? — Agi=1.
If an integral solution is wanted, one way of obtaining it is to substitute
u[v for m where u*— Av*=1, i.e. where u, v is another solution of the
original equation, and we then have

b= + Av?) p+ 24uvg, ¢,=2puv + (2 + 41%) g.

But this is all that we can get out of Diophantus as we have him, and
it will be observed that here too we must have ascertained two solutions of
the one equation, or one solution of it and a solution of an auxiliary equation,
before we can apply the method®

1 It may be observed that, in the particular case of the equation x2-332=1, the
assumption of #, z satisfying the equation will not enable us to obtain from the formula

P= + AD?) p+2Auvg, qi=2puv+ (12+ 42%) g
above given the simpler formula otherwise obtained by Tannery (p. 279 above), namely
21=20+39, @1=p+29;
for, if ( #1, ¢1) 1s to be a different solution from ( p, ¢), we cannot make #=1, v=o, but
must take #=2, 2=1, whence, putting 4 = 3, we obtain
Pr=TP+ 12, 1=4p +79,
which is the same as py, ¢a, the next solution to py =2p+3¢, ¢1=2 + 29.
In order to get the latter we have to take #, satisfying, not x2 -3y2=1, but

x2—3y2= —2.

 

The values #=1, v=1 satisfy ¥2- 33?= — 2, and
_ (+302) p+6uvg _ 4p+6g 4
hE Tn Ea (22 +39),
_mvp+ (+300) g pag :
q9 ER eee EEE — mn = (p+29);

and of course p1= + (22+ 3¢), g1= + (2 +2¢) can be taken, since they equally satisfy
2 -3t=1.
THEOREMS AND PROBLEMS BY FERMAT 281

The Indian Solution.

If the Greeks did not accomplish the general solution of our equation,
it is all the more extraordinary that we should have such a general solution
in practical use among the Indians as early as the time of Brahmagupta
(born 508 A.p.) under the name of the “cyclic method.” Whether this
method was evolved by the Indians themselves, or was due to Greek
influence and inspiration, is disputed. Hankel held the former view!;
Tannery held the latter and showed how, from the Greek manner of
deducing from one approximation to a surd a nearer approximation, it is
possible, by simple steps, to pass to the Indian method®% The question
presumably cannot be finally decided unless by the discovery of fresh
documents; but, so far as the other cases of solution of indeterminate
equations by the Indians help to suggest a presumption on the subject,
they are, I think, rather in favour of the hypothesis of ultimate Greek
origin. Thus the solution of the equation ax — éy =¢ given by Aryabhata
(born 476 A.D.) as well as by Brahmagupta and Bhaskara, though it
anticipated Bachet’s solution which is really equivalent to our method of
solution by continued fractions, is an easy development from Euchd’s
method of finding the greatest common measure or proving by that process
that two numbers have no common factor (Eucl. viL 1, 2, X. 2, 3)? and
it would be strange if the Greeks had not taken this step. The Indian
solution of the equation xy = ax + dy + ¢, by the geometrical form in which
it was clothed, suggests Greek origin®.

The “cyclic method” of solving the equation
xt—Ay’=1

is found in Brahmagupta and Bhaskara® (born 1114 A.D.) and is well
described by Hankel, Cantor and Konen®.

The method is given in the form of dogmatic rules, without any proof
of the assumptions made, but is equivalent to a preliminary lemma followed
by the solution proper.

1 Hankel, Zur Geschichte der Math. im Allerthum und Mitlelalter, pp. 203-4.

2 Tannery, ‘‘ Sur la mesure du cercle d’Archimede” in Mém. de la soc. des sciences
phys. et nat. de Bordeaux, 11° Sér. 1v., 1882, p. 325; cf. Konen, pp. 27-28; Zeuthen,
¢“L’Oecuvre de Paul Tannery comme historien des mathématiques” in Bibliotheca Mathe-
matica, Viz, 1905—0, pp. 271-273. 2

3 G. R. Kaye, “Notes on Indian mathematics, No. 2, Aryabhata” in Journal of the
Asiatic Society of Bengal, Vol. 1v. No. 3, 1908, pp. 135-138.

4 Cf. the description of the solution in Hankel, p. 199; Cantor, Gesck. d. Math. 13,
p- 631. 2

5 The mathematical chapters in the works of these writers containing the solution in
question are contained in H. T. Colebrooke’s Algebra with arithmetic and mensuration
Jrom the Sanskrit of Brahmegupta and Bhaskara, London, 1817.

6 Hankel, pp. 200-203; Cantor, 13, pp. 632-633; Konen, 0p. cit., pp. 19-26.
282 SUPPLEMENT

Lemma.
If x=p, y=¢ be a solution of the equation

AP+is= a2,
and x =p’, y=¢ a solution of the equation
Ar +s =i,
then, say the Indians, x = pp’ + Agqg', y = pg’ + p'¢ is a solution of the equation
Ayr se wma
In other words, if
Ag: +5 =p
: Ag? +s oh
then A(pg £29)" +55 =(pt' + 497)"

This is easily verified’.
In particular, taking s=s', we find, from any solution x =p, y=¢ of

the equation
AR Lts=a2

a solution x = p* + 4g y= 2p¢ of the equation
Ay: + 2 = a,
Again, particular use of the lemma can be made when s=+1 or s=+ 2.

{a Hs=+1,and x=p, y=¢ is a solution of

AP + 1= 0%
then x = p* + A¢>, y= 2p¢ is another solution of the same equation.
lfs=—1,and x=2, y=¢1sa solution of
Ayi—1=27

then x = p* + 44% y= 2p¢ is a solution of
Ar +1=a7
(0) Hs=%2 0nd x=4, y=gis a solution of
dz =
then x= #2 + 44°, y= 257 is a solution of
Ay: + gq =a
In this case, since 2p¢ is even, the whole result when the values of

x, y are substituted must be divisible by 4, and we have x = § (2* + 44),
¥ =pq as a solution of the equation

i

1 For, since s=22— Ag? s'=p'2— Aq’?
ss'=(p%— Ag?) (p?~ 49?
=(22)2+ (49g): - 4 (pg): ~ 4 (9)?
={(20)=2422'99' +(A99)?} — 4 {(27)** 2897’ + (#'9)?}
=(pp' = Aqq) 2-4 (pg £9'9)%
THEOREMS AND PROBLEMS BY FERMAT 283

Solution proper of the equation x*— Ay*=1.

We take two numbers prime to one another, g, 4, and a third number s
with no square factor, such that

AP + s=p
the numbers being also chosen (in order to abbreviate the solution) such
that s is as small as possible, though this is not absolutely necessary.
(This is a purely empirical matter; we have only to take a rough ap-
proximation to ,/4 in the form of a fraction p/g.)

[It follows that 5, ¢ can have no common factor; for, if § were a
common factor of s, ¢, it would also be a factor of 2% and g? ¢* would have
a common factor. But p, ¢ are prime to one another. ]

Now find a number » such that

a= 217 is a whole number.
[This would be done by the Indian method called cwttaca (* pulveriser ”),
corresponding to our method by continued fractions. ]
Of the possible values of » a value is taken which will make »*— 4
as small as possible.
Now, say the Indians, we shall have :

s=:k is an integral number,

 

Sa 2
and Ag? f=) =p

(Again the proofs are not given; they are however supplied by Hankel.)

 

. ror, . j .
1 Since Tim Y is an integral number, all the letters in ¢js=p2+¢» represent
integers.
Further, s=p2— Ag;

therefore, eliminating s, we have
; 71(22- Ag) =p+gr,
or 2 (pq1—1)=q (r+ Agq).
Since p, ¢ have no common factor, ¢ must divide gg; — 1; that is,

21-1
q

=an integer.

We have next to prove that s;= (72 — 4)/s is an integer.
qt — PA oll -2pnsts
A 2 = !
q q
s(q12s — 2pq1 + 1)
7
and, since 5, ¢ have no common factor, it follows that
qis-2pptr I-A
pe
rP—A g¥s—2ppn+1 gi? (p2- Ag?) - 21+ 1 pg1—1\2
5 = 9? = 9? = q = Ag?

Now 72

 

, since s=p2- 49%;

therefore is an integer,

is an integer.

 

Also n=
284 SUPPLEMENT

We have therefore satisfied a new equation of the same form as that
originally taken. :

We proceed in this way, obtaining fresh results of this kind, until we
arrive at one in which s =+ 1 or + 2 or +4, when, by means of the lemma,
we obtain a solution of

AY + 1= a7

Example. To solve the equation 67) + 1 = a

Since 8? is the nearest square to 67, we take as our first auxiliary equation
G67. 12—5=32 so that 7i=8. g=1, 5=—2.

Thus ¢, 227, We put » = 7, which makes ¢; an integer and at the

2 —
same time makes 5, = — rea =6 as small as possible.
Thus h=-5 H=(pn—1)/g=—41,
and we have satisfied the new equation
67.52 + 6= 41%

 

Next we take p=232, and we put s,=35, giving g,=11 ; {thus
72—0
Sa = 7, and py = (p14, — 1)/1 = 90, and
67. (v1) ~ 7 =90%
Next g, =~ a, and we put 7; = 9, giving ¢; = — 27 ; therefore
72-6 —go0.27—1I
=e, i 3 a and

67.(27Y — a= (227)
As we have now brought our s down to 2, we can use the lemma, and
67(z. 27.2210 4 = (2217+ 67.27%,
or 67 (11934) + 4 = (97684)*;
therefore, dividing by 4, we have
67 (5967)° + 1 = (48842)".
Of this Indian method Hankel says, “It is above all praise; it is
certainly the finest thing which was achieved in the theory of numbers

! Hankel conjectures that the Indian method may have been evolved somewhat in
this way.
If 4g%+s=p2 is given, and if we put 4¢2+5" =p", then
A (pg 29 +55" = (pF ~ 499")
Now suppose 2’, ¢’ to be determined as whole numbers from the equation py’ — p'g=1,
and let the resulting integral value of pp’ — Agg’ be 7.
Then A +ss’=72, and accordingly »2— 4 must be divisible by s, or s'=(4 -7?)[s is
a whole number.
Eliminating p’ from the two equations in 2’, ¢', we obtain
¢'=(2+gn(P = AgY)=(p+g7)s,
and, as stated in the rule, »- has therefore to be so chosen that (p+ ¢/)/[s is an integer.
THEOREMS AND PROBLEMS BY FERMAT 285

before Lagrange” ; and, although this may seem an exaggeration when we
think of the extraordinary achievements of a Fermat, it is true that the
Indian method is, remarkably enough, the same as that which was redis-
covered and expounded by Lagrange in his memoir of 1768'. Nothing is
wanting to the cyclic method except the proof that it will in every case
lead to the desired result whenever 4 is a number which is not a square;
and it was this proof which Lagrange first supplied.

Fermat.

As we have already said, Fermat rediscovered our problem and was

the first to assert that the equation

2—-Ay=1,
where 4 is any integer not a square, always has an unlimited number
of solutions in integers.

His statement was made in a letter to Frénicle of February, 16572
Fermat asks Frénicle for a general rule for finding, when any number not a
Square is given, squares which, when they are respectively multiplied by the
gen number and unity is added to the product, give squares. If, says
Fermat, Frénicle cannot give a general rule, will he give the smallest value
of y which will satisfy the equations 61)*+ 1 = a* and 109)? + 1 = a???

At the same time Fermat issued a challenge to the same effect to
mathematicians in general, prefacing it by some remarks which are worth
quoting in full, J

“There is hardly any one who propounds purely arithmetical questions,
hardly any one who understands them. Is this due to the fact that up to
now arithmetic has been treated geometrically rather than arithmetically?
This has indeed generally been the case both in ancient and modern
works; even Diophantus is an instance. For, although he has freed
himself from geometry a little more than others have in that he confines
his analysis to the consideration of rational numbers, yet even there
geometry is not entirely absent, as is sufficiently proved by the Zetetica
of Vieta, where the method of Diophantus is extended to continuous
magnitude and therefore to geometry.

“Now arithmetic has, so to speak, a special domain of its own, the
theory of integral numbers. This was only lightly touched upon by Euclid
in his Zlements, and was not sufficiently studied by those who followed
him (unless, perchance, it is contained in those Books of Diophantus of

1 “Sur la solution des. problémes indéterminés du second degré” in Mémoires de
’Acad. Royale des Sciences et Belles-Lettres de Berlin, t. XX111. 1769 (= Oeuvres de
Lagrange, 11. pp. 377 sqq.). The comparison between Lagrange’s procedure and the
Indian is given by Konen, pp. 75-77.

2 Qewwres de Fermat, 11. pp. 333—4.

3 Fermat evidently chose these cases for their difficulty ; the smallest values satisfying
the first equation are y=1226153980, x=1766319049, and the smallest values satisfying
the second are y=15140424455100, x =158070671986249.

4 Ocuvres de Frermat, 11. pp. 334-3.
286 SUPPLEMENT

which the ravages of time have robbed us); arithmeticians have therefore
now to develop it or restore it.

“To arithmeticians therefore, by way of lighting up the toad to be
followed, I propose the following theorem to be proved or problem to
be solved. If they succeed in discovering the proof or solution, they will
admit that questions of this kind are not inferior to the more celebrated
questions in geometry in respect of beauty, difficulty or method of proof.

“ Given any number whatever which is not a square, there are also given
an infinite number of squares such that, if the square is multiplied into the
given number and unity is added to the product, the result is a square.

“Example. Let 3, which is not a square, be the given number; when
it is multiplied into the square 1, and 1 is added to the product, the result
is 4, being a square.

“The same 3 multiplied by the square 16 gives a product which, if
increased by 1, becomes 49, a square.

“ And an infinite number of squares besides 1 and 16 can be found
which have the same property.

“But I ask for a general rule of solution when any number not a square
is given.

“ Z.g. let it be required to find a square such that, if the product of the
square and the number 149, or 109, Or 433 etc. be increased by 1, the
result 1s a square.”

The challenge was taken up in England by William, Viscount Brouncker,
first President of the Royal Society, and Wallis!. At first, owing apparently
to some misunderstanding, they thought that only rational, and not neces-
sarily integral, solutions were wanted, and found of course no difficulty in
solving this easy problem. Fermat was, naturally, not satisfied with this
solution, and Brouncker, attacking the problem again, finally succeeded in
solving it. The method is set out in letters of Wallis? of 17th December,
1657, and 3oth January, 1658, and in Chapter XcviiL of Wallis’ d/gebra ;
Euler also explains it fully in his 4/gebra®, wrongly attributing it to Pell*.

1 An excellent summary of the whole story is given in Wertheim’s paper ‘Pierre
Fermat’s Streit mit John Wallis” in Abkandlungen zur Gesch. der Math. 1X. Heft
(Cantor-Festschrift), 1899, pp. 557-576. See also Konen, pp. 29-43.

2 Oeuvres de Fermat, 111. pp. 457-480, 490-503. Wallis gives the solution of each
of the three difficult cases last mentioned.

3 Euler, Algebra, Part 11. chap. VII.

This was the origin of the erroneous description of our equation as the ‘‘ Pellian ”
equation. Hankel (p. 203) supposed that the equation was so called because the solution
was reproduced by Pell in an English translation (1668) by Thomas Brancker of Rahn’s
Algebra; but this is a misapprehension, as the so-called ¢‘ Pellian” equation is not so
much as mentioned in Pell’s additions (Wertheim in Bzbliotheca Mathematica, 1113,
1902, pp. 124-6; Konen, pp. 33-4 note). The attribution of the solution to Pell was a
pure mistake of Euler’s, probably due to a cursory reading by him of the second volume
of Wallis’ Opera where the solution of the equation ax2+ 1=3? is given as well as informa-
tion as to Pell’s work in indeterminate analysis. But Pell is not mentioned in connexion
with the equation at all (Enestrom in Bibliotheca Mathematica, 1113, 1902, p. 200).
THEOREMS AND PROBLEMS BY FERMAT 287

Fermat appears to have been satisfied with the actual solution’, but
later he points out that, although Frénicle and Wallis had given many
particular solutions, they had not supplied a general proof? (i.e. presumably
that the solution is always possible and that the method will always lead
to the solution sought for). He says, “I prove it by the method of
descente applied in a quite special manner.... The general demonstration
will be found by means of the descente duly and appropriately applied.”

Further on, Fermat says he has discovered * general rules for solving
the simple and double equations of Diophantus.”

“Suppose, for example, that we have to make

2x + 7967 equal to a square.

“I have a general rule for solving this equation, if it is possible, or
discovering its impossibility, and similarly in all cases and for all values
of the coefficient of x* and of the absolute term.

“Suppose we have to solve the double-equation

2x + 3 = square
2X us il] :

“Bachet boasts, in his commentary on Diophantus? of having dis-
covered a rule for solving in two particular cases; I make it general for
all kinds of cases and can determine, by rule, whether it is possible or not*.”

Thus Fermat asserts that he can solve, when it is possible to solve
it, and can determine, by a general method, whether it is possible or
impossible to solve, for any particular values of the constants, the more
general equation

22 — Ay = B.

This more general equation was of course solved by Lagrange. How

Fermat solved it we do not know. It is true that he has sometimes been

1 Letter of June, 16358, to Kenelm Digby, Ocuvres de Fermat, 11. p. 402.
2 ¢“ Relation des nouvelles découvertes en la science des nombres,” Oeuwres, 11. P- 433.
3 See on Diophantus 1v. 39, and above, pp. 80-82. 3

4 With this should be compared Fermat’s note on Dioph. 1v. 39, where he says,
similarly : ;

‘“ Suppose, if you will, that the double-equation to be solved is

2x + 5 =square
6x + 3 =square :

“The first square must be made equal to 16 and the second to 36; and others will be
found ad infinitum satisfying the question. Nor is it difficult to propound a general rule
for the solution of this kind of question.”

No doubt the double-equation in this case, as in the others referred to in the “Relation,”
would be transformed into the single equation

22—Au2=p
by eliminating x. I think this shows how Fermat was led to investigate our equation:
a question which seems to have puzzled Konen (p. 29), in view of the fact that the actual
equation is not mentioned in the notes to Diophantus. The comparison of the two places
seems to make the matter clear. For example, the two equations mentioned above in

this note lead to the equation 72-3242= —12, and the solution /=6, u=4 is easily
obtained. :
288 : SUPPLEMENT

credited with the very same solution of the equation x*— Ay*=1 as that
given by Brouncker and Wallis; but this idea seems to be based on a
misapprehension of a sentence in Ozanam’s Algebra (1702). Ozanam
gives the Brouncker-Wallis solution as “une régle générale pour résoudre
cette question, qui est de M. de Fermat”; and possibly the ambiguity
of the reference of “qui” may have misled Lagrange and others into
supposing that the “regle” was due to Fermat.

For the history of the equation after Fermat’s time I must refer to
other works and particularly that of Konen'. Euler, Lagrange, Gauss,
Jacobi, Dirichlet, Kronecker are the great names associated with it. I
will only add a few particulars with regard to Euler? as coming nearest
to Fermat.

In a letter to Goldbach® of roth August, 1730, Euler mentions that he
requires the solution of the equation x*— 4)’=1 in order to make
ax’ + bx + ¢ a complete square. He goes on to observe that the problem
of solving x®— 4»*=1 in integers was discussed between Wallis and
Fermat and that the solution (which he already attributes to Pell) was
set out in Wallis’ Opera. There 1s an indication in this very passage that
Euler had then only read the Brouncker-Wallis correspondence cursorily,
for he speaks of the equation 109)*+ 1 =x? as being the most difficult
case solved by them, whereas the most difficult examples actually solved
were 4333+ 1=aand 3132 + 1'= 2%

A paper of a year or two later* contained the proof that the evolution
of successive solutions of ax? + bx + ¢=»* when one is known requires that
one solution of @é®+ 1 =7’ must also be known. Similarly, in his 4/gebra®,
he shows that the solution of the latter equation is necessary for finding all
the possible solutions of the equation ax? + 4 =? the importance of which
remark is emphasised by Lagrange® :

In’ the paper quoted in the last paragraph Euler finds any number
of successive solutions of ax? + dx + c=3? and the law for forming them,
when we are given one value n of x which will make ax*+dx +c a
complete square and one value p of & which will make aé*+ 1 a complete
square, or, in other words, when az? + dn +c=m? and ap + 1 =¢> He
then takes the particular case ax? + dx + d%=3* where (since x=o0, y=4
satisfies the equation) we can substitute o for » and & for = in the
expressions representing the successive solutions of ax? + bx +c=3% Then
again, putting =o and #=1, he 1s in a position to write down any

! Konen, op. cit.; cf. Cantor's Geschichte der Mathematik, 1v. Abschnitt XX., as
regards Euler and Lagrange.

2 Cf. Konen, op. cit. pp. 47-58.

3 Correspondance mathématique et physique de quelques célébres gbométres du XV111inme
siecle, publiée par P. H. Fuss, Pétersbourg, 1843, I. p. 37.

4 “De solutione problematum Diophanteorum per numeros integros” in Commentarii
Acad. Petropol. 1732-3, V1. (1738), pp. 175 sqq.= Commentationes arithm. 1. pp. 4—10.

5 Algebra, Part 11. ch. VI.
6 Additions to Euler’s 4/¢ebra, ch. viII,
THEOREMS AND PROBLEMS BY FERMAT 289

number of successive solutions of a¢’*+ 1 =7%* when one solution ¢=p,
n=g¢ 1s known. The successive values of ¢ are

0, V2 209, 409° — ps son

and the corresponding values of 7 are

I 0 21 4~ 30, ...
the law of formation being in each case that, if 4, 5 be consecutive values
in either series, the next following is 295 — 4.

The question then arises how to find the first values p, ¢ which will
satisfy the equation. Euler first points out that, when @ has one of many
particular forms, values of p, ¢ can at once be written down which satisfy
the equation. The following are such cases with the obvious values of

2 and ¢.

a=e

=; 221, 94
asl+1; p= 2, 0 madt a,
a= 4 alt pe igmaditd a
(where a may even be fractional provided ae’~! is an integer),
a= (od + Ley + 207 + 3B): peg, g=od tly ffiluy,
amid tall; B= fey g=talilt hn,

But, if @ cannot be put into such forms as the above, then the method
explained by Wallis must be used. Euler illustrates by finding the least
values p, ¢ which will satisfy the equation 31£?+ 1 =%?% and then adds a
table of the least solutions of the equation af®+ 1 =7° for all values of
a (which are not squares) from 2 to 68.

The important remark follows (§ 18) that the above procedure at once
gives a very easy way of finding closer and closer approximations to the
value of any surd ,/a. For, since ap® + 1 = ¢% we have Ja = /(¢*— 1)/2
and, if ¢ (and therefore p also) is large, ¢/p is a close approximation to \/a;
the error is not greater than 1/(2p*./a). Euler illustrates by taking ,/6.
The first solution of 62 + 1 = (after £=0,9=1)isp=2,g=35 Taking
then the series of values above given for aé* + 1 = %% namely

£=0, 7, 2/0, YPr—p, ..- 4, B, 208-4,
n=1, ¢ 29°— 1%, 40°— 30, ... &, £5, 20F= FE,
and substituting p =z, ¢ =5, the successive corresponding values 2, Q
of & 7 respectively become
P=o0, 2, 20, 198, 1060, 19402, 192000, 1001193,..-.
Q=1, 5 49, 485, 48071, 47525, 470449, 4656905, ...
and the successive values ()//” are closer and closer approximations to ,/6.
It will be observed that the method of obtaining successive approximations

H. D. 19
290 SUPPLEMENT

to A/a from successive solutions of a+ 1 =7* is the same as that which,
according to the hypothesis of Zeuthen and Tannery, Archimedes used in
order to find his approximations to ./3.

The converse process of finding successive solutions of aé*+ 1 =7* by
developing ./a as a continued fraction did not apparently occur to Euler
till later. In two letters’ to Goldbach of 4th Sept. 1753 and 23rd Sept.
1755 he speaks of a “certain ” method and of improvements which he had
made in the “Pellian” method but gives no details. His next paper on
the same subject? returns to the problem of finding all the solutions of
ax’ + bx +c=y* or ax*+b=y* when one is known, and in the course of
his discussion of the latter he arrives at “the following remarkable theorem
which contains within it the foundation of higher solutions.

“if =a, y=0 satisfies oxi p="
and X=¢, y=d satisfies on + g=717,
then X=bct ad, y=0dt ane satisfies an? + pg = 2."

That is to say, Euler rediscovers and recognises the importance of the
lemma to the Indian solution, as Lagrange did later.

More important is the paper of about three years later? in which Euler
obtained the solution of the equation a®— 4y*=1 by the process of con-
verting ,/A4 into a continued fraction, this course being the reverse of that
which was, according to the hypothesis of Tannery and Zeuthen, followed
by Archimedes, and to the feasibility of which Euler had called attention
in 1732-3. He begins by stating, without proof, that, if ¢* = /* + 1, then
¢g/p is an approximation to .//, and ¢/p 1s “such a fraction as expresses
the value of ,// so nearly or exceeds it so little that a closer approXYmation
cannot be made except by bringing in greater numbers.” Next he develops
certain particular surds, namely ,/(13), /(61) and ,/(67), after which he
states the process generally thus. If /z be the given surd and z the root
of the greatest integral square which is less than z, the process will give

I

JZ=0+— 1
a+ = I

fsa

Cc rr

d + etc.

the successive quotients a, 4, ¢, , being found by means of the process
shown in the following table :

1 Correspondance etc., ed. Fuss, pp. 614 5q., 629 sq.
2 “De resolutione formularum quadraticarum indeterminatarum per numeros integros”’
in Novi Commentariz Acad. Petropol. 1762-3, 1X. (1764), pp. 3 sqq.= Commentat. arithm.
L pp. 297-315. :

3 “De usu novi algorithmi in problemate Pelliano solvendo” in Novi Commentarii
Acad. Petropol. 1765, X1. (1767), pp. 28-66= Commentat. arithm. 1. pp. 316-336. The

paper seems to have been read as early as 15 Oct. 1756.
THEOREMS AND PROBLEMS BY FERMAT 291

 

 

 

 

 

 

Take and It follows that
1. A=9 a=z-Al=z-22 eg?
I Bead p=""Fita(a-5) b522E
HI. C=fi-8 y= aro 5- 0) g2t
IV. D=vye-C 0=""L=pc(c-D) gl
V. E=8-D =F eyia-5) gE
ete. etc. x :

(This 1s of course exactly the process given in text-books of Algebra,
e.g. Todhunter’s.)

Euler now remarks as follows.

1. The numbers 4, B, C, D ... cannot exceed 7; the first, 4, is equal
to 7; since a = (v + A)/a, aa — A= 1B = v, and so on.

2. Unless where one of the numbers a, S, v, 8...1is equal to unity,
none of the corresponding quotients a, 4, ¢, 4... can exceed v.

3. When we arrive at a quotient equal to 2, the next quotients will be
a, b, ¢, 4... In the same order.

4. Similar periods occur with the letters a, 3, y, 6... and the term
of this series corresponding to a quotient 22 is always I. :

The successive convergents to the continued fraction are then investi-
gated and it is shown that, for successive convergents ¢/p beginning
with 2/1,
¢*—sp*=—a, +B, —v, +9, —eetc..in order.
It follows that the problem is solved whenever one of the terms with a
positive sign, 3, §, { etc., becomes I. a :

Since unity for one of the terms a, 3, y, & corresponds to the quotient
27, and each fresh period begins with 22, the first period will produce
a convergent ¢/p such that ¢* — 26> = + 1; and the negative sign will apply
if the number of quotients constituting the period is odd, while the positive
sign will apply if the number of quotients is even. In the latter case we
have a solution of our equation at once; if, however, ¢*— 20° =—1, we
must go on to the end of the second period in order to get an even number
of quotients and so satisfy the equation ¢*—3p°= +1. Or, says Euler,
instead of going on and completing the second period, we can satisfy
the latter equation more easily thus.

Suppose ¢* — zp* = — 1, and assume
Y=2p9, ¢=2¢+1
Then gi — 2p? =49* + 49° + 1 — 43p°F°

= 1d {f= 1)
= 1.
19—2
292 SUPPLEMENT

[This last derivation of a solution of y*—zx*=1 from a known solution
of > — 2a” = — 1 is of course the same as the Indian method of doing the
same thing, for they assumed p’ = 2p, ¢' = ¢* + 2% and ¢* + 28° = ¢*+(¢° + 1).]

We thus see that in Euler’s method there is everything necessary to the
complete solution of our equation except the proof that it must always lead
to the desired result. Unless it is proved that the quotient 22 will actually
occur in the development of the continued fraction in every case, we cannot
be sure that the equation has any solution except x =o, y =1.

I cannot, I think, do better than conclude by a quotation from
H. J. S. Smith’, the first part of which is well known? ‘Euler observed
that [if 772— DU*=1] T'/U is itself necessarily a convergent to the value
of \/D, so that to obtain the numbers 7" and U it suffices to develop /D
as a continued fraction. It is singular, however, that it never seems to
have occurred to him that, to complete the theory of the problem, it was
necessary to demonstrate that the equation is always resoluble and that
all its solutions are given by the development of .,/0. His memoir
contains all the elements necessary to the demonstration, but here, as
in some other instances, Euler 1s satisfied with an induction which does
not amount to a rigorous proof. The first admissible proof of the re-
solubility of the equation was given by Lagrange in the Mélanges de la
Société de Turin, Vol. 1v. p. 41°. He there shows that in the development
of ./D we shall obtain an infinite number of solutions of some equation of
the form 72—DU?=4 and that, by multiplying together a sufficient
number of these equations, we can deduce solutions of the equation
7?—DU?=1. But the simpler demonstration of its solubility which
is now to be found in most books on algebra, and which depends on
the completion of the theory (left unfinished by Euler) of the development
of a quadratic surd as a continued fraction, was first given by Lagrange
in the Hist. de I Académie de Berlin for 1767 and 1768, Vol. XX1IL p. 272,
and Vol. xx1v. p. 236% and, in a simpler form, in the Additions to Euler's
Algebra’, Art. 37.”

1 ¢“Report on the Theory of Numbers, Part 111.,” British Association Reports for 1861,
London, 1862, p. 315= Collected Works, Vol. 1., Oxford, 1894, p. 192.

2 It is given in Cantor, Gesch. d. Math. 1v. 1908, p. 159, and referred to by Konen,
op. ct. p. 51.7

. 3 “Solution d’un probléme d’Arithmétique,” finished at Berlin on 20th Sept. 1768
and published in Miscellanea Zaurinensia, 1v. 1766-1769 = Oeuvres de Lagrange, 1.
pp- 671-731.

4 The references are: ‘“ Sur la solution des problémes indéterminés du second degré,”
read 24th Nov. 1768 and published in the Mémoires de I’ Académie Royale des Sciences
et Belles-lettres de Berlin, Vol. XXIIL., 1769, pp. 165-310= Oeuvres de Lagrange, 1I.
PP: 377-535; ‘‘ Nouvelle méthode pour résoudre les probléemes indéterminés en nombres
entiers,” read 21st June, 1770, and published in AZmoires de I’ Académie Royale des
Sciences et Belles-lettres de Berlin, Vol. XXIV., 1770, pp. 181-250 = Oeuvres de Lagrange,
11. pp. 655-726.

5 The Additions of Lagrange were first printed as an appendix to Zmens d’ Algibre
par M. L. Euler traduits de allemand, Vol. 11., Lyons, 1774; second edition, Paris,
1798 ; they were thence incorporated in Oeuvres de Lagrange, VII. pp. 158 sqq.
293

SECTION 111.
THEOREMS AND PROBLEMS ON RATIONAL RIGHT-ANGLED TRIANGLES.

1. On No. 20 of the problems about right-angled triangles added
by Bachet to Book vi. (“To find a right-angled triangle such that its
area is equal to a given number”) Fermat has a note which shall be
quoted in full, not only for the sake of the famous theorem enunciated
in it, but because, exceptionally, it indicates the lines on which his proof
of the theorem proceeded.

“The area of a right-angled triangle the sides of which
are rational numbers cannot be a square number.

“This proposition, which 1s my own discovery, I have at length
succeeded in proving, though not without much labour and hard thinking.
I give the proof here, as this method will enable extraordinary develop-
ments to be made in the theory of numbers.

“If the area of a right-angled triangle were a square, tiere would exist
two biguadrates the difference of which would be a square number. Con-
sequently there would exist two square numbers the sum and difference of
which would both be squares. Therefore we should have a square number
which would be equal to the sum of a square and the double of another
square, while the squares of which this sum is made up would themselves
[7.e. taken once each] have a square number for their sum. But if a square
is made up of a square and the double of another square, its side, as I can
very easily prove, is also similarly made up of a square and the double of
another square. From this we conclude that the said side is the sum of the
sides about the right angle in a right-angled triangle, and that the simple
square contained in the sum is the base and the double of the other square
the perpendicular.

“This right-angled triangle will thus be formed from two squares,
the sum and the difference of which will be squares. But both these
squares can be shown to be smaller than the squares originally assumed
to be such that both their sum and their difference are squares. Thus,
if there exist two squares such that their sum and difference are both
squares, there will also exist two other integer squares which have the same
property but have a smaller sum. By the same reasoning we find a sum
still smaller than that last found, and we can go on ad infinitum finding
integer square numbers smaller and smaller which have the same property.
This is, however, impossible because there cannot be an infinite series
of numbers smaller than any given integer we please.—The margin is too
small to enable me to give the proof completely and with all detail.

“By means of these considerations I have also discovered and proved
that no triangular number except 1 can be a biguadrate.”
294 SUPPLEMENT

As Wertheim says, it may have been by following out the indications
thus given by Fermat that Euler succeeded in proving the propositions
that a* — »* and «a*+yp* cannot be squares, as well as a number of other
theorems connected therewith (Commentationes arithmeticae collectae, 1.
PP- 24 sqq.; Algebra, Part 11. Chapter xin).

Zeuthen! suggests a method of filling out Fermat's argument, thus.

The sides of a rational right-angled triangle can be expressed as

2+, wt — 97, any.

As a common factor in the sides would appear as a square in the
number representing the area, we can neglect such a factor, and assume
that x? — 3 and therefore also x + y and x —y are odd numbers and that
x, y are prime to one another, so that x, y, x +3, x —y are all prime to
one another. :

We have now to test the assumption that the area of the triangle
2 (© =) (+7)
is a square. If so, the separate factors must be squares, or
w=, y=19%,
BL =0 17-0 =
(“There would exist two biguadrates the difference of which [6 — ov) would
be a square, and consequently there would exist two squares the sum and differ-
ence of which [1 + v*, w® — v*] would both be squares,” Fermat.)
- From the last two equations we obtain
: =p —=(p-9) (+9)
(“We should have a square number which would be equal to the sum
of a square and the double of another square | p* = 20° + ¢*],” Fermat.)

Now p+ ¢ and p—g¢ are both even numbers because, on the above
assumptions, p* and ¢* are both odd; but they cannot have any other
common factor except 2, since #® and 2® are prime to one another. It
follows therefore from the last equation that

2m ”*
ee pre={2 ’ r--10
. where 7 is an even number.

We obtain, therefore,

9 2 2\ 2
: += 5 72°\*
2? L375 (m*)* + @ 3
2 2

i) be
’ ! : { 7 . .
The whole numbers 72* and are therefore sides of a new right-angled

: a mn?
triangle with the square area oe

1 Zeuthen, Geschichte der Mathematik im XVI. und XVI11. Jahrhundert, 19o3, p. 163.
THEOREMS AND PROBLEMS BY FERMAT 295

(“If a square is made up of a square and the double of another square
[2*= 22% + ¢*), its side is, as I can very easily prove, also made up of a

2
square and the double of another square 2 =m? + (2) 2 From this we

conclude that the said side 1s the sum of the sides about the right angle
in a right-angled triangle, the square [7°] being the base and the double of

2
the other square 2 %) | the perpendicular,” Fermat.)

That the sides of the new triangle are less than those of the original
triangle is clear from the fact that the square on its hypotenuse #® or
x 1s a factor of one of the perpendicular sides of the original triangle’.

As now an infinite series of diminishing positive whole numbers is
impossible, the original assumption from which we started is also impossible.

It will be observed, as Zeuthen says, that the proof includes also the
proof of the fact that #* — 2* cannot be a square and therefore cannot be
a fourth power, from which it follows that the equation #* = #* + 2" cannot
be solved in whole numbers, and consequently cannot be solved in rational
numbers either.

The history of this theorem would not be complete without an account
of a “proof originating with Fermat” which Wertheim has reproduced?
In the small paper of Fermat's entitled ““ Relation des nouvelles découvertes
en la science des nombres?” containing a statement of his method
of “diminution without limit” (descente infinie or indéfinie) and of a
number of theorems which he proved by means of it, there is a remark
that he had sent to Carcavi and Frénicle some proofs based on this |
method. And, sure enough, Frénicle gives a proof by this method of
the theorem now in question in his “Traité des triangles rectangles en
nombres?” Wertheim accordingly concludes that we have here a proof
of Fermat’s. A short explanation is necessary before we come to Frénicle’s
proof.
~ We obtain a right-angled triangle z, x, y in rational numbers (2? = «x? + 7)
if, a, 4 being any integers and a> 4, we put

s=+ 0 x= y= 24}.
If a is prime to ¢ and one of these numbers is even, the other odd, then
it is easily shown that the greatest common measure of x, y, z is 1.
In the right-angled triangle a® — 4® and 2aé are the perpendicular sides,

1 Zeuthen’s inference at this point diverges slightly in form from what we actually find
in Fermat’s own statement of his argument. Fermat does not actually say that the new
right-angled triangle is a triangle in smaller numbers than the original triangle and with
the same assumed property, but that its formation gives us two new square numbers the
sum and difference of which are squares, and which are smaller than the two squares
originally assumed to have this property.

2 Zeitschrift fiir Math. uw. Physik, hist. litt. Abtheilung XLI1V. 1899, pp. 4-0.

3 Oeuvres de Fermat, Vol. 11. pp. 431-6.

Y Mémoires de I’ Académie Royale des Sciences, V., Paris, 1729, pp. 83-1060.
296 SUPPLEMENT

and (a? — 0°) ab is the area. a, 4 are called the generating numbers (the
numbers from which the triangle is formed) and if a is prime to 4, and one
of them is odd and the other even, so that x, y, z have no common factor
except 1, the triangle is called a primitive triangle.

If (a®*—0%) ab is the area of a primitive right-angled triangle—and it
is enough to prove the proposition for such—each of the three numbers
a®— 0? a, b is prime to the other two. If, then, the product is a square
number, each of the three factors must be square, and in that case a®— #*
will be the difference between two fourth powers. The theorems

(1) the area of a right-angled triangle in rational numbers cannot be
a square number, and

(2) the difference of two fourth powers cannot be a square,
accordingly state essentially the same fact.

The proof which Frénicle gives of the first of these propositions depends
on the following Lemmas.

Lemma 1. /f the odd perpendicular of a primitive right-angled triangle
is a square number, there exists a second primitive right-angled
triangle with smaller sides which has for its odd perpendicular
the root of the said square number.

XH 2-F=2 it follows that &>=2+~, so that a, 4, ¢ are the sides
of a right-angled triangle. The odd perpendicular of this second triangle
1s ¢, for by hypothesis is odd; consequently the even perpendicular is
0, while a is the hypotenuse. The triangle is “primitive” because a
common divisor of any two of the three numbers a, 4, ¢ would divide
the third, while by hypothesis @, 4 have no common’ factor except 1.
Next, the second triangle has smaller sides than the first, since c<c?
a<al+ 0? b< zal.

By this lemma we can from the triangle 9, 40, 41 derive the triangle
3, 4, 5, and from the triangle 225, 25312, 25313 the triangle 15, 112, 113.

Lemma II. Zf in a primitive right-angled triangle the hypotenuse as
well as the even perpendicular were square, there would exist a
second primitive right-angled triangle with smaller sides which
would have for hypotenuse the root of the hypotenuse of the first,
Jor odd perpendicular a square number, and jor even perpendicular
the double of a square number.

Let the sides of the first triangle be a®+/%, a®*—#%, 2ab. If 2ab were

a square, ab would be double of a square ; therefore, since a, 4 are prime
to one another, one of these two numbers, namely the odd one, would
be a square, and the other, the even one, would be double of a square.
Let a be the odd one of the two, é the even. If now the hypotenuse
a® + 5 were a square number ¢% we should have a second right-angled
triangle a, 4, ¢ which would necessarily be primitive and in which the sides
would be smaller than those of the first triangle; for c<¢’ 4<2ab and
a<a®—bsince a+b>a,a—-0Z1.
THEOREMS AND PROBLEMS BY FERMAT 297

By means of the above two lemmas combined we can now prove that
the area of a primitive right-angled triangle cannot be a square number.

Let the sides of the triangle be a®+ 74% a* — 2% 2ab. If now the area
were square, the product of the perpendicular sides would be double of
a square. But the perpendicular sides are prime to one another. There-
fore the odd perpendicular a®—2* would be a square, and the even
perpendicular 2a/ the double of a square. But, if ¢*—4* were equal to
¢ we could (by the first Lemma) find a second primitive triangle with
smaller sides in which the odd perpendicular would be ¢, the even per-
pendicular 4, and the hypotenuse a. Again, since 2a/ would be double
of a square, af would be a square, and, since a is prime to 4, both a and
4 would be squares. The second triangle would accordingly have a square
number both for its hypotenuse (a) and for its even perpendicular (4).
That is, the second primitive triangle would satisfy the conditions of the
second Lemma, and we could accordingly derive from the second primitive
triangle a third primitive triangle with still smaller sides which would,
exactly like the first triangle, have a square number for its odd perpendicular,
and for its even perpendicular the double of a square number.

From this third triangle we could obtain a fourth, and by means of the
fourth we could obtain a fifth with the same property as the first, and so
we should have an unending series of primitive right-angled triangles, each
successive triangle having smaller sides than the one before, and all being
such that the odd perpendicular would be a square number, the even
perpendicular the double of a square number, and consequently the area
a square number. This, however, is impossible since there cannot be an
unending series of integral numbers less than any given integral number.

Frénicle proves, by similar considerations, that neither can the area of a
right-angled triangle in rational numbers be the double of a square number.

In enunciating Fermat’s problems on right-angled triangles I shall in
future for brevity and uniformity use & 7, { to denote the three sides, while
¢ will always represent the hypotenuse and ¢, 5 the two perpendicular sides.

2. Tv find a right-angled triangle (¢, & n) such that

S

{=
E+9=10° :
[Since £* = & +»? this problem is equivalent to that of finding «x, y such
that /
xX+y=0
a4 yt = Ee ?

which is Question 17 in Chapter xiv. of Euler's Algebra, Part 11.]
First method.

Form a right-angled triangle from the numbers x + 1, x; the sides will

then be
l= tant, E=a2x+t 1, n=2% tox
208 SUPPLEMENT

We have then the double-equation
axl tow ty =f
202+ 48 +1 =107) ;
The ordinary method of Diophantus gives the solution x =-212%; the
triangle will therefore be formed from —2 and —22 or, if we take the

i‘

numerators only, — 5 and — 12, and the triangle is (169, — 119, 120) which
is equally the result of forming a triangle from + 5 and + 12.

But, as one of the perpendiculars is negative, we must find another
value of x which will make all three sides positive.

We accordingly form a triangle from x +5 and 12, instead of from
5 and 12, and repeat the operation. This gives for the sides

{=x + 1024160, &=2"+ 10% 119, B= 24% + 120,
and we have to solve the double-equation
x? + 10x + 169 = 2?
Bhqnt T=
Making the absolute term the same in each, we have to solve
x? + 10X + 169 = of,
169x% + 57462 + 169 ="
‘The difference is 168x* + 5736x, which we may separate into the factors
14%, 12x + 2858 (the sum of the terms in x being 26x or 2. 13x).
Equating the square of half the sum of these factors to the larger

expression, or the square of half their difference to the smaller, we find
in the usual way

The triangle is therefore formed from 2139995 12, or from 2150905,
246792, and the triangle itself is ;
4687298610289, 4565486027761, 1061652293520,

the hypotenuse and the sum of the other two sides being severally squares.

Second method.
This is the same as the first method up to the forming of the triangle
from x + 5 and 12 and the arrival at the double-equation
2% + 10% + 169 = 7,
tant  1=17
Multiply the two expressions together, and we must have
x* + 445° + g108% + 57562 + 169 = a square
= (13 + 2808 x— 2), say;
this gives, as a matter of fact, the same value of x, namely

2048075
X= 50566 7
THEOREMS AND PROBLEMS BY FERMAT 299

and the triangle is the same as before’.

In his note on Diophantus vi. 22 Fermat says that he confidently
asserts that the above right-angled triangle is the smallest right-angled
triangle in rational numbers which satisfies the conditions. :

[The truth of this latter assertion was proved by Lagrange® Lagrange
observes that, since £ +5 =? £2 + »*= x4, say, we have, if we put 5 for £ — 7,

224s onl;
or : 2t—gimst
and, if x, y is any solution of the latter equation,
= (P+) m= Pr)

! For comparison we may give Euler’s solution (4 /gebra, Part 11., Art. 240; Commen:-
lationes arithmeticae, 11. p. 398).
We have to solve the equations
2 arch grey?
' x24 y2 =A | :
First make x2+ 2 a square by putting x =a? — 42, y=2ab, so that
: x24 y2= (a2 + 62)
To make the last expression a fourth power put a=p2— ¢2, 6=2pg, so that
2 + R=(g2+ 2),

and accordingly x24 y= (p2% +g?)

We have now only to make x + y a square.

Now x=a2- =p -6p2%% +4, y=1ab=4p% - 407%;
therefore Pr 4%) — 629% — 45° + gt =a square.

In solving this we have to note that #, ¢ should be positive, must be >g¢ (for other-
wise y would be negative), and a> 4 in order that x may be positive.

Put Pap — 6022 — 4p + r= (P72 - 289 + 2
and we obtain 42%) — 69% = — 4 pPg + 6p%¢%, whence ply =3 .
But, if we put p=3, y=2, we find x= — 119, a negative value.

To find fresh values, we can substitute for p the expression 2g +7 and solve for the
ratio ¢[r; then, by taking for ¢ the numerator and for » the denominator of the fraction
so found, we find a value for p and thence for x, y. This is Euler’s method in the
Algebra. But we avoid the necessity for clearing of fractions if (as in the Comment.
arithm.) we leave 2 as the value of ¢ and substitute 3+2 for 3 as the value of 2.

. We then have = 81+108v+ 5402+ 1228424,

: +4p%0= 216+2160+ 7202+ 823,
— 6292 ="— 216 — 1447 — 2422,
~4fp’=~ 96-32,
Sgt=" "10,
whence x+y =1+ 1480+ 10202 + 2003 + vt =a square =(1 + 747 — v2)%, say;
and we obtain Ma
1343 = 429, OF nib, and p= 3+o="1%9, while ¢g=2.
y 4

Taking integral values, we put p= 1469, ¢=84.

Therefore a=1385.1553=2150905, b0=1068.1469=246792,
and x=4505486027761, y=1061652293520,
which is the same as Fermat’s solution. :

2 N. Mémoires de I Acad. Royale des Sciences et Belles-léttres de Berlin, année 1777s
Berlin, 1779 = Ocuvres de Lagrange, 1v. pp. 377-398.
300 SUPPLEMENT

He sets himself therefore to find a general solution of the equation
2x*—y*=2% and effects it by a method which is a variation of Fermat's
descente, one of the most fruitful methods, as Lagrange observes, in the
whole theory of numbers. The modified method consists of two parts,
(1) a proof that, assuming that there exist integral values of x, y greater
than 1 which satisfy the condition 2x*-)*=3? there are still smaller
integral values which will also satisfy it, (2) the discovery of a general
method of deducing the latter from the former. This being done, and
it being known that x =1, y =1 are the minimum values, the successive
higher values are found by reversing the process. Lagrange found that
the four lowest values for x, y give the following pairs of values for &, 7,

namely
(1) é=1 y N=0,
(2) £{=u120 y N= =TT0,
(3) £=2276953 » N=-473304,

(4) &=1061652293520, 7=4565486027761,
so that the last pair (4) are in truth, as Fermat asserts, the smallest possible
values in positive integers. |

3. 70 find a right-angled triangle {, & n such that

Jae
é=ny=of

[This is of course equivalent to solving
Eee)
+= wt
Form a triangle from the numbers x + 1, 1; the sides will then be
{exitontas fualtor penta,
We have then to solve the double-equation
Bt an ta=’
a2 a= | :
Solved in the ordinary way, this gives x =—11; consequently the
triangle is formed from — 2%, 1, or from — 3, 12.
We could proceed, as in the last problem, to deduce a new value for x,
but we observe that the triangle formed from 5, 12, Ze the triangle 160,
119, 120, satisfies the conditions.

4. To find a right-angled triangle {, & vn such that
{=u
E+my=02]"
where m is any number.
Fermat takes the case where 7 = 2.
Form a triangle from «, 1; the sides are then Z=a%+ 1, £=a%— Hi
7= ax.
THEOREMS AND PROBLEMS BY FERMAT 301

Therefore
La | must both be squares.
at gqx—1
The difference = 2 — 4x, and by the usual method we find x = £5.
But £ = a? — 1 is negative unless x>1. We therefore begin afresh and
form a triangle from x +5, 12.
The sides of this triangle are
A+ 10x + 169, A+ 10X— 119, 24% + 120.
We have therefore to solve the double-equation
x? + 10x + 169 = 2°
x*+ 58x +121 oy
Fermat multiplies the two expressions together and puts
xt + 68x% + 8704% + 11012X + 20449 =a Sums
= (143+ 5300 ~ SHERI, say;
therefore oN
and the triangle is formed from 103447257961, 17749110120.
The double-equation could also have been solved by the usual
Diophantine method, as in the next problem to be given.

5s. Zo find a right-angled triangle {, & mn such that
¢ or
E—mn=0" >

Form a triangle from «x + 1, 1, so that the sides are

where m is any number.

Suppose that 7 = 2.

{=F rants S=aton in=avda.
Therefore we have to solve
EE Ln
x? — wa
Solving in the usual manner, we obtain x = — 1%, so that the triangle is
formed from — J, 1, or from — 5, 12, and is ieee (169, 119, — 120).
We have to replace the value of x by a value which will avoid the
negative sign. Form a triangle, then, from x — 5, 12.
The sides are a?— 10x + 169, x” — 10x — 119, 24% — I20.
The double-equation now becomes
x2 — 10% + 169 = #*
22 — 58x + 121 Tr

Multiply the second equation by 189, and we have to solve

6

x — 10% + aii]

il

9 9802 oop
Shar — 302 x 4 169=0
SUPPLEMENT

The difference = 554° — 83220 = 4 x (fra 4298),
Equating the square of half the difference of the factors to the smaller

expression (or the square of half the sum to the larger), we have

302

x= 4593455
= Te 01I0

and the required triangle is formed from 4363225, 552552, the sides being
19343046113329, 18732418687921, 4821817400400.
Or again in this case we can multiply the expressions x*— rox + 169
and «® — 58x + 121 and put their product :

xt — 68x + 870x%— 11012% + 20449 = a square
5506 4. AY
208 x + x*)?, say,

E=Eisos
6. Zo find a right-angled triangle {, & n such that
{=

E+ mm =v?

and the result will be the same as before,
- 4593455

where m is any given number.
Form a triangle from x + 1, 1; its sides will be

Yetm=3
{=altamta f=afton poay ia
We have therefore to solve the double-equation

alt an=7~7
A +8x+6=0°)

Solving this in the ordinary manner, we shall find x = J
Hence the triangle is formed from }3, 1, or (in whole numbers) from

13, 12; the sides are therefore 313, 25, 312.
Fermat also finds the solution by multiplying the two expressions and

making the product a square;
J

xt + 105% + 224% + 12x = a square

, E{2 5 an 3),

SEE |

12)

say.
and the triangle

This gives the same value of x as before, x

18,313, 25; 312
7. Tv find a right-angled triangle {, & n such that
E=u’
E—mm=0v*|"

where m is a given number.
Fermat takes the case m = 3.
Remembering that in. the corresponding problem with a plus sign we
found the triangle 313, 25, 312 which is formed from 13, 12, we form the

triangle in this case from & — 13, 12; its sides are
{=a—26x+ 313, £=a"— 200+ 25, n=24v— 312.

We have then 22-26 + 25 = 2
x* — 98x + 961 = 7° :
THEOREMS AND PROBLEMS BY FERMAT 303

Multiplying the first expression by £51 we have to solve the double-
equation
ox? — 24986 5 t g61 =o

x2 — 98x + 961 = 7* 1

The difference = #2 x? — 22538 5 — 28 5 (30 5 112868)

Proceeding as usual, we find x=2 : 38513L; the triangle is formed
from x — 13, 12, or (in whole numbers) ok 23542921, 3320440, and the
sides are

568864891005841, 539673367418641, 179888634210480.
The same result is obtained by multiplying the expressions a — 26x + 25
and a — 98x + 961 and making the product a square ; we put
at — 1245° + 35344% — 27436% + 24025 =a square
=(af ~ 13718 + 135),
and the result is x = 27881731 ag before.

8. 7v find a right-angled triangle {, & vn suck that
=u?
{+mmp=v"]"

where m is any given iuimnber.

Suppose m = 2. Form a triangle from x + 1, 1; the sides are

[=f 4 ants, E=stton n=0%+32,
We have then to solve the double-equation
Aton
’+6x+6=90" |

The usual method gives x=}, and the triangle is formed from 2, 1, or

(in whole numbers) from 3, 4, being the triangle (41, 9, 40).

Since {+n =a+4x +4 =a square, we have actually solved the problem
of finding a right-angled triangle {, & vn such that

£=u
{+9=tF
{+ 29 =n’

9. 70 find a right-angled triangle {, & vn such that

=u’
{—my=2v*]"
Suppose m = 2.

Since the corresponding problem with a plus sign just preceding has
the solution (41, 9, 40) formed from the numbers 5, 4, we form a triangle
in this case from x— 5, 4; the sides are

where m is a given number.

[=x'—10%+ 41, £=2° — 10x +0, 9 = 8x — 40.
304 SUPPLEMENT

We have then to solve the double-equation
x? — 10x + 9=12u")
x? — 26x + r21=2%]

De Billy (or Fermat) observes that this double-equation “seems to admit
of solution in several ways, but it will be found that it is hardly possible to
find a practical solution except by the new method” (expounded earlier in
the Znwentum Novum) of making the absolute terms equal (instead of using
the equal terms in x? which method gives, in fact, the value x=o0). That
is to say, we make the absolute terms in the two expressions equal by
multiplying the first by 121, and the double-equation becomes

 

EL a¥ 220004 121 = 0?
x?— 26x + 121 = 7° I

The difference =112x2—-2T8 x = Sx (Rta — 122),

Equating the square of half the difference of the factors to ¢? or the
square of half their sum to #?, we find x = 38.

Therefore the triangle 1s formed from 42% 4 or (in whole numbers)
from 493, 132, and the sides are 260473, 225625, 130152.

Since { —n=a*— 18x + 81 =a square, the above actually amounts to the
solution of the problem of finding a right-angled triangle i, & n such that the

three conditions

=i
= y=of
{—2n=12"

are simultaneously satisfied.

De Billy (or Fermat) observes however that, while the above oz¢ solution
satisfies the conditions of both problems, it is not so with all solutions of
the problem involving the two conditions only; but only primitive triangles
satisfying the conditions of that problem satisfy the additional condition.
Thus the triangle (624, 576, 240) is such that one of the perpendicular
sides is a square and the difference between the hypotenuse and twice the
other perpendicular is also a square, but the hypotenuse »zzus the latter
perpendicular is not a square.

10. 70 find a right-angled triangle {, & vn such that

p=s
E+ n=1
E—fy=tF
= Zy=af

Assume x, 1 — « for the sides &, n about the right angle respectively.

This supposition satisfies the second condition.

Again, since ¢n=a —«% the third and fourth conditions are satisfied,
for ’=x% w=1—-2x+ x2
THEOREMS AND PROBLEMS BY FERMAT 305

It remains to satisfy the conditions
p=l n=l

and P=Fitf=1~28+387=0n ws
The difference = 2a2?—x=4x (4x —2), and we find, in the usual way,

= 10
=a

The triangle is (£3, +9, %)

11. 70 find a right-angled triangle {, & m such that
E=a cube
E—Ln=a square

Fermat assumes {=1, p=x, so that the first condition is satisfied,
1 being a cube.

We must now have 1-x=4"
and also P= tfi=1+ai=f
The difference = * + {x = }a (4x + 2), and we find x=— 272.
In order to derive a positive value for x we substitute y — £22 for x in

the equations, which gives

2 544
Yn r hen

Make the absolute term in the first equation equal to that in the latter

ig 12460
by multiplying by and we have to solve
361. 225

24609..124609.,._,,/2
“foczE 169150)" ¥%

9 544 124609 .__,2

Y ~s3%) ¥iser =0
9

: 2 39 26815
The difference =y* - 1353360 =r (V — 153130):

We find accordingly

Y=-"806970928200
A ee 272 __.90032607119
and X=) = 555 = 80970928200

12. 70 find a right-angled triangle {, & n such that
{+ Lén =r
Form a triangle from the numbers x + 1, x ; the sides are
l=202+ax +1, f=2x ti, n=2x+2%.
Thus {+ 3én= 24% + 52? + 3x + 1 must be a square.
Suppose 2 + sa + 3x + 1= (3x + 1)%
and we have x= 31

H. D. 20
306 : SUPPLEMENT

Now substitute y — 1 for x in the expression to be made a square, and
we have
2)’ — 32)" + 33y + $35 = a square
19 17)\2
= (38 + 15)" say,
15389 15389 _ 11 67:
4

= oS
whence y = 4%, and x = 155 - 4 = 52

wes

The triangle is accordingly found.

13. 70 find a right-angled triangle {, §, n such that
§+ fin =1A
Form a triangle as before from x + 1, x, and in this case we shall have
22% + 30% + 3x + 1 =a square
= (2x + 1), say,
whence x= — 2.

Substitute y — 3 for x in the expression to be made a square; thus
28+ 492+ 81y + AS =a square
= ($55 + 1) say,
whence y=4%33%, and x= $933 - 8 = 143%
the triangle being therefore
549 3
122 1

4337
9312)

14. 70 find a right-angled triangle {, &, m such that
(§+n)"—$én =u

let =x, 9=1; then {3 =a + v =a square

Also, by the condition of the problem, x*+ 2x + 1 =a square]

The usual method of solution gives x=— $2.

Substitute therefore y — 25 for x in the two expressions, and we have
the double-equation

rz Bore

5¢

Or, if we make the lg term in the first expression the same as in
the second by multiplying by $323

69)
5329,2_ 101251

5329
1389) — 32856) t e302 =?
2 55 5329 __ 2
Y= Vt iS =
% —89960 24638 ,,-.36 3110 2163
The difference =$3§35°— 3138y = 33 (BY — 338d)
39655

and we find y= 32392, so that x =y — 35 = 29085.
Therefore the two perpendicular sides of the triangle, in whole numbers,
are 39655, 129648, and the hypotenuse is 135577.
THEOREMS AND PROBLEMS BY FERMAT 307

15. 70 find a right-angled triangle {, &, m suck that
(€ +m) + $n =a square.
This problem is mentioned in Fermat's letters to St Martin of 31st May
and to Mersenne of 1st September 1643". The result only is given (in

the letter to Mersenne), and not the solution ; but it can easily be worked
out on the lines of the solution of the preceding problem.
Let £= x, y=1; we must therefore have
adic both squares.
X+Sx+1
Solving in the usual way by splitting the difference §x into the factors
5.2x we find x=-33.
Substitute y — 33 for x in the two expressions, and we have to solve
Yt (3iyr=2,
P+ 23y wif =07
Multiply the last by ($3) so as to make the absolute terms the same;

and we have to solve
2

as

y-2yel
G8 5+ GDF. y+ (
The difference = {(83)*— il +2803,

= 100 44879
2p {iy +482),

89
80
89
80
39

A
it

We therefore put (y— iy =p: -33y+ (88),

4879  39\__14870N2  /830\2
whence Y\Erio" 23) = (3&7) ER)
39 __190281
and y=34231 so that x =y — 33 =130281,
2

The ns triangle is therefore (325552, *A%22L 1) or, in whole
numbers, (205769, 190281, 78320).

16. 70 find a right-angled triangle {, &, n such that
GC +m. Lén=a square.
Fermat takes the case where 7 = 2.
Form a triangle from the numbers «, 1; the sides are then
I=a?11, E=al~14, n=2n,
Thus we must have (x? + 1)? + 2x (x*— 1) a square, that is,
xt + 24° + 2x2 — 2x + 1 =a square

= (2% + x + §)?, say,

Bf
.

whence x

v Qeuvres de Fermat, 11. pp. 260, 263.

20=—>2
308 SUPPLEMENT

But this value makes x®—1 negative; so we must seek another by
putting y + } for x in the expression to be made a square.

We have 3*+ 3° + 39" — fy + 333 =a square
= i — 26) + $1577) say.
This gives y= {43££ and v=y + }=H311
Therefore the triangle is generated (in whole numbers) from 571663
and 436440.

De Billy adds that there is one case in which the problem is impossible.
Tannery observes in a note 6 thi this remark seems to refer to the case in
which 7 = 8.

17. 70 find a right-angled triangle {, & vn such that
— én =a square.

Form a triangle from x — 1, 4; the sides will then be

{=ai-ont1y, Z=x'—-28-~15, 9=3x-8
Thus (x? — 20 — 15)? — (44 — 4) (x*— 2x — 15) must be a square, that is,

—8x®— 144% + 112x + 165 =a square :
= (x*— 4x — 15)", say.
This gives x =— 45, and accordingly, to find another value, we substitute
y—242 for x in the expression to be made a square.

We must therefore have
Y'—38 +1900, 541g 4 BAERS =a square

= Oo == Nay Jee 235), say.
This gives y =5423 and x=y — 15 = 8571,

The triangle is therefore formed from ¢%%%, 4, or (in whole numbers)
from 6001, 2280.

The sides are therefore 41210401, 30813601, 27364560.

18. 70 find a right-angled triangle {, & n such that (if & > 3)
(§—n)*— 2m = a square.
This problem is enunciated in Fermat's note on vi. 22. He merely

adds that the triangle (1525, 1517, 156) formed from 39, 2 satisfies the
conditions, but does not give the solution.

The solution is however easy to obtain by his usual method, thus.
Form a triangle from x, 1, so that

{=afty1, C=a'-73, y=2a.
Then ((—n)t—2n?=(x?—2x0—1)*—

= at — 42% — 62% + 4x + 1.
THEOREMS AND PROBLEMS BY FERMAT 309

* . .
This has to be a square ; let it be equal to (x*— 2a — 5)? say ; this gives
4% +1 = 20% + 25,
or x=—3

The triangle formed from —32, 1, or from —3, 2, will have one side
negative. To avoid this, we proceed as usual to form a triangle from

y—3, 2.

Thus {=P—-06x+13, C= —-6y+5, n=4y-1s,
and (E—nP—2f=(F~10p + 17) ~ 2 (4p — 12)?

=y'—20)2 + 10297 —148y + 1.

In order that this may be a square, suppose it equal to (y*— 10y — 1)? or
y—201 +98’ +20y +1.

It follows that 102y* — 148y = 98) + 209,
and y = 42.

The triangle required is formed from y-3, 2, that is, from 39, 2,
and is accordingly 1525, 1517, 156.

Fermat does not tell us in the note on vi. 22 what use he made of this
problem, but the omission is made good in a letter to Carcavi', where he
says that it was propounded to him by Frénicle (who admitted frankly that
he had not been able to solve it), and that it served to solve another
problem which had occupied Frénicle. The latter problem is the
following.

19. 70 find a right-angled triangle {, & n such that

{
{—n tare all squares.
$=

~ Fermat does not actually give the solution, but presumably it was
somewhat as follows.

Form a triangle from two numbers x, y ; the sides are then
{=f +p f=, y=auy
Now {—n=x+*— 2xy and is ps0 facto a square.

The other conditions give
x? + y*=a square,

and x? ~ py? — 2xy = (x — y)* — 2p” =a square.

These conditions are satisfied by the two perpendicular sides of the
triangle of the last problem, that is, by x = 1517, y= 156.

1 Qecuvres de Fermat, 11. p. 2635.
310 SUPPLEMENT

The triangle required is therefore formed from 151%, 156 and is

(2325625, 2276953, 473304):

The present seems to be the appropriate place for a problem contained
in a letter from Fermat to Frénicle the date of which was probably
15 June 1641.

20. 70 find all the right-angled triangles in integral numbers such that
the perpendicular sides differ by 1.

If a right-angled triangle is formed from x, y, the difference between
the two perpendiculars is either a?—3?— 2xy or 2xy—(x?—3?), that is
to say, either (x —y)?— 23? or 2° — (x —y)%. As this difference is to be 1,
we have to find all the integral solutions of the equation

i= (a= =t 1

Those who are familiar with the history of Greek mathematics will here
recognise an old friend. The equation is in fact the indeterminate
equation

28—-fF=11,
which the Pythagoreans had already solved by evolving the series of
“side-” and “diagonal-” numbers described by Theon of Smyrna, the
property of which they proved by means of the geometrical theorems
of Eucl. u. 9, 10.

If x, y are two numbers such that -

25% —- i= + 1,
then the numbers x + y, 2x + y will satisfy the equation

252 P=—1 ;
fresh numbers formed from «x+y, 20 +» by the same law will satisfy the
equation

24° p= "> = 7,
and so on.

Take now the equation
8 = (r=,
where x, y are two numbers from which a right-angled triangle has been
formed. We can deduce a right-angled triangle formed from x’, y' where

Wey P=;
for by the above law of formation we have only to take
Y=y+(x-y)=x
x —y =2p+ (x—p)=x+y,
whence also &' = 2x + y.

1 Oeuvres de Fermat, 11. PP: 221 sqq.
THEOREMS AND PROBLEMS BY FERMAT 311

Fermat gave two rules for the formation of this second triangle. The
first rule is in the letter above quoted.

First Rule. 1f h, p, b be any right-angled triangle satisfying the con-
dition (% being the hypotenuse, p > 4 and p —& = 1), then, if a triangle be
taken in which

the least side =24+p + 24,

the middle side =24 +p + 26+ 1,
the greatest side =3% + 2 (p + 0),
this triangle also will be a right-angled triangle satisfying the condition.
To verify this from the above considerations we have to consider two
cases, according as 2xy 1s greater or less than a? — 3%
Take the case in which 2xy > a? — »?; then
2)? — (x —y)2=+ 1,
and accordingly
2) («' - 7) B= ly
or I EB
The least side, therefore, of the second triangle
2x’y = 2x (22+ y)=2(x? + 1) + (2p) + 2 (4° — 17) ;
the middle side
ay many + 1s
and the hypotenuse
x24 3%= (2x +p): + af = 3(2% +7) + 2(x?—y* + 2p).
The expressions on the right hand are those given by Fermat's rule.
Second Rule.

This rule is given in a letter of 31 May 1643 probably addressed
to St Martin.

Fermat says: Given any triangle having the desired property, then, to
find another such triangle from it, “subtract from double the sum of
all three sides each of the perpendiculars separately [this gives two of the
sides of the new triangle], and add to the same sum the greatest side [this
gives the third side].”

That is to say, the sides of the new triangle are respectively
2 (2x + 2xy) — (2 — 3%),
2 (24% + 2x9) — 2x,

2 (25% + 2x9) + (x2 + 32).

L Oeuvres de Fermat, 11. p. 259.
312 SUPPLEMENT

In fact the three expressions reduce as follows :
2 (22° + 2p) — (8° — J) = 38% + 4xy + yi = (2x + y): — 2%,
2 (22% + 2p) — 2x = 22(2% +p),
2 (22% + 229) + 8% + 2 = (2x + ¥)* + 47;
and the result agrees with the formation of the triangle from «/, y’ above.

From the triangle (3, 4, 5) we get (20, 21, 29); from the latter the
triangle (119, 120, 169), and so on. The sixth such triangle is (23660,

23661, 33461).

21. 70 find all the rational right-angled triangles in whole numbers
which are such that the two perpendiculars differ by any given number.

To his explanation of the First Rule above, applicable to the case
where the given number is 1, Fermat adds in his curt way: ‘same method
for finding a triangle such that the difference of the two smaller sides is
a given number. I omit the rules, and the limitations, for finding all the
possible triangles of the kind required, for the rule is easy, when the
principles are once admitted.”

He adds, however, to his Second Rule?! its application to the case
where the given number is 7.

There are, he says, two fundamental triangles with the desired property,
namely 5, 12, 13 and §, 15, 17. [In the case of the former 2a > 22-57,
and in the case of the second x? —y* > 2xy.]

From the first triangle (5, 12, 13) we deduce, by the Rule, a triangle
with the sides 2. 30-12, 2. 30-5, 2. 30+ 13 or (48, 55, 73); from the
second a triangle with the sides 2.40-15, 2.40-8, 2.40+ 17, or
(65, 72, 97)-

And so on, ad infinitum.

Next to the explanation of the first of the above Rules Fermat
mentions, in the same letter, the problem

22. To find right-angled triangles in integral numbers {, & mn (§> 7)
such that
{~7
=
He observes that alternate triangles of the series in which the two
smaller sides differ by 1 satisfy the conditions, those namely in which the
smallest side 7 is 2ay and not x%—y*; for x%+3®—2xy is a square, and
¢ —n, being equal to 1, is also a square.

| are both squares.

1 Oeuvres de Fermat, 11. p. 259.
THEOREMS AND PROBLEMS BY FERMAT 313

Thus, while 3, 4, 5 does not satisfy the conditions, (20, 21, 29) does,
and, while the next (119, 120, 169) does not satisfy the conditions, the
triangle after that, namely (696, 697, 98s), does.

Frénicle naturally objected, in his reply, that the triangles should not
be limited to those in which the smaller square representing the difference
between the perpendicular sides is 1, and proposed the problem in the form

To find all the triangles ({, & n) such that
{rv
£9

and one square does not measure the other.

| are both squares,

Fermat seems to have, in the first instance, formed the triangle
from two numbers x, y where
x=72+1, y=27r-3,
and then to have given the more general rule of forming a triangle from
x=2 45 yzalr-—1s)s,
where 7, 5s are prime to one another (Letter from Frénicle of 6 Sept.
1641)".

It appears from a letter of Fermat’s to Mersenne of 27th January
1643% that St Martin propounded to Fermat the problem, apparently
suggested by Frénicle?,

Given a number, to find how many times it is the difference between the
(perpendicular?) sides of a triangle which has a square number for the
difference between its least side and each of the two others respectively.

The number given was 1803601800, and Fermat replies that there are
243 triangles, and no more, which satisfy the conditions. He adds “ The
universal method, which I will communicate to him if he asks for it, is
beautiful and noteworthy, although I doubt not that Frénicle has already
given him everything on the subject of these questions.”

23. 70 find two triangles, {, &, qn and {, &, v' (§> 0, & >) such that
{i=l ~ 1)
ym l~g
Suppose the two triangles formed from (x, y) and («', y) respectively,
the sides being
{=a+), E=z2ap, g=2"-)~
Z Le x’? +9? & = 2x, 7 = x? —
1 Ocuvres de Fermat, 11. p. 233.

2 Jbid., p. 250.
8 bid. p. 247.
314 SUPPLEMENT

Then we must have
(=F = 252 7
and gf =n yy ={r yy :
which equations show that y =’, and that
(=F 0 =F
are three squares in arithmetical progression.

Suppose that these squares are 1, 25, 49 respectively; thus y =5;
B-y=3, so that x=1+95; 2’ ~y=7, so that ¥’'=3+7%.

Fermat accordingly gives the rule: Find three squares in arithmetical
progression ; then form the first triangle from (1) the sum of the sides of
the first and second squares and (2) the side of the second, and the
second triangle from (1) the sum of the sides of the second and third
squares and (2) the side of the second’

In the particular case, the triangles are formed from (6, 5) and from
(12, 5) respectively; the triangles are therefore (61, 60, 11) and (169, 120,
119) respectively.

For solving the problem of finding three square numbers in arithmetical
progression Fermat seems first to have given a rule which was not general,
and then in a later document to have formed the sides of the three squares
as follows : :

—25, Prars+2s P+ 4rs+ 2s

Frénicle formed them thus?:
B—29-9 +9 P+200-0
the latter form agreeing with Fermat's if p=7+ 5, and ¢ =.

Frénicle expresses his formula neatly by saying that we take for the
side of the middle square the hypotenuse of any primitive triangle formed
from p, g¢, i.e. p+ ¢% for the side of the smallest square the difference
between the perpendicular sides of the same triangle, i.e. p*— ¢*— 2p¢, and
for the side of the largest square the sum of the perpendicular sides of the
same triangle.

Suppose the primitive triangle is (28, 45, 53) formed from (7, 2).
Then the sides of the three squares in arithmetical progression are 17, 53
and 73, the squares themselves being 289, 2809, 5329. The triangles
derived from these squares and having the above property are formed from
(70, 53) and from (126, 53) respectively, and are therefore (7709, 7420,
2091) and (18685, 13356, 13067).

L Qcuvres de Fermat, 11. p. 225.
% Jbid., 11. pp. 234-5.
THEOREMS AND PROBLEMS BY FERMAT 315

24. 70 find two right-angled triangles (¢, & 0) and ({, &, v') such that
£ rs 7 = £ J 2 :
§={
Form the triangle { & » from the numbers «, 1; then
{zal +1, il=22=—1, n=on
Thus ¢=x%+1; and, since &€—y' =¢—-np=a"—2x—1, it follows that
7 =2x+ 2.

It remains to secure that £% + #2 = (x? + 1)? + (2x + 2)* shall be a square,
that 1s,

x*+ 62% + 8x + 5 = a square
= (a+ 3)" say ;
therefore x = 1. ;

Hence the triangle  §& 7 is formed from §, 1 or from 1, 2; but this
solution will not do, as it gives a negative value for & Accordingly
we must find a fresh value for x, which we obtain by forming the triangle
from x + 1, 2.

The sides are then
{=sisan ts, Emalton-—13, modo ty;
thus E=xt+2x+5, P’=—(a—20—7)=4x+ 12.
Therefore (x? + 2x + 5)? + (4x + 12)* must be a square, or
xt + 44° + 30x? + 1164 + 169 = a square
_ (13+ $a — 2%), say,
from which we obtain x=— 1525 and the triangle is formed from — 313, 2,
or (in whole numbers) from — 979, 1092.

“We can use these numbers as if both were real and form the triangle
from 1092, 979. We thus obtain the two triangles

2150005, 2138136, * 234023,
21635017, 2150905, 246792,

which satisfy the conditions of the question.”

25. 70 find two right-angled triangles ({, & n) and (§, &, v') such that
£iy=24 oo
di
§={
Form the triangle {, ¢, » from the numbers x + 1, 1; then
{=a tanta, t=nliom n=ox+ta,

Thus £=a2+ 20+ 2, and y= {+9 & = 2x.
316 SUPPLEMENT

We must now have 2 + n?= (2? + 2x + 2)? + (2%)? a square ; that is,
xt + 42% + 124% + 8x + 4 = a square
= {af + am + 4), say;
whence x = — 3.

Accordingly we substitute y — 2 for x, and we must have
Fa 3P-20y + 388 =a square
=f - 209 +7, wy,

This gives y=21, and x=2}- 3 = 2.

The triangle , &, 7 is therefore formed from 22, 1, or from 29, 26, and
is therefore 1517, 165, 1508 ; the triangle {, £, % is 15235, 151%, 156.

Or again we may proceed thus from the point where we found x =—3.
The triangle ¢, £ n may be formed from — 4, 1 or from —1, 2.

We therefore form a triangle from x — 1, 2 and start afresh.

The sides are

{=x~2x +5, C=xi-ax-3, N=qn—4

Thus § =x’—2x+5,and p= +n —§ =4x— 12.

Hence (x? — 2x + 5)* + (4x — 12)* must be a square ; that is,

at — 44% + 304% — 1162 + 169 =a square
= (13 - $5 + 2), say.

This gives x=4%%, and the triangle {, §, 7 is therefore formed from
2%, 2, or from 29, 26, as before.

The remaining problems on rational right-angled triangles in the
Inventum Novum are cases given in Part 11. of that collection to illustrate
the method of the Triple-Equation due to Fermat and explained by him on
Diophantus v1. 22 as well as, at greater length, in the Znventum Novum.
An account of the method will be found in a later section of this Supple-
ment ; but the problems applying the method to right-angled triangles
will be enunciated here.

26. 70 find a right-angled triangle {, & vn such that

{E+m=2)
P+al=u
E+é=2"
+n=2w’

By Problem 2 above find a right-angled triangle %, p, 4 (% being the
hypotenuse) in which #4, p+ 4 are both squares; the first condition is thus
satisfied.
THEOREMS AND PROBLEMS BY FERMAT 317

To find § §, 7, put {=kx, =x, p=20x
The three remaining conditions thus give a ‘triple-equation” in x.
[The numbers would of course be enormous. ]
27. 70 find a right-angled triangle {, & m such that
Ett Fetes
(bone FP rm=7),
(E+n+ 0 +mi=w’
where m is any given number.
Fermat supposes 7 = 2.

Assume for the required triangle (3x, 4%, 5x); we have then the
triple-equation
T4454 3x=1"

T4423 + Ax=92 Ls
144%3 + Tox = 70
the solution of which gives x = 33535, and the triangle is
507 6786 845
38016» 38016» 38016"

28. 70 find a right-angled triangle {, & n such that
uy
(+++ é=ut
(E+9tiey=0F

(E+ +0)? + ml=w?
Suppose 7 = 2.

Find a triangle (Problem 3 above) in which {, £— 5 are both squares,
say the triangle (119, 120, 169). Put 119x, 120%, 169x for the sides of
the required triangle, and we have the ‘triple-equation ”

1664642 + 119x = 2?
166464x% + 120x = 7?
16646422 + 338x =u?

29. 70 find a right-angled triangle {, &, n such that
E—=0*
grniratait |
(+9 tf +y=0"
(E+ + 0+ ml=w?
where m ts any given number.

* The enunciation has £ (£ — £7) instead of £2 — 37 ; but £ (£ — 3&9) is inconsistent with
the solution given, and I have therefore altered it so as to correspond to the solution.
318 SUPPLEMENT

Take a right-angled triangle in which 1, p are the sides about the right
angle and are such that 1 —}p is a square (Problem 11 above).

Let ¢ be the hypotenuse of the triangle so taken, so that g= \/(2* + 1),
and take as the sides of the required triangle x, px, gx; we thus have
the triple-equation

(1+p+ gl +x=0"
(1+p+g)}2a+px=1°
(1+p+9) x + mgx =u’

30. 70 find a right-angled triangle {, &, n such that
E(E+n)=12
(Sry iY ti=0r
(Etn + Ot y=df
+++ {=F
First find a triangle in which one of the perpendiculars is a square, and
the sum of the perpendiculars is also a square, say 40, 9, 41.

Take gox, gx, 41x as the sides of the required triangle ; and we there-
fore have the triple-equation

S100%% + qox = 1/7
Siooa + gx =F

8100x% + 41x = 20?

SECTION IV.
OTHER PROBLEMS BY FERMAT.

31. Zo find two numbers &, m such that
(1) Ef)
(2) n— (8 — 1?) are all squares.
(3), (4) Sin)
Leté+n=1—2% £—n=2x,50 that =, y=4— 2x, and
E—n?=2x— 4a
Thus (3), (4) are both satisfied.
The other conditions (1) and (2) give
45% — 2x + § = o*
a
The difference = 2x =4x.1; and, putting (2x + })2=4a?— 2x + 4, we
find x= 5%.
THEOREMS AND PROBLEMS BY FERMAT 319
The required numbers are therefore §, 2%.

Another pair of numbers satisfying the conditions will be obtained by
substituting y + 5 for x in the expressions to be made squares, and so on.

32. 70 find two numbers & n such that
E+nx(E-m)
+n (E—7)

Let the numbers be § + x, 4 — x; therefore £ — 1, as well as £— 7»? is
equal to 22. The sum £4 yp=1,

} are all squares.

Therefore 1 + 220 must be a square, or we have the double-equation
I t2x=00
I—-2x= :
Replace x by $* +» so as to make 1 + 2x a square ; therefore
I —2y—¥® =a square

={1 = 39), sy,
whence y=2%, and x =§y* + y= 12.

’ 1 49 l
The required numbers are therefore $3, 3% -

33. 720 find two numbers &, n such that
(+7) (£2 +9) = a cube.
Assume £= x, 3=2 — x; therefore
(E+n) (E+) =2 (2x7 - 4x + 4)=a cube
= (2 — 2 x)’, say.
This gives x =— 3; and to get a “real” value of x we must substitute
y— 3 for x in the expression to be made a cube.

[Thus 4)" — 44p + 125 =a cube
= (5-130) say
= 125 — 449 + 97%)" — Sere
and y=122825 so that x =y —§ = 24133.
The required numbers are therefore 2$ 73:
Cor. We observe:

(1) that the numerators 26793, 15799 satisfy the conditions ; ,

(2) that we have in fact solved the problem Zo divide 2 into two parts
such that twice the sum of their squares is a cube;

(3) that we can solve in the same manner the problem 70 find two
numbers such that any multiple of the sum of their squares is a cube. Thus
suppose that the multiple is 5; we then assume x and 5-—x for the
numbers and proceed as above;
320 SUPPLEMENT

(4) that we can also deduce the solution of the following “ very fine
problem ”’

To find two numbers such that their difference is equal to the difference of
their biguadrates or fourth powers.

In other words, we can solve the indeterminate equation
Sym fi
For we have only to take the two numbers found above, namely 26793
and 15799, and divide by (as a common denominator) the root of the cube
formed by multiplying their sum by the sum of their squares.

This common denominator is 34540, and the two required numbers are

26793 15799
348340) 31540"

This latter problem is alluded to in Fermat’s note to Diophantus 1v. 11
in these terms: “But whether it is possible fo find two biguadrates the
difference between which is equal to the difference between their sides is a
question to be investigated by trying the device furnished by our method,
which will doubtless succeed. For let two biquadrates be sought such
that the difference of their sides is 1, while the difference between the
biquadrates themselves is a cube. The sides will, in the first instance, be
-o and 13. But, as one is negative, let the operation be repeated, in
accordance with my method, and let the first side be x —%; the second
side will be x + 3, and the new operation will give real numbers satisfying
the condition of the problem.”

34. 70 find two numbers &, m such that
& + 30 = a square.

Fermat (or De Billy) observes that it must be required that the first
biquadrate (£%) shall not be unity, for in that case the problem would be
foo easy, since 1 + 3.1=4 and 1 + 3. 16= 40.

Assume =x, p=x — 1; therefore

4x*— 124° + 182% — 12x + 3 = a square
= (24? — 3x + 1)? say.

This gives x=1!, x—1=2; and a solution in whole numbers is
f=11,7=3 In er I+ 3.3'= 14041 + 243 = 14834 or 122%

We can also take any equimultiples of (11, 3), as (22, 6) and (33, 9);
and the latter pairs of numbers severally satisfy the condition of the
problem.

20703 15799

10994” 10994
we can obtain the same solution of the main problem as that given above

(gn 15799).
34540 34540

 

1 It gives in fact as a solution of the subsidiary problem, and from this

 

 
THEOREMS AND PROBLEMS BY FERMAT 121

SECTION V.
FERMAT’S TRIPLE-EQUATIONS.

Fermat's own description of his method of “triple-equations,” which is
contained in his note on vi. 22, is as follows :

“ Where double-equations do not suffice, we must have recourse to
triple-equations, which are my discovery and lead to the solution of a
multitude of elegant problems.

If, for example, the three expressions

X + 4, 2: + 4, 5X + 4
have to be made squares, we have a triple-equation the solution of which
can be effected by means of a double-equation. If for a we substitute a
number which when increased by 4 gives a square, e.g. * + 4 | Fermat says
x* +4), the expressions to be made squares become
: Fra+s, +r, spitaoptas
The first is already a square ; we have therefore only to make
2+ Syr+4)
sy +2004 4)
severally squares.

That is to say, the problem is reduced to a double-equation.

This double-equation gives, it is true, only one solution; but from
this solution we can deduce another, from the second a third, and so on.
In fact, when we have obtained one value for y [say y =a], we substitute
for y in the equations the binomial expression consisting of y plus the value
found [Ze y +a]. In this way we can find any number of successive
solutions each derived from the preceding one.”

The subject is developed in the Doctrinae Analyticae Inventum Novum
of De Billy already mentioned so often.

It will be observed that the absolute term in all the three expressions
to be made squares is a square. It need not be the same square in the
original expressions ; if the absolute terms are different squares, the three
expressions can, so far as necessary, be multiplied by squares which will
make the absolute terms the same, when the method will apply.

We may put the solution generally thus. Suppose that

ax + p*

bX +g%

ox
have to be made squares (a, 4, c or some of them may be negative as well
as positive).

Put : ax =y* +2py,
which makes the first expression a square (or of course we could put
ax = a¥y* + 2apy). :

H. D. 21
322 SUPPLEMENT
Substitute (* + 2py)/a for x in the second and third expressions.

b
Therefore ” (+ 229) + &°

S(t + apy) +r

must both be squares ; or, if we multiply the first expression by 7° and the
second by ¢* (so as to make the absolute terms the same), we have to solve

the double-equation

b

Rr apg) r=,

gs a
Stra) rr =o
— cq? i cq?

Lyrap Ty

 

The difference = i

This has to be nr into two factors of the form Ay, wy + v, where
v must be equal to 2¢7 (in order that, when % {(A + n) y + v} is squared and
equated to the first, or when 1 ian is equated to the second, of
the two expressions, the ats terms ¢*#* may cancel each other).

A different separation into factors is possible if 4/a and ¢/a are both
squares ; but otherwise, as Fermat says, the method gives only one
solution in the first instance ; the above difference must necessarily be
split into the factors

(or —cq?) qr
gE fy and y+ 207.
agr it » A q
Half the sum of these factors
JEU ray,

= Tw Ler

a evga +g7.
Te apqr

Squaring this and equating it to 4 (+ 2py) + ¢*7%, we have

1 rT F-Zy 2007 En
{15( apgr hye IEEE ea hg

=1

 

therefore
an or
Sl tes + (ag*7 + 0° — pg)’ 4
oy {22 Ah
=} = ap } ’
ot to ag®r? + bp — p29) — 4abpig*ri)

2 > (07°p* + cp?g* — ag’)
THEOREMS AND PROBLEMS BY FERMAT 323

that is,
I(@¢H + rip + Apigt — 20cp*g*r — 2cap’y'r — 2abp*g*r*)
= 4apg’s (021 + pg — gr),
qapg*r* (orp? ro A
TEP Br + EP9t = 20cp' yr? — 2cap’y'r? — 2abp’g
whence =~ 7 2) 1s found.

Exx. from the Znventum Novum.

(1) 2x44
3X +4 + to be made squares.
6x + 4

or

 

Here a=2, 4=6,2=3, p=g=7=2; therefore
4-2.32(6.16+3.16—2.16)

 

 

> 16°(4+ 36 +9 re 3.2-2.2.0)
16.7
pe
Lg r {112° TT? 6 1120
and =llrs nll 100 2 = Blom -a, 23) = or
(2) X+ 1
3x +4 to be made squares.
2x +9

Here a=1,6=3,¢=2, p=1, g= 2, r=3; therefore

4:36(3.9+2.4-36)

 

i 36°+9.81+4.16-12.36-16.36-6.36.9
ol Td 90 Yad
—36.46+9. ie 16 363°
and « = 3+ ay = (44)" + 2 (334) = 222838.

~The disadvantage of the ions is that it leads so soon to such very
large numbers.

Other examples from the /Znventum Novum are the following, which,
like those above given, can be readily solved ab initio without using the
above general formula.

(3) To solve 1+ x=
Itz2a=1" 7
I + 3u= ut

Put x =? + 2y, and substituting in the second and third expressions we
have only to solve the double-equation
2+ apti=08
5+ 1op+r =r)’
21—2
324 SUPPLEMENT

The difference = 3 (J* + 23) = 3v (1 + 2).
Equate the square of half the difference of the factors to the smaller

expression ; thus
(y—1¥=2+4y +1,

whence y = — 6, and x =)? + 2 = 24.
(4) Equations Sd om
3x +9=0

5% +9= 202

In this case we put x=)’ + 6y, and we have to solve
{rtp +g=e }
50°+6y)+9=%")

The difference = 2 by + 6y) = 2y (y + 6); we then have

Qy-3 =r rade,

and y = — 84, so that x =3* + 6y = 452%.

(5) Equations i smi)
F—3 X=

1+ 5x =f J

If we assume x=)? + 2y, we find y = & and v= 2A.

There are two other problems of the same sort which are curiously
enunciated.

(6) “To find three cubes such that, if we add their sum to numbers
proportional to the cubes respectively, we may have three squares.”

What Fermat really does is to take three cubes (a #, ¢*) such that their
sum 1s a square (this is necessary in order to make the term independent
of x in each of the three expressions a square) and then to assume
ax, x, Fx for the numbers proportional to the cubes. He takes as the
cubes 1, 8, 27, the sum of which is 36. Thus we have the triple-equation

36+ x="
360+ Sx=09%
36+ 27x = 2 J

We put x=3°+ 12y in order to make the first expression a square.

Then, solving the double-equation
30+ SF raap=v |
36+27()+129)=20)°

we obtain y = 204 and x =? + 12y = 229320

(7) “To find three different square numbers such that, if we add
to them respectively three numbers in harmonic progression, the three
resulting numbers will be squares.”

Fermat first assumes three square numbers 1, 4, 16 and then takes
2x, 3a, 6.x as the required numbers in harmonic progression. (He observes
THEOREMS AND PROBLEMS BY FERMAT 325

that, of the three numbers in harmonic progression, the greatest must be
greater than the sum of the other two.) We thus have the triple-equation

I+2x =u
4+ 3x=mefin,
16 + 6x = 70?

16 + 32x =u
16+ 120 = 97
16+ 6x =2t]

or, if we make the absolute terms the same square,
~ 2,12

Ty for x, we solve

Making the last expression a a by putting y
-2and x=1(1*+8)=

as usual and obtain y =
Fermat observes that ei lem of the form
= ed

£4 x=
rong Ll
2 +s = uf

that is to say, of the form
PX tax =u
Flt dn=af

a A

can be similarly solved, because they can be reduced to the above linear
form by putting a = 1/y and multiplying up by y=
Lxamples.

To solve the triple-equation

(1)
qx an =
4x + 6x =F
4x° + gx = WW?

If x = 1/y, this is equivalent to
y+ 4=u"
6y+4=0"
gy +4 ="
z and solving as usual, we find
293%, and x=32438,

Putting y= 12° + 2
s=—-2%8 y=12+25=3%03%
(2) Equations AE + pal
4+ mw = 98
9x? + 200 = 70
I'his is equivalent to
r+ I=" |
Wtd=9C |
2y +g =u? J
We put y = z* + 25 and, solving the double-equation
27 (22+ 9 + Sh )
= 144769

0
9s SO Tr x= Sans
326 SUPPLEMENT

(3) “To find three square numbers such that, if we add their sum to
each of their roots respectively, we obtain a square.”

Choose, says Fermat, three squares such that their sum is a square and
such that the root of the greatest is greater than the sum of the roots of the
other two (the reason for this last condition will shortly appear); eg. let
the squares be 4, 36, 81, the sum of which is 121.

Let 44% 364% 81a? be the three square numbers required; therefore

121x2 + 2x = 2°
12152 + 6x = 72
1214% + one = wf

The solution, arrived at as above, is a =;%2%5%.

Fermat actually used his triple-equations for the purpose, mainly, of
extending problems in Diophantus where three numbers are found
satisfying certain conditions so as to find four numbers satisfying like
conditions. The cases which occur are in his notes to the problems
Hl. 15, IV. 10, 20, Vv. 3, 27, 23; they are referred to in my notes on
those problems.

De Billy observes (what he says Fermat admitted he had not noticed)
that the method fails when, the absolute terms being the same square, the
coefficient of x in one of the linear expressions to be made squares is equal
to the sum of the coefficients of x in the other two. Thus suppose that

Trex Tax nhsy
have to be made squares. To make the first expression a square put
x = 2p’ + 2y. The other expressions then become
I+6y+6)% 1+ 10)+ 10°
The difference is 4)* + 4 = 2y (2y + 2), and the usual method gives
(2y+1)’=10)" + 10¥ + 1,
or 6y* + 6y = o,
so that y = — 1, and consequently x = 2y* + 2y =o.
It does not however follow, says De Billy, that a set of expressions so
related cannot be made squares by one value of x. Thus 1 + 5x, 1 + 16x

and 1+ 21x are all squares if x= 3, the squares being 16, 49, 64. He
adds (§ 11) that “we must observe with Fermat” that the triple-equation

I+ xX=7%°
I+2x=0"
Tk Sut

not only cannot be solved by the above method, but cannot be solved at
all, because “kere cannot be four squares in arithmetical progression,” which
however would be the case if the above equations had a solution and we
took 1 for the first of the four squares.
THEOREMS ‘AND PROBLEMS BY FERMAT 327

The subject of triple-equations has been taken up afresh in a recent
paper by P. v. Schaewen!. The following are the main points made.
(r) The equations ax +p? = u*
bx + gt = 17

x+r=ur
can be reduced to the form

1 tax =u
1+ Ba =
yh =f
by substituting 7.x’ for x, where m is the least common multiple of 2% ¢% #2

(2) The method of Fermat has the disadvantage that, with one
operation, it only gives one value for x and not by any means always the
smallest solution. From this point of view there is a better method, namely
that of finding the general solution of the first two equations, substituting the
general value of a so found in the third equation and solving the resulting
equation in a new unknown. Consider the equations

I+ax=1u
I1+6x=19"
I+ex=ut

Suppose 1 + ax = p* some square. Therefore
1+6x=1 +211),

and, multiplying by a? we have to make
abp® + a* — ab a square.
This is a square if p=1 ; and we therefore substitute ¢ + 1 for p. Thus
abg® + 2abg + a* = a square

772 2
= (2 s+a), say.
Therefore (at — ) g=2 ( a a0) .
3% n

2an (m — nb)
J=

“and Na
abn® — m*

)

. _2amn— abn’ — w*
whence pope ma
2-1 1 ((2amn — abn’ — m®): — (abn —m?)Y)
Te Ta { (abn® —m?)? }

— gq mn {im — (a+ b) mn + abn?)
(abn® — m*)? :

Substituting this value of x in the expression 1 + cx, we have a biquad-

)

and X=

 

: eal
ratic expression in ~- which has to be made a square, namely
7"
mt — genin + {4 (a + b) c — 2ab} m*n* — qabomn’ + a*b*n’,

1 Bibliotheca Mathematica, 153, 1909, pp. 289-300.
328 SUPPLEMENT

Example. Find x such that
I—x, I+4x, 1+ 7x are all squares.
First find the general value of x which will make the first two ex-
pressions squares ; this is
_— qmn (Mm — 3mn — 4n)
= (m+ gn)?

)

or, if we substitute £ for 2x/m,
pallet 3b 2)
(+ 1)?
We have now to make 1 + 7x a square ; that is,
A+ 144° + 234% — 144 + 1 =a square.

The first solution of this is # = + 1, and by means of these values we get
the further values 4=% and 4 = 1% (cf. Euler's solution of the problem of
making x +1 and x +1 simultaneously squares quoted in my note on
pp. 84, 85). The corresponding values of x are respectively

120 120120
2 —~ — and

A Teg 427°

 

 

Fermat’s method gives, as the next solution after #, the value

Znoysous
X=1go0e552T

(3) v. Schaewen observes that the problem of finding x such that three
different expressions of the form zx + 2 are all squares can always be solved
provided that we know one solution ; in this case the absolute terms need
not be squares. I doubt however if he is right in supposing that the
possibility of solution in this case was not known to Fermat or De Billy.
I think it probable that Fermat at least was aware of the fact; for this case
of the triple-equation is precisely parallel to that of the double-equation

2X +5=1

6x + 3=9 } 2
given as a possible case by Fermat in his note on Bachet’s conditions for
the possibility of solving double-equations (cf. note on p. 287 above).
Fermat says that the square to which 2x + 5 should be made equal is 16
and that to which 6x + 3 should be made equal is 36 (corresponding to
x = 5%), adding that an infinite number of other solutions can be found.

(4) Lastly, v. Schaewen investigates the conditions under which the
equations
Itax=v 1 téasth, t+l@+d)x=w'
which cannot be solved by Fermat’s method, are nevertheless capable of
solution, and shows how to solve them when they /4ave a solution other
than x =o.
SOLUTIONS BY EULER. PROBLEM 1 320

SECTION VI
SOME SOLUTIONS BY EULER.

PROBLEM 1. 70 solve generally the indeterminate equation’
+t Aol

Vieta solved this equation on the assumption that two of the four
numbers are taken as known.

[I noted on p. 102 Euler's remark that, if 3°+ 4% is turned into the
difference between two cubes by the direct use of Vieta’s second formula,
the formula gives 3° + 4* = (452) - (3%2F but not 3° + 4°=6"—-5% I ought
however to have observed? that the latter can be obtained from Vieta’s fi»s¢
formula if we multiply throughout by a® + 4% The formula then becomes

&(@ +0 = (@® + 0°) + &* (a — 20 + 0? (24° — PP)".

Putting a=2, 6=1, we have 18 =9°+12°+ 15% which gives (after division
by 3%) 6°=3%+ 4* + 5% The next solution, obtained by putting a = 3, é=1,
5848 =28 +353 + 95%, if n=3, §=2, we have 105%= 33° + 70° + 92%; and
so on. Similarly Vieta’s second formula gives

; ad (a® at 248) a? (@ Na oP)? SL 03 (a — 0)? A bo? (2a° + 2%,
and we obtain other integral solutions ; thus
a= 2, 0=7, we have 208= 7 +44 175
ifa=3, b= 1, we have 87°=26" + sst+ 78%:
and so on.]

(1) A more general solution can be obtained by treating only one of
the three numbers x, y, z as known.

To solve @ + 3 +p =v",
put X=pu+7, y=qgu—r7r;
“therefore :

437 (pro us 3 (P=) + (P+) i =v
7 3
= {a + — (2+) say ;

a

and we obtain, after dividing out by (pg + ¢) «2,
2 ALL 7’ »
yD {Ff -prt Pu=tplpt Ne nf oY,

VV. Comment. Acad. Petrop. 1756-57, Vol. VI. (1761), pp. 155 sqq.= Comment.
arithm. 1. pp. 193-206. Cf. pp. 101-2 above.

* See Nesselmann’s ‘“ Anmerkungen zu Diophant” in the Zeitschrift fiir Math. wu.
Physik, XXXVIL (1892), Hist. litt. Abt. p. 123.
330 SUPPLEMENT

Er (prg)-30r(p 9)

EF P= pte)

3@pr (p+g)—ar (2p — 2p) =r (P+)
(ppg +9) -r (+9)

“ epipr gar {fue tpg)

J FF =2g +P) -P (P+)

a (p= pg + ¢) — 3a (p*— 4°) + 2a” (P+ 9)
ad (pr —pg+g)—7" (Pp +9)

where a, » and the ratio p : y may be given any values we please.

or

X=pu+r=

3
J

J

v=a+’ (p+ y=

(2) A more general solution still is obtained if we regard none of the
first three cubes as known.
Suppose that, in the equation

B+ ri=0v

X=! + pu, y=nttgu, s=—nl tru.
Therefore
tp Bagi anip) Put smh | hiv pt ou
+ 37% + 3ng* + 4
+ 378%7 J — nr? + 7
ep +c g-r
Put now v=mt + pnt]

m?
and we have, after division by #7,

timp +n (GF — 7) + u (p+ §F r= 301 nip + nt (g + 1)f
+ wp +o (g +7),

whence, neglecting a common factor which may be chosen arbitrarily, we
have ;
t=mb (p++) — mp + 12 (9 + 7),
w= 3m {mp + n® (q+ 7)}* — 3m {mp + n (4° — #*
or, if we divide by the factor ¢ + 7,
=m (¢* — qr + 7°) = 3m*n’p* — 3mPnip (g + 7) — n° (g +7)?
u=— 3mn (gq — 7») + 6m°n’p + 3m°n* (q + 7),

41

so that x, y, z and v can be written down.
The solution is, however, still not general.
{3) . General solution.
70 find generally all the sets of three cubes the sum of which is a cube.
Suppose Ar BL C= or 43x B=) C2
and assume d=ptg B=p—gs Cmpr—9, D=pts
Then AR + B= 28° + 6pgt, IP — CP = 25® + 6#%,
so that 2(F + 379) = (s+ 37°).
SOLUTIONS BY EULER. PROBLEM 1 331

This equation cannot subsist unless p* + 3¢% s*+ 37° have a common
divisor. Now it is known that numbers of this form have no divisors
except such as are of the same form.

To find them, we introduce six new letters to take the place of
bs D1, 5, thus: let

p= ax + 30y, s=3cy— dx,
g=bx—ay, r=dy+ex,
whence 2% + 39° = (& + 38%) (a2 + 37), 8° + 37° = (&* + 36) (2% + 3),
and our equation, divided by x* + 35 becomes
(ax + 38) (a + 30%) = (3p — dx) (&* + 3¢%) ;
s0 that 2 re ys a ay
y a (a® + 30%) + d (d* + 3¢°)
and we may put x=—3nb (a* + 30°) + 3nc (d* + 36%),
y= na (a* + 30°) + nd (d* + 36).
Hence the values of p, ¢, 7, s are found to be
2 = 3n (ac + bd) (@* + 36%),
g =n (36c — ad) (@* + 36%) — n (a + 30°),
r=n (d+ 32 —n (36 — ad) (a* + 30°),
§ = 3n (ac + bd) (a® + 30%),
and A =n (3ac + 3bc— ad + 36d) (d* + 3¢%) — n (a + 30°)’,
B=n(3ac— 3bc+ ad + 36d) (d* + 36) + n (a® + 36°),
C=n(d?+ 32 —n (3ac + 36c — ad + 306d) (a + 30°),
D=n(d*+ 368) +n (3ac — 30c + ad + 36d) (a® + 36°).
These values satisfy the equation
A+ B+ C= 7,
and, since no restriction has been introduced, the solution is capable of
giving all the sets of three cubes which have a cube for their sum.

More special forms for 4, B, C, D can of course be obtained by putting
zero for one of the letters a, 4, ¢, 4, and still more special forms by com-

-bining with the assumption a =o or 4 =o the assumption & = * ¢, or com-
bining with the assumption ¢=o or 4 = o the assumption 4 = + a.

Two cases are worth noting.

First, suppose =o, d=¢, and we have

 

 

 

A = 8na — nat, B=16nac® + na', C= 16nc*— 2na’, D=16n + qna’e.
If further we write 2a for a and 7/16 for z, we have
d=na(l- a), B=na(2é+a®), C=m(@—-d), D=n (+24),

which is equivalent to Vieta’s solution of his second problem.

Secondly, suppose d =o, é =a, and we have

A =18nad —16na', B=16na', C= gnc'—24na*, D = gnc’,
or, if we write 1a for a,
A = gna — nat, B=na*, C= ognct— sna’, D=gnc,
332 . SUPPLEMENT

which, if # = a = ¢ = 1, gives the simplest solution of all
A=8, B=1, C=6, D=9g, and 1° + 0° + FP=¢’.
In proceeding to other solutions we have to remember that, while
A, B, C, D must be integral, they should all be prime to one another ; for
those solutions in which 4, B, C, 2 have a common factor are not new
solutions in addition to that from which the common factor is eliminated.

Thus, while giving any values, positive or negative, to the numbers
a, b, ¢c, d in the formulae
x= 3nc (42 + 36%) — 3nb (a + 30°),
y=nd(d* + 36%) + na (a* + 30°),
we have to choose for # such a fraction as will make x, y prime to one
another. We then form
p=ax+ 30, g=0x—ay, r=dy +x, s=3y—-dx;
and, after again eliminating any common factor, we put
A=p19, B=prg, C=r—3, Darts,
and we shall have Wr Br Cl=
(The cases in which one of the three cubes 4°, /% C? is negative will
give the solutions of the equation «® + 3° = 5° + 2%.)
While any values of a, 4, ¢, Z may be taken, it is necessary, if we want
a solution in which 4, B, C, D will be small numbers, to choose a, 0, ¢,
so that a*+ 30% &?+ 3c may have a common factor. Euler accordingly
sets out a table of all numbers of the form 7? + 3#* less than 1000 (giving
m values from 1 to 31 and = values from 1 to 18), and then chooses out
cases in which a* + 30° 4* + 3¢® have a tolerably large common factor.
Now, assuming that at + 30% = mk,
at + 3c = nk,
we have (supposing further that ac+ dd =f, 36c — ad = g)
A=n(3/+2) —wk,
B=n(3/-g) + mh,
C=nwk-m(3f +g),
D=wk+m(3f-9).
In these formulae /, ¢& may be either positive or negative, the signs of
a, b, ¢, d being ambiguous ; and we may put
S=x(ac+ bd) ! i f=2x (acid) }
g=12(30c— ad) g=+ etal) }”
But, if f changes sign while g remains unaltered, we get numbers of the
same form, only in different order; therefore we may confine ourselves to
the positive sign in f.

either
SOLUTIONS BY EULER. PROBLEM 1 333

Example 1. let
+30 = 10, sothata=4, b=,
it 37 =96 50 thatd=1 | Ser ment)
r= c= i= 2
Then m= 1, 2=4., £=19,

The following values for /; ¢ result, viz.

lL. f=21, 11,0 f= 10; nL. 7-19,
g=ti1, g=+.10, =X 10,
Iv. i fms, VY. f=16, Nl. /=o,
g=g 37, S=%26, S=o 38,
while, since m=1, n=4, £= 19,
A= 12f+ 4g 19,
B=12f-4g+ 19,
C=304-37~9
D=304+3f—-¢

The values (VI) f=o0, g=1 38 are excluded because, if f=o0, A=-5
and C=.0.

The values (I) give

 

 

 

A=233+44, thatis, | A=997 | Ad=180 or d=3
B=271F44 li. B=osy lp Beye Bes
C=241 +11 | C=230 C=2e2 C=4
HE — 2
J=367 301 | D=356 D=378 J=06
The values (II) and (III) give, after division by 9
A=11% 4, that is, | A=150r 4= ; A= 9
B=13%F4 | B=9 = B=17
C=13%1 | C=17 = C=14
D=10F1 D=1 P=0 D=20
The values (1V) give
A= 411148, thatis, | A= 180 or A= 65 =—107
B= 7097 143 B=-069 B==123 B= any
C=2897 37 C= 252 C= 34 C= 326
= 310F 37 D= 282 Er 04 | = 126

Lastly, the values (V) give
Ad=173 Laos, | that is, A=29y. | d= 69 or d=23
B=211F tog B= 107 B=318 B= 105
C=1256T 26 C=230 C=232 C= 94
D=352F 26 D=326 | D=378 D=126

 
334 SUPPLEMENT

Thus from the one assumption for a®+ 36% d?+ 3c we have the
following solutions :
227° £50 anni 2563 107° + 3562 = 227° + 336°
107° + 230° + 277% = 7268 253+ off= 632+ 34°
23° + 04° + 105° = 1207
7° + 14° + 15= 20°
T+ £4 ga @&
Lxample 2. Assuming

2+ 37 =28, so that Ey or el or 257)
)

J=3 $=2 O=1
d+ 38=84, so that sid or er or Sd
c=g £=4 envy!
we have %2=28, m=1, n= 3, and the following solutions will be obtained :
+ 624 S=. gf + 12°= o°+ 10°
34° + 39° + 63° = 72" 10° + 27° = 19° + 24%
20° + 54° + 79° = 87° 17° + 30° = 26° + 36°

3+ 43 i Si= 6°
35 248° + 707 = 35°

PROBLEM 2. 70 find three numbers x, y, z suck that
X+y, X+% y+3

X=, X—3 y—23
are all squares.

First solution’.
Assume that E-p=ff N—gmgh pgm
therefore =x —4°, 2=2—-7% and = +2,
The first three formulae now become
Xty=2x~f% x+a=28~0% y+a=233x—- pF -4

Suppose that 2x—p°>—¢*=7# so that 2x =2#+ p+ ¢*; therefore we
have to make # + ¢* and #* + p* squares, while in addition ¢* =p? + 7°.

Let gE B+ pund~F 7r=0ap;
then £4 (+0) =1r+a' +0" + 2020 }
and + (@-8Y=2 +a + 8 — 24%

must be made squares.

Comparing now #’+a'+4' with 2+4d* and 20%* with 20d, let us
suppose = a8 = ICR, c=1¢", d=/PR, a=, F=g# (or a=}
b=gk); then the assumption #2 + @* + 4* = 2 + 4* will assume the form

72 + fi +g = fig + 2a
or P= figt— [0+ IR — a = (1) (g*— 7).

v Algebra, Part 11. Art. 235.
$ Bs
SOLUTIONS BY EULER. PROBLEM -: 335

Hence the problem is reduced to finding the differences of two pairs of
fourth powers, namely /*— 4* and g¢* — 4%, the product of which is a square.

For this purpose Euler sets out a table of values of m*—n' corre-
sponding to different values of =, », with a view of selecting pairs of
values of s*'—»* the products of which are squares.

 

 

 

 

| m? I | net =n? | om? +a? | ned — at
ST
4 I 3 5 3:5
9 1 8 10 16. 5
9 4 5 13 5.13
pos 10 1 I5 17. 4 Ty
| 16 9 wo aliiny 25.
25 ¥ iia 26 16s 3.13
25 9. 16 34 26. 2.1%
49 Ty] 48 50 25 X10. zl T
49 16 33 65 3. LCL TY Tg
64 1 liaibg 65 Ol niin. TY
81 49 1 3» 130 64. 5.13
I21 4 1) any 125 25.50. 5.73
121 9 } Cui 130 16. 2. 3. 5.93
721 od 40] 72 170 i144. 5.17
144 23 | aio 169 160. 7.17
169 I | 168 170 {16, 3. 5. 7.1%
260. i Sr i. 38 250 256.006. 5. TT
225 64 | 161 289 239. 7.23

 

 

 

One solution is obtained from f*=g, #* = 4, g*= 81, /* = 49, whence
= (f'~A)(g'-#)=5.13. 64.5. 13 = (520) = 270400.
Therefore
~a=fh=21,0=gk=18,p=a"-F=117, g=a* + B*= 765, » = 2ab = 756 ;
therefore 2x =72+4p°+4*=869314, Or X= 434657
y=x—f= 420003
Z=x—¢g'=-150568
The last number z may be taken positively; the difference then becomes
the sum and the sum becomes the difference ; therefore
v= 434657, x+y =855625 =(925), x —y= 13689 =(117)},
JY =420968, x +2=7585225=(765)%, x —z=284089 =(533)%
5=150568, y+2=7571536= (756), y— z= 270400 = (520)
We might also have taken f2=g9, #2 =4, g?=121, /2= 4, which would
equally have given a solution.
336 SUPPLEMENT

Second solution’.
This later solution (1780) of Euler’s is worth giving on account of the
variety of the artifices used.
We can make x+y and x —y squares by putting x= p*+¢% v = 2p.
Similarly x + z, x — z will be squares if x=+*+§ z= 27%.
Therefore four conditions will be satisfied if only p* + ¢* =7* + 5%
Now [cf. Diophantus 111. 19 and pp. 105-6 above] if we put
= (RD + AY,
a can be made the sum of two squares in two ways; in fact
p=ac+ 0d, r=ad+ be
g=ad—be, s=ac— bd,

and
y=209=2 (@%cd + abd? — abc® — Bed), 5= 27s = 2 (aPcd + abe — abd? — bed),
so that y+z=g4cd (a — 0%), y—z=4ab(d*= 0).

These latter expressions have to be made squares.
First make their product * — z*> a square; this means that
ab (a®— 8%). cd (d* — *) must be made a square.
To effect this, let us assume that «@(d*—¢*) =»ab (a* — 4%); we may
further, since the question depends on the relations between the pairs of
letters a, 6 and ¢, &, suppose that Z= a.

 

 

 

We have then e(d~ O) = hia 3),
. 220 — J : :
whence a* = 3 , which fraction has accordingly to be made a square.
ih
1 z
Suppose that @ =4 —¢, so that Sgr 0 —20c + and we have
o=~ (21+ 1) Bc + (1? + 2) b*;
fd #tz
therefore TE.
£ 2 +y

Put 6=7+ 2 and c= 27° +1 ; therefore a=1—7*=d.

As we have now made the product of the expressions ab (4% — ¢*) and
cd (a — 0%) a square, it only remains to make either of them singly a square,
say ab (d*—c%).

But ab (d* =) =ab(d—c) (d+ c)=3n* (n* — 1) (n+ 2)%

We have therefore only to make 3 (#21) a square, which is easy, since
n® — 1 has factors ; for we have only to put

2
2 I) = 7 2
3 —1)=" (n +1),
S
2 2 2
+
L (n+1), or n= fry
Sr
L Mémoires de I Académie Impériale des Sciences de St Pétersbourg, 1813-14, VI. (1818),
PP. 54 sqq.= Commentationes arithimeticae, 11. pp. 302-5.

which gives 3(n—1)=

a
02
S
SOLUTIONS BY EULER. PROBLEM = 337

[Euler had previously tried the supposition a= 4+ ¢, which would
require 3(#°+ 1) to be made a square, which is impossible. ]

All the conditions are now satisfied, and we have to find a, 4, ¢, @ etc.
in terms of /, ¢:

12/7 ded

ee —— b = 7”? + 2 =
(337 =F) Be J
3/0467 bs ayet

(38 I

As the whole solution depends on the ratios of a, 4, ¢, 4, we can
multiply throughout by the common denominator, divide by 3, and put

pade—yg fi bsfi-of i togh em fit ofiott ock
whence 2=—3 (f+ or), r= +30f'0 + 810%
g=—f + 2figi— 810%, s=—10/%s

[Euler took @ to be 7°— 1 instead of 1 — #® and consequently obtained
positive signs for the values of p and 5; he also has ¢ = — (f*— 9¢*)>, which
appears to be a slip. |

Assuming therefore any values for #, ¢ in the first instance, we first find
values for a, 4, ¢, 4, then values for pg, ¢, 7, 5, and lastly values for x, y, 2.

It is to be observed that it is a matter of indifference whether we get negative
values or not ; for positive values can be substituted without danger.

A=d=1—7=—

c=21+1=

Euler gives four examples.

If f=1, g=1, we find that x, y have equal values; this solution there-
fore does not serve our purpose.

The same isthe case if f= 3, g=1.

Suppose then that f=2, g=1; thereforea=d=-106, £=17, c= 33; and
(taking positive signs) we have

2=800, g= 305, v=317, s= 250,
and x =1733025, y= 488000, z= 418304,
: X+y=1105% &—y=495"
x +z=1073% ¥—12=5017
y+2= 052%, y—5=204%
Iff=1,g=2,wehave a=d=16,0=137, c= 153, and
2 = 4640, g= 20705, 7= 21217, = 256,
leading to large numbers for x, y, z.

Euler adds that, if x, y, z satisfy the conditions of the problem, another
solution is furnished by X, ¥; Z where

X={Ff+7-2%), Y=2{ +2217), Z=2(#+)— 7)

H, D. 22
338 SUPPLEMENT

PROBLEM 3. 70 find three squares such that the difference of any pair
is a square, or lo find x, y, z such that
— 3 xP- 2% y' — 5° are all squares.
Any solution of the preceding problem will satisfy this, but the numbers
would be large and we can get smaller solutions’.

2 2

Dividing by 2% we have to find three squares, Es and 1, such that

x? JE x2 2
G5 G-1) G-9)
are all squares.

The last two conditions are satisfied if we put
8 rts yx:
Zz pP-1 and 2 1’
2 2 2 2 2 2
and we have only to make S- Cay en) a square.

=F) Fey

 

 

 

 

Now
(Zt) eiy ty Ft Zin gtr
Sl att a tl «Ll fe 0)
_i(Pr=0 (=).
(Z= 171)

Therefore (°° — 1) (¢*— 2°) or (p*%*—1) % go 1) has to be made a
square.

(1) The latter expression is a square if
7 +g qg < 72 iL 2

= as an
And pg 7 = ¢% a square ; therefore

 

 

Foot 2+ : . 5
wor or fy (f* +g? . hk (4? + £°) must be a square.
 f=a+o, g=a—4, /i=c+d /k=c—d the expression becomes
4 (a* — 0%) (c¢* — #4), which must be a square.
From the Table to the last problem we may take the values

a*=9, 0’=4, &=81, d*= 49,

 

which make the expression a square.
Then f=35,g=1, 2=16, k=2, pg=22, g[p= 258 = 83, so that >= 182

g=4% and Reine r=4

2
AL ary Soin 2188 is the solution.

=i" 9% g=iam
1 Algebra, Part 11., Arts. 236, 237.

 

 

Tr herefore = =
SOLUTIONS BY EULER. PROBLEM 3 339

To obtain whole numbers, we put z=153 and then x=-697 and
y=18s.

Thus a7 = 485809) and a®—y?=451584=(672),
P= 34228 PP —2= 10816 = (104),
£m 27400 x% — 22 = 462400 = (680)>.

7*
(2) Without using the Table, we may make ( %?— 1) i - 1) a square
in another way.

Put ¢/p =m or ¢=mp, and (m**— 1) (m*— 1) has to be made a square.

This is a square when p=1; substitute therefore 1 +s for # and we
have
(m*— 1) (mm — 1 + 4s + 6m°s® + 4m’s® + m?s*).
Dividing out by (#®—-1)* and, for brevity, putting a for m?*/(m*— 1),
we have
I + 4as + 6as® + 4as® + ast,
which has to be made a square.

Equating this to (1 +/5 + g5?)? let us determine f; ¢ so that the first
three terms disappear ;

 

therefore af = 4a, Ori f= 2a,
and b6a=2¢+f or g=1(6a-,3)=3a— 24"
Lastly, the equation gives 4a + as = 2/g +g’, so that
Lae 4o—=120°%84% | 4-120 488 "© 41 (2a=-T1)
g£—-a 4a*—128°+9a*—a 4a*-128°+9a—1 4a —8a+1’

Now in the expression for @ may have any value.

Ex.1. letm=32 sothata=4;

 

Leg. 3 400 = Te
therefore alii wv 23’ 7 2= 23’ q 23°
whence > 200 if 290.

Zz 470’ 7 4047

Ex.2. letm=%,50thata=1;

3. 260 2
therefore s=g. B02 = 29, g=-11,

whence «x/z, y/z are determined.
Euler considers also the particular case in which a =m?/(m?-1) is a
square, &* say. :
The expression 1 + 40% + 64%? + 40° + 6%* is then equated to
(1 + 20% + 05%),

; — 20 — — 24?
and we obtain ss IE Re nd p="— 2

 

22——2
340 SUPPLEMENT

Ex. ais a square if m = §, and in that case 6=%. Therefore p= —35
g=mp=—211 and accordingly
y 5% 433

27 911’ 7 14g

PROBLEM 4. 70 find three square numbers such that the sum of each
pair is also a square, i.e. to find numbers x, y, z such that
BLP Br Pe?
are all squares’.
Dividing by 2% we have to make

x2 7 x? 7B
Ths, mae 1, =+t7
Zz Z Zz Zz

all squares.
The second and third are made squares by putting

a | 2—1
= ang asi ;
and it only remains to make
2 TV CP]
wy + oy = , or ¢*(p*—1)*+p% (9 — 1), a square.
This can hardly be solved generally, and accordingly we resort to
particular artifices.

1. Let us make the expression divisible by (+ 1)3 which is easily
done by supposing p+ 1=¢—1, or g=2+ 2, so that ¢ + 1 becomes p + 3.

Thus (p+ 2)? (p—1)*+ 2° (p + 3), or 2p* + 8° + 6p — 4p + 4, must be a
square.

Suppose 200 +82 + 6p" — 4p + 4 =(gP* +P + 2)3,
and let us choose f, ¢ such that the terms in p, #* vanish ; therefore f= —1,
and 4g+1=6, or g=4.

We now have 2p +8 =g% + 2f¢

-#42-4,
so that p = — 24, and ¢=— 22, whence
gL li=r U3 y.0-1. 453

 

z 9p Bla oy 44
Making z=16. 3. 11, the least common multiple of 48 and 44, we have
the solution
X=11.23.25=6325, vy=12.21,23=5706, 3=3.11.16=523,
and +): =23(275+ 252%) = 23°. 373%
B+ E=11° (575% + 48%) =11. 5775,
Pr2=122(482% + a4) = 12%, 4822

1 4lgebra, Part 11., Art. 238.
SOLUTIONS BY EULER. PROBLEMS 3, 4 341

2, 3. Euler obtains fresh solutions by assuming, first, that
g—1=72 (2 =k 1),
and, secondly, that g—1=%(p-1)
4. Lastly, he makes our expression divisible by both (p+ 1)* and
(p—1)* at the same time.

For this purpose he takes

 

prt
g = p+ ’
whence prr=22 0D nq p-r=teznllon)

Substituting in the formula ¢*(p*-1)*+ p*(¢2~ 1)? the value of ¢ in
terms of p, Z# and then dividing by (2*- 1)? we have the expression
(p+) p=
(E+ TN pudy ?
and we have to make (p+ 1)’ (p + £)* + p* (> — 1)* a square,
or pit (P+) +282 + (2+ 1) + (2-1) p+ 28 (P+) p+

must be a square.

 

We now equate this to {+ (£2 + 1) p — 2}},
whence we have
22+ (+1) + (P=) p+2t (+1) ={(£2+ 1) — 22 p— 2f (£2 + 1),

 

which gives 42+ (2 — 1) p+ 4 (2+ 1) =o,
4
= TR
therefore prea t iN
? iia 24 Fay
—-3£2+
and ot
&— 3c

where # can be chosen arbitrarily.

Ex. Let#=2; thenp=—3 g=~3} and
w Lovey gen ng

z. 2p So I 2g 44

 

Putting 2=4.4.5.11, the least common multiple of 80 and 44,
we have i
X=~ 0001.13 == 420,
Y==4.5:9.13=—2340,
Z2=. 4.45112 380,
342 SUPPLEMENT

and Sih = 32. vat{van + 3600)= 37. 137.61",
a+ =11 (1521 + 6400)  =11°. 30),
F+7#=20"(13630+ 1936) =207.125>
ProsLEM 5. (Extension of Dioph. 1v. 20 to five numbers.)
70 find five numbers such that the product of every pair increased by unity

becomes a square’.

Euler had already shown (see pp. 181, 182 above) that, if mn + 1 =/2
then the following four numbers which we will call a, 4, ¢, 4 have the
property, viz.

a=m, b=n, c=m+n+2/, d=4.(/ +m) (+n).
If now z is the fifth required number, the four expressions

1+az 1+063 1+c¢z I+dsz
must all be squares.

If, says Euler, we had to satisfy these conditions singly, the difficulties
would be insuperable. But here too it happens, as in the former case,
that, if we make the product of the four expressions a square, the
expressions are all severally squares.

Let the product be I+ pz +g + 72° + 524
where accordingly
p=a+b+c+d, g=ab+ac+ad + bec + bd + dd,
7 = abc + abd + acd + bcd, s= abcd.

Suppose now that
I+p2+g8t +728 + 52t = {1 + Lp + (Rg — 1°) 2°);
therefore, since the absolute term and the terms in z, 5° vanish, we have
r+s52=p(39— 32°) + Qe — 32%
—p (lpg — 1p
whence a=ol 00 30)
(g—32°)—s
Now it will be found (see the proof lower down) that
0-1 =-%{ 471);
the denominator of the fraction will therefore be } (s — 1)?; that is, the said
denominator fortunately turns out to be a square; if it were not so, the

single expressions 1 +az, 1 +63, 1 +¢z, 1 +ds could not have been made
squares.

As’ it is however, substituting for 1g — 12? its value in the numerator
and denominator of the fraction for z, we have
47 +2p (s+ 1)
el
G-1)

Y Commentationes arithmeticae, 11. pp. 50-52.
SOLUTIONS BY EULER. PROBLEMS 4, 5 343

and all the conditions will be fulfilled, that is, all the expressions
ab+31, ac+i, ad+n, Stl bd,

d+1, az+1, bz+1, c+, dz+1
will be squares.

Lemma. To prove the fact (assumed above) that
g-Lf=—1(s +1)
For brevity, put m+n +l=f, I +m) (/+n)=F, so that k=/f+/lmn;
and, since a =m, b=n, c=f+/, d= 4k, we have a + 6 + c= 2f, and therefore
p=2f+ 4k
Again, since g=(a+db+c)d+ (a+d)c+ ab,
g=8fk + (m+n): +2, (m+n) + mn;

and, since mn + 1 = /% the latter expression becomes

g=8%+f*- 1.
Moreover, s=abcd = gmnk (f+ 17);
therefore 1+qg+s=8k+/f+4gmnk (f+),
and we have to see whether the right-hand expression is equal to }2
Now I= + 4% + 44.

Assume then, as a hypothesis, that
8fk +f + gmnk (f+ 1) =f + 4 fk + 44%,
ic. afk + amnk (f+ 10) = 442,
or, if we divide throughout by 44,
[4 mn(f+1)=rk=fI*+Ilmn, from above ;
that is, f+ fmn =f,
which is of course true, since mn + 1 = /*
Consequently it is proved that
1g+s=if% or (lg=3) ==} +1)

Ex. 1. Assume m=1, 2 =3, 50 that /=2; therefore
an, O=3 ¢=8, d=120,
whence pP=132, g=149%, 7=4224, $=12380,
and we deduce that
gad 4224 +264.2881 777480

 

2879? 8288641
The conditions are satisfied, for
a+ 1=72% wtr=_3% ad+1=11%
fe 1= 55, bd = 10° od + 1= 31%

are1=(SH), Geers (HE r=

do + 1= (OOTY
344 SUPPLEMENT

Ex. 2. - To get smaller numbers (since we must put up with fractions)
let us put m=1, n=25, so that /=3; therefore
a=l, s=%, ¢=0, dAd=48,

whence p=27, J=4513, 7=031%, $= 3bo,
and pmol OIB EI af gle
359° 128881

ProsLEM 6. Euler has a general solution of the problem of Dioph. 111.
15, viz,
70 find three numbers x, y, z such that

ytety, xztxts, yEty+z
are all squares.

(1) Put xt1=4, y+1=8,2+1=C, 50 that 4B~1, 4C—1 and
BC —1 have to be made squares.
Let AB= +, AC=0°+1, BC=r+1;
therefore ABC= J{(*+1) (+1) (#2 +1)}.
To make this expression rational, let us regard p, ¢ as given and put
(+1) {(F+1)=nmP+7% so that m=gg +1, n=pTg; therefore
ABC = Jim? +n?) (7 + 1)} = J{(mr + n)* + (nr — m)*}.
Put the latter root equal to m7 + 2 + ¢ (nr — m); therefore
nr —m=z2mrt + 2nt + nrt* — mt?
_m(i-1)— 2nf
Tn(f—1) + 2mt’
2 2 2 2 2 2 2
Therefore ~A+1= {ini ay and 48C = sles)
{n (2-1) + 2mt}? n(£—1)+ 2mt
thus, since ZC=7>+ 1, we have

and

Ln Led + 2mt
+1

and, since m* + n*= (p> + 1) (¢* + 1),
lh 1)(£+1)
Cn(P—1)+2mt’

(+1)? +1)
n(t—1)+2mt’

)

where m=pg+ 1, n=2 +4.
This solution is very general, inasmuch as we may choose p, ¢ as we

please, thus equating 45 —1, AC—1 to any given squares; and, as #

can be chosen arbitrarily, we have an infinite number of square values for
BC~1.

(2) Euler adds two methods of obtaining solutions in integers, the
second of which is interesting.

1 ¢‘Considerationes circa analysin Diophanteam,” Commentationes arithmeticae, 11. p. 577.
SOLUTIONS BY EULER. PROBLEMS 5, 6 345

Take two fractions 5 and 550 related that ad —éc=+1; and form a
third fraction 3 which is similarly related to either of the former

fractions.
Then the following three numbers will satisfy the conditions :
A=d+8, B=>+4% C=lta¥+ d+)
For, since ad —bc=+1,
AB = (ac + bd) + 1,
AC= (act d® + bd + Fy +1,
BC=(2+ac+d*+ bd) + 1.
(Cf. Dioph. 11. 19.)

Simple solutions are seen thus :

 

 

ac a+c

Z 9 iv A B C

2 : os 1 Sit S2+af+2
1 = sh 2 2f2-af+1 2f2+af+1
: a £2 5  3%-art1 #fP Girth

and so on.

(3) If two of the numbers 4, B are given such that 458 —1=72° we
can find an infinite number of values for a third, C, which with 4, B
will satisfy the conditions.

For, since 4C—1 and BC — 1 have to be squares, take their product
ABC? — (A + B) C+ 1 and equate it to (mC + 1)?; we have then
_A+Brm (4 +m)? _(B+my

CEP or whence AC-1="J and BC=1= 7 5

Therefore we have only to make 4B —m? a square ; that is,
+1 —-m=a square =" say, so that wm’ + 2 =p +1.

* Take now two fractions @ and a such that e+ a?=1, and let m= ap + a,
n=ap— a; then
c+ BLz2(ap+a)

lp=aR. 7

where a, a are determined by giving any values whatever to f; ¢ in the
expressions
me a

Fie TPE

 

 

a
346 SUPPLEMENT

PROBLEM 7. 70 find four numbers such that the product of any pair
plus the sum of that pair gives a square; or, in other words, to find four
numbers A, B, C, D such that the product of any pair diminished by 1 is a
square, that is, such that ]

AB, AC—1, AD—1, BC —1, BD—-1, CD73
are all squares’. (Cf. Diophantus 1v. 20.)
Let us regard two of the numbers 4, B as given, being such that
AB—1=2% or AB=2+ 1.
Let a, a be such fractions that ¢* + a*=1, and put
A+ Biratlap ta)
rm
Similarly let 4° * B*= 1, and put for the fourth number
A+B+2 (p+ B)
—E-
Thus five conditions are satisfied, namely, that
AB~-1, AC~1, BC-1, AD—1, BD ~—1 are all squares.
The sixth condition, that CD — 1 shall be a square, gives
(A+B) +2(A4+B){(a+b)p+a+p}+4(ap+a)(bp+pB)
— (ap — a)? (Bp — 6)* = a square,

where 42 has at the same time to be equal to p* + 1.

e

DD

Regarding a, a, 4, 3 and p as given, we have
tr Alt

A A 2
and the expression to be made a square becomes the following expression
in powers of A,

Ate (a+) p+ 243 + 1) +24 (p+1)(a+d)p+ (p+ 1)
+24 (a+ B) +4d(ap +a) (bp +B) +24(p +1) (a+ p)
: = A*(ap— a)’ (Bp — 0)"
: Equate this to the square of
A+Ad(a+b)p— (+1)
+4 (a +B),

A+ B=4+

 

and we have
A (a+b) +2(a+d)(a+P)p+ (at Bf —4(2 +1)
—4 (ap +a) (8p +B) + (ap — a)’ (Bp — 0)*}
=44 (PF +1){(@a+8)p +a+pl

whence 4 is found.

Euler goes on to some particular cases, of which the following may
be given.

L Commentationes arithmeticae, 11. pp. 579-582.
SOLUTIONS BY EULER. PROBLEMS 7, 3 347

Suppose 6=—a and B=-a; we then have
A+B +2(ap+a) pA+tB-z2(apta)
pa 0. pmiR
and the expression above in 4 which had to be made a square becomes
A+ 282+ 1) + (p+ 1)
— 44% (ap + a)?
— A*(ap— a)
This can be put in the form
(= (2 + Df + A (ap— a) {4 — (ap— a)

by virtue of the relation a? + a? = 1.

C=

Our expression is clearly a square if 4 —(ap—a)’=o, or ap —a=2,

that is, p= (2 +a)/a, and

: (2+a)a 2a + 1

ap ra=>—rt ter
a a

and in that case

 

pr Ce A+B + 2(2a+1)/a ls Arad:
cad 4 4a
je(Ad+DB)—aa=2
Tr

where 4 can be chosen quite arbitrarily.

Putting a =(/%—g¢%)/(f* + g%), a= 2/g/(f* + &%), we obtain the following
as a solution, where 7, 7 can be any integers whatever.

 

mfr 12 (of? + &%)
r— ; B = ;
Confy 2mfg
(m+ 3n)* f+ (m— Tere (m—3n) f+ (m+ n)g®
C= ie - D=>—2"L 2,
: 8mnfy Smnfe
Ex. Suppose f=1, g=2, wm=3 7=6;
therefore A= B=if,. C=513 i D=453,
ad AB-1=(),  AC-x=GEf AD-1-(

~1={13¥, BO-v=(ily, CD-1=(Gi3%
PROBLEM 8. 70 find four numbers such that the product of any pair
added to a given number n gives a square’.

(1) A particular solution is found in this way. Tet A,B, C,D he
the required numbers, and, since 45 +» has to be a square, put

A=n?~-F, B=n’—d5
so that AB = (nac— bd)? —n(ad—bc)®. [Cf. the Indian formula above,
p. 282.]

L Commentationes arithmeticae, 11. pp. 582-3.
348 SUPPLEMENT

The condition that 4B + zis a square is therefore fulfilled, provided

; a ¢
that ad — bc=+ 1; therefore we have to take fractions +, — such that

oa
ad — bc = + 1; and, when this is done, the fractions 5 oC and pi will have

the same property in relation to either of the former fractions.

We accordingly put
A=na®- 2, B=nf—42
C=n(a+c)—(+dy, D=n(a—c)*—(b-d)>
Thus five conditions are satisfied, and it only remains to make CD +n
a square ; that is,
rn? (a® — 2)? — 2m (ab — cd)? + (6° — 4%)?
— 27 (ad — bc)? | =a square
+n
or, since (ad —bc)*=1,
7? (a® — 2)? —n{2 (ab — cd)? + 1} + (* — d*)* = a square.
(2) We obtain a general solution by the same method as that applied
above (p. 345) in the problem of making 48 — 1, BC -- 1, etc. squares.

Put 45 =p*—n; then, to make AC + n, BC + » both squares, take the
product of these expressions and equate it to (z + Cx)?; therefore
W+n(Ad+B)C+ ABC =n + 20Cx + Cx?
n(A4 +B 2x) uw(A=x)
SEE A Ang a
so that (x? — 4.B)/n must be a square.

whence C=

Letthen 2*—-AB=a-p+n=nf or 2*—nf=p"—n

Similarly let us put 2 —722=p’—», so as to get

RE Co Do gs
and it remains to make CD + 7 a square,
that is, (Ad +BP—2(x+20) (4 +B) + ny’? + 4x0

must be a square.

But, since B 2g and 4 + partisan
(after multiplication by 42)
At —2A% (x +0) + 24% (2 —n)— 24 (p*—n) (x +0) + (p*—n)?
+ nA%*s?

 

the expression becomes

+ 44%v,
SOLUTIONS BY EULER. PROBLEMS 8, 9 349

which must be a square ={4?- 4 (x + v) — (p* — »)}* say; therefore
A(x +0)? — 4 (p2—n)— ny*s® — 400} + 44 (x +0) (pF — n) =o,
so that
4 (x+2) (=n) - 4 (x +2) (=n)

Tne + 4 (Pn) = (x +0) +430 n()P—1)(F—1)+2x0 + 20° = 30’

of a A Np)
ny*z — 2n (2+ 2°) + (0 + x)
(3) A particular solution is obtained by assuming that z = — x, so that

z=y, and :
A+ Bax poArBrex

3 ) i
while 45 =p* — n= x — my*

C

For then we have to make
A+ 42 {2 (p2— n) + ny* — 4a} + (p* — 2)’ a square ;
that is, (42 -p* + n)* + nA%* (»* — 4) = a square.
This is satisfied if we put y= 2, so that pF =a? — 3a.

Suppose p = x —¢, and we have

 

 

 

ie p-z Le 3m + 2
gw = and p= = yor p=? a and » = 27,
and hence d= (m2 — 22) (gnae® — 2%)
42%0° :
We may therefore put
fll 5 Slome=#)
= 297 J = of J

cnr arg (fog
= 8fatu ? ir 8fgtu :
It will be seen that in this solution C+ D=1(4 + B).

PROBLEM q. 70 find four numbers such that the product of any pair
added to the sum of all gives a square’.

First find four numbers 4, B, C, D such that the product of any pair
increased by a number 7 gives a square (Problem 8).

Take as the numbers sought mA, mB, mC, mD, and, since 7? (AB +»)
is a square or m°ALB + m’n is a square, we have only to make m*: equal to
the sum of the four numbers or 7 (4 + B + C + D), whence

A+B+ C+D
metallia

L Commentationes arithmeticae, 11. pp. 583-5.
350 : SUPPLEMENT

But, since in the other problem C+ D=}(A4+ B), this gives
_3(4+ B) _gn(foe)ui=3(f +e)

27 4njfgtu ?
where 7 as well as /, ¢, # and # can be chosen as we please.

nm

Since » may be chosen arbitrarily, take p®=a®— 37, as in the last
problem, so that z=1(x*—-p?%, and 4B =p*—n=} (4 — x7).
Accordingly we may put
fren p.m
3 J

J

 

Apt rap LE ae

 

 

 

 

 

therefore
3/¢
and hence Clap (0s - 3h
12fg 2
D2 3p (Rr G38)
12/% 2
aid | dBc p=-tUrap rane
2/¢ 2
2 02 yo
therefore m=(4+ B+ C+ D)/n= 6(/° +30) : 3(/ Pr
2g (** — 2)
Now two of the numbers, 4, B, can be chosen arbitrarily, and
p+a=3%, pull;
S
r+ or 34g — Bf?
therefore =a TY ond ve ;
2 4/8 2fg
2 (94g*~ Br) (dg - B®
so that nmi p= Af NA) ;
and cB _ yg-Br lt Br py
= 42 4 i
while m= 3(4 +B)
29

If, in order to get rid of fractions, we put 4 = aft, B = 44/g, we have
C=(a+0)fg—3a8* +4 D=(a+b)jg+3a8® bf?

ER TE CRI Ls A
n= (9ag® — tf*)(ag’— 4%), m (9a = bf) (ag — oF?)

Ex. Let f=2,2=1; therefore
A=8a, B=8) C=66—a D=s5a-20,

12{a+d) = 12{a+d)
(92a -40)(a~ 46) (40—09a)(4b—a)’

 

m=
SOLUTIONS BY EULER. PROBLEMS 9, 9A, 10 351

The following are simple cases : :
(2) a= 3, b=1, whence d=4o0, B="3, C=31, D=323, m=il.
{2) a=11, &=2, whence 4=33, B=16 C=1, D=z1, m=z
If f= 5, £=1, we can obtain integral solutions, thus.
d=g00, B=200 C=30b+tza JD=3a- 200,
paaaieiats ;
(250 — 9a) (256 — a)
Assuming then a= 19, 4 =7, we have
A=3%0, B=140, C=248, \D=1z, m=4,
so that the required numbers are

475, 175 310, 13,
the sum of which is 975.

We can also solve the corresponding problem :

9A. 70 find four numbers such that the product of any pair minus the
sum of all gives a square.

For we have only to give 7 a negative value.

PROBLEM 10. 70 find three numbers x, y, z such that
xX+y+3
yz + zx + xy are all squares’.
xyz J
(This may be expressed as the problem of finding 2, ¢, # such that the

equation &—p&+9¢f—r=o0 has all its roots rational while p, ¢, » are
all squares.)

Take nx, ny, nz for the numbers required, so that
n(x+y +2)
n* (xy + xz + yz) + must all be squares.
xyz
If the first and third conditions are satisfied, we must have, by

multiplication,
xyz (x + y + 2) = a square.

Put therefore xyz (x+y + 2) =? (x+y + 2)},
aan _P (x +)
whence ayz=0(x+y +2), and z= ah
0
Since xyz = npn we must have, in order that zxyz may be

a square,
n=mxy (x+y) (xy - 2°).
L Novi Commentarii Acad. Petropol. 1760-6 1, Vol. VIIL (1763), p. 64 sqq.= Commen-
tationes arithineticae, 1. pp. 239—244.
352 SUPPLEMENT

If the values of z, z thus found be taken, the first and third conditions
are satisfied, and the three numbers will be
nx = mx (x+y) (xy — 0%),
ny = m*xy* (x +3) (xy = 7°),
nz = mv*xy (x + y)~
The second condition requires that
xy +z (x sp)=ay + ZEIT a square.
Suppose for this purpose that xy —2°=2* (this introduces a restric-
tion because there are doubtless plenty of solutions where xy —2* is not

a square); therefore

 

+ of (x +3)

Senn JEU,

x u
2 2 2
x2 +04 u
and xy=02+4, x +y= CE that we must make
a
2 +08 + yt a SUE,
7* + ou?
Put x= 79, so that y= , and

(0? (22+ 1) + 2°}

V+? + es =a square
£ pare
or Lu + Put + 0 (2 + 1) + 20°07 (2% + 1) + «=a square,
Ze UH (E+ 1) + 0%? (382 + 2) + ut (£2 + 1) = a square

= {0? (££ + 1) + su7), say.
Therefore (382+ 2) +2 (£2 + 1) = 250° (2 + 1) + $F,

Erg — 2

= 25 (P+ 1) 30-2 =a square.

and =

Further, let s = #— 7», and we shall have
ot 27-71 + 1
# 28—-(27+ 3) Prat —2(r+1)
Multiply the numerator and denominator by 2#¢ — #2 + 1, and we have
oe (27t—7*+ 1)?
0 grt —2(37°2 + 3r—1)B+(277 + 372+ 27 —3)P—2(37 — 1) (r+ YEH 2(r—1)(7 + 1)*
The problem is accordingly reduced to making the denominator of this
fraction a square. If we suppose this done, and Q to be the square root,
while # and 7 are determined as the result of equating the denominator to
@?% we shall have

 

Vv  2rt—r +1 22 + 2
3 Trapt a X=dy v= or
)

whence we can derive the numbers required.
SOLUTIONS BY EULER. PROBLEM 10 353
Now the denominator to be made a square is easily made such if the
coefficient of #4 or the absolute term, is a square; and the absolute term
1s a square if 2 (» — 1) 1s a square.
Case I. Suppose »=1; the coefficient of #* is then a square and the
absolute term vanishes.
We have 42% — 108% + 482 — 8¢ = Q? while vfu=2¢/Q.

Suppose Q = 222 — £¢, and we have

9 xX 25 49 39 2 4 — 36
P-Brmilf adt==-92 ‘=f =T.
rons ows 1”

we therefore put v=—36, #=173, ¢=-3%, and x= fw= 128; further

JO? 31228 35.1240

 

tv 128 123. >
: 30%. 47609
A es

and, since xy — 2° => the required numbers will be
123%: 25. 1240 747609. 173

 

 

ny 128.128 /
oy = 128:25°.1249°. 47609 . 173°,
LA hd 128%. 128 5
© 367.123. 28. 1249. 47609
nes 128. 128? a

In order to get rid of fractions, put 72 =128, and we have
ux = 1282, 173%. 1249. 47609 = 490356736. 50463641,
ny = 5°. 173%. 1249". 47609 = 934533025. 59463641,
27. = 306". 1240.. 476007 = 61701264. 50463641.
The product of the three numbers is obviously a square; their sum is
found to be 25. 59463641%; and the sum of the products of pairs
=173%. 594636417. 18248924559376
=(173.59463641. 4271876)"
Case II, Put v= 3; then Pe 5 and

12

 

6-4 + ZT L354 BE = (F
= (§ — 34 + 157, say,

66 —41p= LAT and #= 43.

whence
: se 2
Accordingly Q = 435%, i 12, and we put v = 19, # = 14.
: +k 4
Therefore x=ly=060, y= 2 =551  x+y=4}8%,

 

H. D. 23
354 SUPPLEMENT
-and the required numbers are

2, 557. . 196
pn 90: 257-4137. 10

i= 14%. 60%. c57. 4157
ay = SSS AIST 006 pe ggpn. gas if m= 60.
ng = $2.00 390 £137 a 107.1557 - 4157
That is, nx = 705600.2315449 = 1633780814400,

ny = 109172.2315449= 252782198228,
nz = 1500677 . 2315449 = 3474741058973.

The product of the three numbers is clearly a square; their sum will be
found to be 2315449; and the sum of the products of pairs

 =14%.2315449%. 6631333489
= (142315449. 81433)"
These numbers are much smaller than those first obtained.

If fractional numbers are admitted, we may divide those found in the
last solution by 2315449? and the solution will Le

705600 196 361

PROBLEM 11. 70 find four numbers x, y, 3, u such that

X+y+z+u

xy + yz +...

XYZ + YRU + ZUX + UXY
xyzu

are all squares’.

A general solution being apparently impossible, some particular as-
sumption simplifying the problem had to be made. Euler therefore
assumed as the four numbers

Mab, Mbe, Mcd, Mda,
which assumption, although five letters are used, involves the restriction

that the product of the first and third numbers is equal to the product of
the second and fourth. :

We must therefore have En

M (ab + bc + cd + da)

M? (abc + bd + eda + da*h + 2abcd)
MP (abd + abcd? + bed? + abcd)
Marca?

all squares.

\ Novi Commentarii Acad. Petrop., 1772, XVIL (1773), pp- 24 $4q. = Commentationes
arithmeticae, 1. pp. 450-5.
SOLUTIONS BY EULER. PROBLEMS ro, 11 355

The above assumption therefore has the advantage of making the.
product of the four numbers automatically a square and also of making the
third formula take the form

M3abcd (ab + be + cd + da).

Since the first formula 47 (ab + bc + cd + da) must be a square, it follows
that abcd must be made a square.

In order to make the first and third formulae squares, take
M=ab+be+dd+ da,
or, if the latter expression has a square factor, say /2 put
M= (ab + bc + cd + da)|f>.
We now have only two conditions remaining to be satisfied, namely
00d = A SQUATE: ols ois eshscutinsenten ravines (1),
ab’c + bd + cd*a + dab + 2abcd =a square ............ {2).

The expression in (2) reduces to
(a® + 2) bd + ac (6* + d?) + 2abcd,
or bd (a* + ¢*) + (b+ d)* ac = a square.

We have therefore only to find numbers a, 4, ¢, & satisfying these two
conditions. It is further to be noted that a, ¢ are connected by a relation
similar to that between 4, 2, and the whole question depends on the ratios
a:c and 4:d. We may therefore assume a, ¢ prime to one another and
likewise 4, prime to one another, for, if either pair had a common divisor,
it could be omitted and the relation would still be satisfied.

Consider now the second condition as being the more difficult. Although
two ratios a : ¢, 6 : d are involved, neither can be arbitrarily assumed. For
suppose eg. that 4 : d= 2 : 1; then 24” + 222 + gac would have to be made
a square; this however is seen to be impossible, for, if we put a =p +g,
¢=p —¢, we obtain the expression 13p°— 5¢% which cannot be made a
square. The same impossibility results if we put 4: d=3 : 1. Therefore
the ratios a : ¢ and 4 : & can only be certain particular ratios.

Obviously the first class of ratios adapted for our purpose are square
ratios. Assume then that 4:4 =p%:¢% and put

Poa) vac(p +) = (pga + 5 o), says

therefore 7 (P+ 7°) a + n’p’qc = 2mnpga + mc,
a m® — n*pq?

so that = Te
¢ (P+ §F — amnpy

or, if m=+ kpg,

ey (#7) pg’ in

Ql >

23—:2
356 SUPPLEMENT

Now, if values could be found for %, 7, p, ¢ such as would make ac or
n (£2 =n?) {n (p> + 9°) + 2Ap’°} a square,

we should have a solution of the problem, since, #2 being already a square,
abcd would then be a square.” Euler however abandons the investigation
of this general problem as too troublesome and as certain, in any case, to
lead to very large numbers; and, instead, he proceeds to seek solutions
by trial of particular assumptions.

Particular values of a/c in terms of p, ¢ are the following, which are
obtained by putting 2=2, #=1; k=3, n=1; etc.

1.2. hee ll ih. 2 mal
c (ZF Tn FFL 4° 7’ c (P+¢)>P+ 60°¢>’
a = 50°7° a 35 150%7°
ML =F rr TFL
en 12° vi 2. 242°)"
Cc (PH raat Le (Brgy
2,2 272
VIL fe oil un Ge gn
¢ HZ +o) 208 £ olf +o) L300
a 92°¢*
xX. = ———2 1
: T16(7 + fy racks oe

Taking now the simplest values of 4/d = p*/¢% let us write down the
simplest corresponding values for a/c:

it? =, ? becomes Bod bh 4 Yn
if S—4, 2 becomes §, 4, %2, 3, fr #0 — 20 Bs
155 = 812 oromes 1.0500. 98, 90 py 220)
0a 0,0 © pevomen 302, 47, 20, 08, 5

The last assumption gives, “praeter exspectationem,” two cases in which
a/c becomes a square ; and these give two solutions of our problem.

t. Putting a= 64, i= gy c= 49, d= 4, we have
M=ab+bc+cd+da=576+441 +196 + 256
= 1469,
and the four numbers are
1469 . 196, 1469 . 441, 1469. 576.
Putting a =64, 6=9, c= 280, d=4, we obtain
M=7576+ 2601 + 1156 + 256 = 4580,

1469. 256,

2.

and the four numbers are

4589 .256, 4589. 576, 4589. 2601.

4589. 1156,
SOLUTIONS BY EULER. PROBLEM 11 357

Again, the form of the expressions abcd, bd (a*+ 2) + ac (b + d)* to be
made squares shows that any values for a/c obtained by the above process
may be taken as values for 4/d. Also a and ¢ may be interchanged. Euler
accordingly sets down as values of 4/d to be tried the following:

8 12 13 20 28 32 33
$F OF 1 5 FF, Sel

) y UG

|

4

3

He obtains no solution from the assumption 4/d = 4, but he is more
successful with the assumption 4/d=%.

Putting 4/d = 4, we have to make
20 (a® + ¢%) + 81ac =a square.
This is satisfied by a/c=r1; let therefore ajc=1 + x, and we have to
make 20{(1 + x)*+ 1} +81 (1 + x) a square; that is, :
121 + 121X + 20%" = a square = (11 + xy)’, say;

121 — 22) a )*-—-z22y +101 m+ 22mn + 1017°

therefore «=-—"—-——=, and . 3
JE 20 . C = 20 mm: — 2072

 

and, by putting m =35, »=1 we obtain a/c=L.

This solution serves our purpose, since it makes abcd a square.
Putting a=16, b=35, ¢=35, d=4, we have

8o+25+20+064 139
ar a

and, if f=3, M=21; the four numbers are therefore

 

M

21. 20, 21.25, 21.064, 21.80,

This is a solution in much smaller numbers ; and
the sum of the numbers =g.21%
the sum of products of pairs = 110%, 215%
the sum of products of sets of three = 120%. 21%
and the product of all four = 10087. 21%,
When one solution is known, others can be found. Take, for example,
the last solution in which, for 4/d=%, we found that
Y—-22y+ 10%
y»—20

 

 

a
c

In order that abcd may be a square, we must have
5 ()®—20)(y*— 22y + 101) =a square.
This is satisfied by y =5. Substitute z + 5 for y, and we have
5 (2*+ 102 + 5)(2* — 122 + 16) =a square,

or 400 + 500% — 4952 — 102° + 52* = a square.
358 SUPPLEMENT

   

 

Equate this expression to (20 + 222 —%%%5*? and we have
2
(= Nes BA io;
32 32
hie an 32 12708 32.1355 _In-10§ 3380
266321 24211 2201 2201’

therefore y=z+5=214365. and the resulting values of a, ¢ are large
numbers which Euler does not trouble to develop. As a matter of fact,

@ 33696. 4.118

27 100468205 © 5.46997

 

It follows that
SL M=5.55606 + 5.109465205 + 4. 109465205 + 4. 5601
= 278480 + 547326025+ 437860820 + 222784
= 985688109;
and, putting f= 9, we have
M = 121689809.
The four numbers are therefore
12168989. 278480, 12168989. 547326025,
12168989 . 437860820, 12168989. 222784.
PROBLEM 12. 70 find three numbers x, y, z such that the expressions
842 aR aE 4 af

are both squares’.

In order to satisfy the first condition, we have ool to put

a= pt + ¢* — 77 y=2p7, 3= 297,

for then B+yr+r=(pP ++ 72)? = 73, say.

The second condition requires that

> x? + a5? + t= O;
therefore, since 3 + 22 = 47° (£* + 4%),
QL =47 (P+) (£2 + ¢ - °F + 165°,

or Far’ = (8 + FUE + 7 — 7°) + apg

In order to get rid of the sixth power of # and so make p* the highest

power of p, suppose that »=p —ng (which Introduces no  tesuictiony;
therefore

Q14 (2 — ng)’ = (P* +4) any +(1 - VI + pF (op — mg)
or ag (p—ng)l= (p+) {2np + (1 — 7°) g}* + 4p* (Pp — ng)™

L Acta Acad. Scient. Imp. Petropol., 1779, Vol. 111. (1782), pp. 30 sqq.= Commenta-
tiones arithmeticae, 11. pp. 457 Sqq- He
SOLUTIONS BY EULER. PROBLEMS r1, 12 359
—-ng) R;

. Let the latter expression be denoted by &% so that Q=2¢(p

and
R= 4 (1+ 7%) p*— gn (1 + n®) pg + (1 + 62° + n*) p*¢*
+ 4n (1 — 22) pg® + (1 — n°)’.
This may be made a square in two ways, either (1) by taking advantage
of the fact that the last term is a square, or (2) by making the first term,

Z.e. making 1 + 77% a square.
{(v) Put B:={{1 ~n +20pg +o’, and make the term in ¢f
I+ 62+ i =4r + 20 (1 — 7B), OF

disappear by choosing a so that

ey Ll we then have
R= Gg

Lo eae, 21470) ,
4p" — 4np Tao ny 3 Sy 29,
whence (15—357°+ 137° — 2%) p= 8n (1 — un?) (3 — 7°) ¢;

this divides throughout by 3 — #* and
2. 5» (1 —n%)
g 5-107 +m

Let p=8n(1—n%), g=5—107°+7*; then » =p —ng =n (3 + 272° — n*),

while
2\2
reteset

 

and x, y, z can be expressed in terms of 7.

Suppose n=2; then
QO=3%0. 7038;

Bx 1.
p=—48, ¢g=-19, r=-10, R=7035,
x = 2565, y = 960, z= 380, or (dividing throughout by 5) x =513, y=192,

z= 76 (in which case Q=106932, £P=553).
Ex. 2. Suppose n= 3; then p=—192, ¢g=—4, =— 180, or (dividing

by —4)
; 2=48, g=1, 7=45 R=14120, (=1270800;
~ x=280, y=090. 48, -~zr=90, :
Dividing the values of x, y, z by 10, we have
and ‘O= 12703, P=433

x=128, y=432, 5=09,
(2) To make the first term in &? a square, suppose 1 + 7°= 7° which

is the case whenever zn = (a*- 0?)/2ab.

We have then : :
R= q4m*p* — anm’*p’q + (m* + 40°) p°¢* + 4n (1 — 7°) pg° + (1 — n*)* ¢*.

Euler solves this in three ways.
360 SUPPLEMENT
First, he puts R= 2mp*—nmpg + (1 —n*)¢* and from this, by taking
a=2,b=1,so that n=2, m=+3%, he obtains the particular solution
x=-—392, y=1386, z=-1056,
oF : x= ‘196, y= 693, z= 528
Secondly, he puts R = 2mp* + 2mpg + (1 — 7°) 4%, and deduces, by the same
particular assumptions,
x¥=936, y=74, %=3552,
or x=4068, y=37, z=1770.
Thirdly, he supposes
mt + 3m

z= 2 TY
2p” — mnpg + % 7%

m* + 4n*

where however the last term should apparently be - FE 2

Euler's son, J. A Euler, gave, in a Supplement to his father’s paper,
another solution as follows.
We know that

(= 1F taf = (4 1F and (fF ~ F245 = (5 2 2)
Multiply the first equation by 4¢* and the second by (£* + 1)?; this gives
49° (7° — 1) + 160°" = 4g" ($* + 1)°
=A SIE (FAP = (2 + 18 = 2,
oF (mF ag (Fa) 4 16 = (0)
Therefore the three numbers
F-)(Fr1), 29(Ff~1) apy
satisfy the first of the conditions.

The sum of the squares of the products of pairs of these numbers must
now be a square; after dividing out by 44% this gives :

(= TR (A= 1 = 457 (5 = 1 (FE 4 1) 4 160707 F ~ 1) =n spare.
But (£i= 18+ 42 (S43) = ( S0)s
therefore (2°+ 1)*(¢*— 1)* + 16%" (#*— 1)’ must be a square.
For brevity, let A2=(2* + 1)}, B*=16p*(p*—1), and
LP = IV + BY, or Ligt + (Ba AD) + AL

must be a square.

Put Ag+ (B= 24%) §* + A = (Ag* + 0),
A? — 2?
ts
wisn T= Bx 2A
SOLUTIONS BY EULER. PROBLEMS 12, 13 361

Now both the numerator and denominator of this fraction are squares
if 22 = 4% — B® for the numerator becomes A? and the denominator
2A: ~ DB + ad J L2~ BY),
which is the square of 4 + /(4*— B?).
But, putting for 4, B their values in terms of p as above, we have

A=-B =p — 120° + 380 — 1207+ 1 = (p* — 6p + 1)%;

 

therefore :
Boe ae) | Caplin) ag
7 A+ J(A*=B) (PP +1)2+p—6p +1 2p —4p’+2 pP—1
Thus FI SE,

and the numbers required will be
Ade lL EN
{f—1) ? in
or, if we multiply by (2*= 1)}
(6 ~~ (+3), apd L—1), 82 (F-)

The sum of the squares of these numbers turns out to be (2° + 1)f
which is not only a square but a sixth power, while the sum of the squares
of the products of pairs is found to be

160° (p— 1)? (2° — 42° oo 220 — 4° I ty
or 16 (22 — 1)*{(p*— 1)* + 16p*}%

Ex. Put p= 2, and we have

 

R=3.7=3% J=78 2=00, Ar Lri=iss=3,
2+ a+ VR =16.4.9. 337% = 30833,

the solution being in smaller numbers than Euler’s own.

PROBLEM 13. 70 find! three positive integral numbers x, y, sz such that
X+ty+tas=uf :
a
To make x® + 3* + 5* a square, put x =a? + 6*— 3 y= 2a¢, z= 20¢ and we
have 2% + y2 + 22 = (a® + 5* + OO)
We have now to make a* + #* + ¢* a square, and we put similarly
a= Rh PRB=9%, Isopr, c=29r;
we have then ++ E= (p++ 7A,
Now let us express x, y, z in terms of p, ¢, 7; this gives
x=p +P +r + 2p + 207° — 64°F,
I= wf ag 1%,
2=8pg7°;

1 Commentationes arithmeticae, 11. pp. 399-400.
362 SUPPLEMENT

therefore
X+y+a=p +g +7 +2020 + 20° — 6427 + 4 27 + 4977 — i 8pgr.
(1) Arrange this according to powers of p, and >
Boyne piy 2(g+7) p+ 8907" +4 + 4g’ r — 641 — 4g + A.
In making this a square, we have to see that p, ¢, are all positive, and
also 2 + 42> 72. Also &® + 2° must be > 2 :
Equate the expression to {#* + (¢ + 7)?}?, and we have
8p + + 4 r — 642 — 4g + r= (g +7),
whence 8pgri=129°7 + 847°, or p=3g+7.
Therefore a=1202+ 397, b=397 +277 c= 297,
where both the letters ¢, » may be given any positive values.

Ex. 1. Suppose ¢ = 2, »=1; therefore :
2=4, a=19, 6=8, s=q;
accordingly the numbers are ion
Xx =d00, y=1i52, 7=04, :
and x+y+z=0625 mas bundy re 198 = gq itm ory
Ex. 2. letg=2,7= 2: therefore
2=5, a=3235, b=20, ,=3,
and : X=961," y=400, 7=7320;
therefore x+y+2=1081=41°, F+PE+7F=1183021=33"
(2) Arrange the Spas for x +y +z according to powers of ¢;
this gives
x+y +z=¢"+ 497+ 29° (fm 37) + aged? + 207-7) + (PP +P)
In order that the terms in ¢* and ¢* and the absolute term may vanish,
equate the expression to
{7* + 297 — (p+ A)},
_2pr(p +7)
: Tipe!
Ex. Suppose g=1, = 1; therefore

whence we obtain

: g=4, a=106, b=a ¢=8
or (if we divide by 2) a=8 ¢=1;, eoe=4;
therefore ’ x=49, y=64, z=8
and Xtytz=11% a+ + P= h56r=8=0%
These numbers are no doubt the smallest which satisfy the conditions.

The case of three numbers is thus easier than that of two (see p. 290,
note). Euler solves the same problem for four and five numbers, and shows
how the method may be extended to six numbers, and so on indefinitely.
SOLUTIONS BY EULER. PROBLEMS 13, 14 363

PROBLEM 14. 70 find three numbers x, y, z, positive and prime to one
another, such that both x+y +s and X* + y* + 3° are fourth powers’.

As in the above problem, put

xd + PF y=, z=1d

and further a+ =~ b=afy, czar,
and make the expression in pg, ¢, » for x+y + 2 a square by equating it
to {#*+ (¢ + 7)’}* as before. This gives p =32g +; but we have now, in
addition, to make p*+ (¢ + 7)? a square.

Put P+(g+7r)?i= {# LEED)

therefore g++) =2fgp+/2(g +7).
Substitute 3¢ + » for p, and this becomes
F-Pt lr+e=o,

2 Be
whence g lof

ro Caf
The problem may therefore be solved in this way.
Take g=f2+2fg—g* and r=g'-3 0-13

so that p=1(f?—g%, then find q, 4, ¢ and then again x, y, 2, in terms

off &
Y¥x. Let f=1, = 3; therefore

gE—L gm=y, fm=4,

or g= 2 h r=, p=,
Thus a=1g, &=8, i=4
and X=400, y=182, 7=04;
so that X+y+5=0625=5% 42+ 24+ R=194481= 21%

To find limits for the values of £, g, change the signs of ¢, 7, putting
g= 7 irs es fog, p=llci- 0)
In order that ¢ may be positive,
Zl 1eNa> 2.414 ...,
and, in order that » may be positive,
glf <x (3+ 413) <3. 302...
Suppose e.g. that f= 2, g¢=5; then

9=1, r=9, =
or in integers g=2 r=18, Jp=oy:
hence a=121, O=750, cma

x=580993, y=17424, z=108864,
x+y+s=707281=20% x*+3°+35°=349707832321=769"

Y Commentationes arithmeticae, 11. p. 402.
364 SUPPLEMENT

ProBLEM 15. (Problem in Fermat’s note on vI. 13.)

«To find a right-angled triangle (in rational numbers) such that either
of the sides about the right angle less the area gives a square’.

; 2%
Let the perpendiculars be il wa so that the area is 5 ; and

2% xy
= — =: Or 2%z— XY
z

g 2
Y lg or y2—xy have to be made squares.
2 7 :

x2 + 2 :
as well as 4 . Yor 45% + 3°
2

 

Since the first two SApresshuns are to be squares, their product must be
so also ; therefore
2xy2® — 2x%yz — x22 + x%p* = a square
: 2
(rom
and, after dividing by. yz, we have

 

 

2p »
2x5 — 23% —xy=—"L xp +< yz,
HE se
whence gL te pp
29” Xx — Py
Ea + Py (295 = py)
Thus 22—Y= 2x — py = 20% — py ;
Byte -apgny ay (pq)
2g°x — py 20°% — py
and xr —Xy= & (2g - py = 27 (29x - pp)

 

2ix—ply 2 play’
WE) lp =gF
Jena oe: 2) _ en
Lx -py fd Puy
Therefore the two expressions are squares if 2¢%x*— p*xy is a square.
Put “ 2°57 — Play =a’;
therefore {22 -Da= poy, or xly= leg 7).
It is sufficient for our solution to know the ratio x/y, since a common
denominator z has already been introduced.

Therefore we may put
x =p = 20075

L Novi Commentarii Acad. Petrop., 1749, Vol. 11. (1751), pp. 49 sqq. = Commentationes
arithmeticae, 1. pp. 62-72.
SOLUTIONS BY EULER. PROBLEM 13 365
Lr =-rl=9
X= 77 ’
Ee MAY dios 1
7? :

whence g—

and z=

It only remains to make 4a? + 3? a square; that is,
4p + 49* — 4¢°7* + #* must be a square.

A general solution of this equation giving all possible values of p, ¢, »
is impossible. We must therefore be satisfied with particular solutions.

Particular solutions (1) and (2).

Put 4+ 4gt — ag’ + = (28 FF;
therefore 40° — 4327 =F 4p*7
- 2
we £=35-)

that is, either

(0) 2=Ly@-») or 2) p=LJ0*-¢).

1) Now p=2 J(2—#) is satisfied b g = + &'r= 20d, whence
2 EN y.

Laie a)
= 2cd 2
or we may put

P= (E+ a) (E—a), g=2d (+d), 7=4ld",

and we thus find values for
=f, yg=af-7 JUt+PD=20+7 sun LESTE,

Ex. 1. Suppose c=2,4d=1; then

p=%.3=13, g=4.5=20, 7r=4.4=16,
$91.28 28 By 220%
256 8 g
(4x2 +37) = 2p + #* = 706,

 

X=228, YES544, 5=225 4%

and the triangle is
a 144 ¥ 4352 Jay) _ 5648

Zz 8’ 275.30" 2 25-39
Ex. 2. lf ¢=3, 4=1, we get the triangle
sw 32.9 yp 381.41 Jax +57) 9.881

z 135° 27. 25.135 2 25.185"

 

And so on.
366 2 ged SUPPLEMENT

(2) In this case p= g J (#7? — 4%), and we have to put

r=c2+d% g¢=2d, whence ny
or p= 20d (E—dY, g=200(F+d%, v=" +d,
while:
=p, y=ur-s, JArLP)=of-r smn 720 Zz

Here, since 2¢® must be > #2,
SEE (P+ A and aad J2> A + 4%

therefore @> (c —d /2)% and
either d>c—d Jz, so that 2

r+. J2’

I
ay

If therefore d=1, either c< Ja +1 or ¢> Ja—1. The second
alternative is satisfied by ¢> 1.

or d>d. J2—g¢ so that 2 mt

Ex. letc=2, d=1, and we have
274.312, g=4.5=20, r=5.5=25;
therefore

175.64 _ 4048
625 28

 

 

x=144, y=175 J@*+3)=337, z=144+
The triangle is therefore

2x 288.25 df, gp 25-178 4378 Jax? + 3?) 223-337 3473

2 7 qo48 253° 2 1048 = A043’ 2 4048 4048 °

 

 

 

Particular solution (3).

Put 45 + 40° — 40° PR + 7 = (2p + 29°);
2 en
therefore 7 —4riri=4 5%, and F=+ SR :
164°

therefore either p=  J(2r2 — 8° , OF = 35 222)
p= il 0% prim ns 0)
The first value is however useless, since 2¢%— 72> o, or 2¢* > 72
We have therefore i= 2 (8% — 27%),
while i :
x=p% y=20"—7, J+) =2p—24% and z=x +

Since 84% — 272 must be a square, put

C 2
8g — 277 = {5 (2g + »} 2

y=)
nd
SOLUTIONS BY EULER. PROBLEM 13 367

therefore a7-2r="5 (20 4) or 4d%g — 2d%r = 268% + Pr,
whence vs g=2+4 24% r=qdi- ad

2g +» =8d? ,/(8¢*— 27%) = 8cd, and therefore p= a :
Multiplying by 242 + ¢%, we have in integers
2=4d (2d? 8), g=(2d?+32P, r=20d*-) (2d +5),
while «x, y, z have the values above stated.
Ex. 1. Pute=1, d=1; therefore

P=4, g=9, r=6] 2£=16, y=126, Jar +P=130,
126.25 207

and 2=16 + 0 | >

 

The triangle is therefore
ow 64 iy 252 Jat) 2060
2 207’ z 207! z T2097

This is the triangle in the smallest numbers satisfying the conditions, as
Euler proves later.

Ex. 2. Since 24°> 7% it follows that ¢/d> 2 — Jz; but it does not
matter whether 24% > ¢* or not, since p, ¢, » may be negative as well as
positive.

Put then d=2, c= 3; therefore 2d’ ~F=~1,and 2d: +A = 17.

Wethen have . p=-24, g=280, 7r=-34;

28118255
aE.

x=576, y=2.7.41.17% JU**+))=2.5.53.313 z=22LY
The triangle is therefore

2% 2304 y_ 28.41.97 J+") 4.5.53.313
28118255 ° 7 28118255

 

r 28110255 7

It is to be observed that in all the above examples it matters not
_ whether ¢, 4 are negative; it will only result in the values of p or ¢ or »
becoming negative, but the values of x and y will not be thereby changed.
Only z will vary, since z may be either

ws SEE os x +

Yip ¥

7

After remarking that the problem of making

4p + 49° — 4°77 + #* or 4p* + (24 — 7%)? a square
may be solved generally by equating it to
2
(2 —-77+ = 7) ;

Euler passes to his general solution.
368 SUPPLEMENT

General solution.
iH = ; 2 are the perpendicular sides of the triangle, let x = ad,

y=a®—0*; the triangle is then

ab 0 +P

7) gig
ab (a* — ¥)

72

 

 

and the area is

Now we found above, at the beginning of the investigation, that
Se Eran WF

20°% = py 207% ~ Fy’
or, since ¢ can be taken positively as well as negatively, «
2
=X + 2 (P29) ’
20°x — py

where x = ab, y=a®— 4%
And we took 20202 — Poy = 1247,

whence z=x+ y@Eef ab + (eM iy ;
72 7°
We have therefore only to satisfy the equation
20°x% — pray = ra,
- w= ar);
and, since xy=ab(a*—0%), we have to find such numbers for a, b that
ab (a*— 0%) may be of the form 2f*—g* or (2f*—g*) I>
Suppose now that such numbers a, 4 have been found that
ab(a*— 8°) = (2/2 —g* A
Then, since x = a¥,

9 g 2 al’ 2 9
&~e)#= TF opr,

and a natural inference is suggested, namely that

 

abg abr
Fe a
Let now p = ab, and accordingly

3 a 2
g=tl rah snaps Y

 

 

3 J)
the triangle is then
2ab 2abg*/i?
3 abd + (& =P) (ab + fh)
C= (a® — 6%) ge?
2 abel? + (a — 7) (ab + fo)’
+B (a + 8%) gh

 

5 abgh + (a — 0%) (ab + fh)
SOLUTIONS BY EULER. PROBLEM 1; 369

Also from any particular values of @, 4 any number of triangles can be
derived satisfying the conditions.

For, if p = ab, and
20 (FF —18) = (2f% = 09) £3,
we have (2f2— gd) A= 24° — 1,
or 2 (f=) = 0 =

Put 2(fh+g) =" (gh+7), and fh—g=" (gh -7);

therefore

 

_2mngh — (27° + m®) fh
g= 2n* — m? ?
(27 + mi?) gh — gmn fh
= 2 2 5
21° —m
or, in integers, p= (2n2— nw?) ab,
g = 2mngh — (271 + m*) fA,
r= (20% + m®) gh — amnfh,
2 3 zs 2
while 7 = nls let ay 2) (229) :
7
2_ Ja 2 21
Thus the triangle (2, = ; zr is known.
z 2

Lastly, to find suitable values for a, 4, Euler writes down all the
numbers from 1 to 200 which are of the form 2#2— 2? including all the
squares arising from the supposition that z=¢ and all he doubles of
squares corresponding to # =o. Inspection shows that the table contains
(within the limits) all the prime numbers of the form 8 + 1, and no other
primes, the doubles of the primes, the products of the primes into all
squares and into one another, and the doubles of those products.

Now, since the product a. é(a +4) (a—0) is to be of the form 2#*— 2?
and the factors a, 4, a+b, a— 06 are either prime to one another or at the
most have 2 as a common divisor, while 2 is itself contained in the form
2#*—1*, the several factors must all be of that form, in which case the
- product will be of that form.

We have therefore first to take some value of 4 in the table and then see
whether there are in the table three other numbers a —24, a, a + ¢ differing
by 4. Euler gives a second table showing values of a corresponding to
values 1, 7, 3, o, 16, 17 etc. off d

The values of a in the table corresponding to é=1 are 8, 17, 63,
#2 127,

Ex. 1. ‘lake ¢=1,a=23; therefore

ab=8, #=0=63  abla~iN=8.9.7=4.9.14,
and 4.9.14=74*(2f?—g?), so that A=6, 2/*—g?=14, and accordingly
f=3, g=2.
H. D. 24
370 SUPPLEMENT

We have therefore in this case
2=8(2n*—m?®), g=24mn—18 (2n*+m?), r=12(2n’+m’)—72mn;
or, dividing by 2,
p=8m—am?, g=12mn—18n—gm?, r=12n"+6m*— 36mn,
whence p+¢=12mn—10n*—13m* —p+q=12mn—26n*—sm’
while z2=8 + Sopa .

Thus there are any number of values of z from which triangles may be
obtained satisfying the conditions.

The simplest value is found by putting 72=1, z= 0, whence

7=6, ptg=—-13, p-g=g3,

and either z=8+1.25=297,
or z2=8+71.169=21215,

The first value gives the triangle in smallest numbers above found
(p- 367),

agb 64  al=/? 350 ALB 260

Zz 20%’ Z.i 20%’ z 207"

 

 

Substituting 1215 for 207, we have the sides of the triangle corre-
sponding to the second value of z.

The particular triangles are also directly obtained from the values of
a, b, f, & #% without bringing in m, zn; for
(a®— 8) (ab + fh)?
7
2
528218) o 207, anny
12 4 4

z=ab +

’

that 1s, z2=3

Ex. 2. Take s=41, a=112; therefore
ab=9.16.41, A ~-F=71.9.17,
and —0=16.9.7.17.41. 71 = (3 ~ 2D A,
whence A=12, and 7.17.40.91= 2/2 ~¢~
The simplest solution is f= 417, ¢= 37.
(a= #) (ab 2 Shy
or
is easily found, and consequently the triangle
sabi (t= gt PP
Zz} zo z =
[Euler finds values for /, ¢ by using the formulae
(20°— PB?) (272 — 8) = (2ay x 83) — 2 (By + ad)?
and F—2pf=2(xty) (xt 29)

Thus z2=ab +

 

 
SOLUTIONS BY EULER. PROBLEMS 15, 16 37

He does not actually give the steps leading up to the particular
solution f= 417, ¢= 37, but it can be obtained thus.
Since 7=2.27-1 and 13=2.3%~ 71, we have
7.17=(2. 2.31.10 ~2(3. V+ 2.1)
=13~2. €=ol13- 3 ~(13=2. 5V=2, 52m 38
Again, since 41=2.52—3? and 71=2.6%- 1% it follows that
41. 71={2.5.6—3.10—2(3.6=5. 1)
=57°-2.137=2 (57-13 ~(37~2.130=2. 44-30%
Therefore, by multiplication,
7.17.41.71={2. 8-39 (2. 44 -319=(2.8.94 + 3.31) ~2 (8.31 + 3.44)
=797°~ 2. 380° =12 (797 — 380)*~ (797 ~ 2. 380)’ = 2. 417° - 37]
ProBLEMS 16. “De problematibus indeterminatis, quae videntur plus
quam determinatal.”

We have seen that by means of certain “Porisms” stated without
proof Diophantus is able to obtain relations between three numbers
x, y, # which have the effect that, when they are satisfied, a quite
appreciable number of symmetrical expressions in x, y, z are auto-
matically (as it were) made squares.

It is clear therefore, says Euler, that, if a general method of finding
“porisms” of this kind can be discovered, the whole subject of Diophantine
analysis will be appreciably advanced. Accordingly he proceeds to discuss
such a method.

The method depends on a Lemma the truth of which is evident.

Lemma. Jf values have been found for the letters z, y, x etc. which
satisfy the equation W =o, where W is any function of those letters,
and P, Q, R etc. are other functions of the letters such that P+ W,
QW, R+W etc. are squares, then, if the values of z,y, x elc. are
taken which satisfy W=o, the resulting values of P, Q, R etc. will
also be made squares.

Cor. 2, Q, 2 etc. will similarly be made squares if P+alW, Q+ BW,
R +y W etc. or, more generally, if
PraW+{W?  Q+BW+nW? R+yW+0W?* etc.
are squares.

+ Conversely, If such values for z, y, x etc. have been assigned as will
satisfy IW =o, all formulae such as P* +a WW, Q*+ BW, R* + yW etc. will
at the same time be made squares.

1 Novi Commentarii Acad. Petropol., 1756-57, V1. (1761), pp. 85 sqq.= Commenta-
tiones arithmeticae, 1. pp. 245-259.

24—2
372 : SUPPLEMENT

And, as the number of such formulae is subject to no limit, it is clear
that an unlimited number of conditions can be prescribed which are all
satisfied provided that the one condition /# =o has been satisfied.

The same Lemma can be extended to the case of cubes or any higher
powers ; for, if =o has been satisfied by certain values, all expressions of
the form 2° + a W will thereby be made cubes, all expressions of the form
P* + a IW will be made fourth powers, and so on.

While it is plain that, if values for z, y, x etc. are found which satisfy
the condition = o, all the expressions P?+ a W, Q*+ BW, R*+y WW etc.
will be made squares by the same values of x, , z, the difficulty will be,
when a number of expressions 2? +a I, Q*+ BW etc. are given which are
capable of being made squares in this way, to identify and separate the
expression WW the equating of which to zero will make the rest of the
several expressions automatically squares. It would indeed be easy so to
hide away the composition of the expressions as to make this separation
itself a most arduous problem. On the other hand it is easy and in-
teresting to begin with /# =o, and then investigate the simpler formulae
which can by this means be made squares. Before proceeding to the
particular cases, Euler observes further that it is convenient to take for I”
an expression in which z, y, x etc. enter symmetrically and can be inter-
changed ; for then, if 2? is such a square that P*+al¥ is a square, and
2,' 7, X etc. are interchanged so as to turn 2? into QF, Ri etc, Qt all;
R*+al¥ etc. will also be squares. Also, since solutions in rational
numbers are required, z, y, x etc. should not enter in any higher power
than the second into the expression 7. Euler begins with expressions
containing two unknowns z, y only.

Problem (1). Given W=y + z— a =o, to find the more simple formulae
which by means of this equation can be made squares.

When the equation y +z —a =o is satisfied, it is clear that the general
formula P? + M (— a +y + z) will become a square whatever quantities are
put for 2 and AM. Accordingly Euler, by giving 2, M various values,
obtains without difficulty 44 different expressions which become squares
when y+ z—a=o.

He supposes M=2, —2, 2, =p, —2—y, y+z+a, n(y+z+a),
(y+s+a)(y—2+a)(z—y +a), and 3 and 2° — 1 times the last expression
respectively, and with each of these assumptions he combines one or more
forms for Z. I need only quote a few expressions which are thus made
squares, e.g.

(-1)+2(-a+y+2)=y*+22+1—2a
(z—1p+2(—a+y+2)=s+29+1—2a)
SOLUTIONS BY EULER. PROBLEMS 16 373

(y—z+1)—2(—a+y+2)=(y—2)1—42+1+2a
(z—y+ i
(y—n)l+2n(—a+y+2)=y"+2nz +n’ —2na,
—z—nl+2n(—a+y+2)=(y—2) + 4nz + n*— 2na,
VF ~yl-atytay=-yitay,
(+22 —(y+2)(—a+y+2z)=ay+az,
(y+z—a)y)—(y+32)(—a+y+2)=a’—ay— az,
(r=) +(y+z+a)(—a+y+2)=32+y +2 + 1-4’,
(yz—n)l+n(y+5+a)(—a+y+3)=1"3" +n)’ + nz’ + n* — nat,
(P+ 2 +a) + (+z +a) (y—z+a)(z—y+a) (—at+y + 2)=4(1*2+ a’y* + @’5°),
and so on. Wherever a new expression can be got by interchanging z
and y, this may be done.

Taking the more particular case of #=y + z—1=o0, Euler obtains the
following expressions which are thereby made squares,

Fran Pyrogen yan,
2+4y, 22—3+y, z—y3
YE +2 +27 202+ 222-1,
which indeed are easily seen to reduce to squares if we put y+z=1 or
y=1-2

The fact that 22+» + 2? is a square if y= 1 —2z or, more generally, if
y=%1%2 1s included in the Porism in Dioph. v. 5. Similarly

V2 + a¥? + a’?
is made a square if we put y=+a zz.

The last expression but one in the first of the above lists, namely
922% + ny’ + nz® + n® — na’, becomes a square whatever value ~ has. If a=,
it becomes

VyE+ny +n +n’—n
or (+ n)(2* +n) —n.

That this is a square when z=y + 1 is part of Diophantus’ assumption
in v. 4 (see p. 104 above).

Euler's Problems (2) and (3) similarly show how to find a number of
formulae which are all made squares by values of y, z satisfying the
equations W=yz—a(y +s) +b=oand W=y'+22—2nyz—a=o.

He then passes to the cases where there are three unknowns.

Problem (4). Given W=x+y +z —a=o, to find the more noteworthy
Jormulae which can be made squares by satisfying this equation.

In this case the general expression P?+ M (x+y + z—a) becomes a
square.
374 SUPPLEMENT

Put M =2n and P=one of the expressions x — 7, y — 7, 2 —n, or one
of the expressions y+z—n, z2+x—n, x +y—n.

These assumptions make the following expressions squares :

z®+ 27 (x +) + n*— 2na, and the two other similar expressions,
(y + 2) + 20x + 7° — 2a, % ” 5
If M = 2ny3, P=yz— ny — nz, and 50 on,
Y32 + 2nxyz + 2y + nt + 2m (n— a) yu
and the two other similar expressions are squares.
If M=—(a+x+y+3), P=y+z—x and corresponding expressions,
a’ — 4xy — 4x2
a — 4yz — 49% are all squares.
a — 43x — 4zy

In particular, if #z=2a, a=1, the following six expressions are made

squares by putting x+y +2z=1%:
B+y+z, (y+20+x,
YV+z+x, (+x)}+y,
Z+x+y, (x+y)7+2
If a = 2, we make the expressions
I—Xy—X2
I—Y%—yx
1—-2X—Xy
all squares by putting x+y +z=2.
Problem (5) finds expressions which are made squares if
W=yz+zx+ay—a(x+y+s)+b=o0
is satisfied.

Problem (6). Given W=x*+y* +22 — 2y5 — 23x — 2xy—a =o, fo find
the more simple formulae which can be made squares by means of solving that
equation.

Here the general formula will be

P+ M(x* +9 + 2° — 2yz — 220 — 289 — a).

UM=—1,P=xv+ty+s

4Y% + 42% + 4Xy + a = a square.
YY M=—1, P=y+s-x, clc,

4Y5 + a

42x + a [ are squares.

4xy +a |

If =—1, P=y—g7 cic,

a+z2(y+2)x—a°
a+2(z+x)y—y° | are squares.
a+z(x+y)z—2
SOLUTIONS BY EULER. PROBLEMS 16 375

In the particular case where a = 47, so that
K+ y+ 2 = 292 + 225 + 28 + 4,
yi+n
2X +n
xy +n
Yitax+xy+n

are all squares ;

or our formula gives the means of solving the elegant Diophantine
problem :

Given any number n, to find three numbers such that the product of any
pair added to n gives a square, and also the sum of the products of the pairs
added fo n gives a square.

By solving the equation

W=x+y*+2%— 292 — 230 — 2x9 — 4n =o,
we obtain z=x+y*2./(xy +n).
We assume, therefore, such numbers for x, y as will make xy +7 a

square ; suppose xy +7n=12% and we then have two values for z namely
z=x+y+2u, each of which along with x, y will satisfy the conditions.

In fact, if z=x+y + 26, while #=,/(xy +n),
+t Jxz+n)=%(x+z2-y)=xtu,
+Jz+n)=3(y+z—x)=y tu,
+ Jz +axray+n)=f(x+y+a)=x+y+u

(Cf. Euler's solution of Dioph. 111. 10, p. 160 above.)

Problem (7). Given
W=x*+y +22 —2yz— 228 —2ay—2a (x+y +2) —b=o,
to find the more noteworthy expressions which can be made squares by
satisfying this equation.
The general expression here is
P+ Mix: +y +3 yen — 0}.
If M=-1, P=x+y+2z+a, we have
(a) 4yz+ 42x + 4xy + 4a (x+y + 2) + a’ + b=a square.
If M=—1, P=x+y+2—a,
(0) 4yz+ 42x + 4xy + a* + b=a square.
fM=-1, P=y+2—x+a, etc,
wr+4a(y+z)+at+b
(¢) 42x + 4a(z+x)+a’+0b | are all squares.

4xy+4a(x+y)+a+b
376 SUPPLEMENT

I M=~1, P=y+2—5—a,etc,
4yz+4ax+a* +o
(@d) 42x + 4ay +a®+b | are all squares.
4xy + 4az +a’ +b
Cor. 1. In order to solve the problem represented by (¢), equate the
expression 4xy + 4a (x +») + a’ + & to a square #? whence
4(x+a)(y+a)=ur-b+ 3a’
or (x+a)(y+a)=1(—b+ 3a");
x and y are then determined by splitting } («2 —& + 3a?) into two factors

and equating x +a, y +a to these factors respectively. Next, solving,
for z, the equation

X+yt +t —2yz— 22x —2xy-2a(x+y+2)-b=o,
we find, since 4xy + 4a (x + y) + a’ + b=, that
z=x+y+atu.
Cor. 2. If 6=— a then, by solving the equation
x+y +28 =2yz + 220 + 2xy + 2a (X + y +2) — a?
we make all the following formulae severally squares,
ye+a(y+32), yz+ax,
zx +a (z+x), 2x+ay,
xy+a(x+y), xy+az
ya+ zx + xy,
yE+ax+ay+a(x+y+ 3),
by assuming 3
z=x+y+at2f{ixy+a(x+y)l=x+y+at ou,
where (x +a) (y + a) is put equal to #*+ a>
An interesting case of this last problem is that in which a=1; and
from this case we can deduce a solution of a new problem in which the
corresponding expressions with a? 3? 2? in place of x, y, z are all squares.
The problem is
To find three square numbers such that (1) the igh of any two added
to the sum of those two, (2) the product of any two added to the third, (3) the
sum of the products of pairs, (4) the sum of the products of pairs added to
the sum of the numbers themselves, all give squares.
We have to find values of «7% 3? z* which will make all the following
expressions squares,
P2442 it al
2p tat, ay
Ta OL
V2 + 22x + xP,
VE + 2x + x2 + a+ y+ 52
SOLUTIONS BY EULER. PROBLEMS 16, 14 377

As we have seen, all these will be squares if
£=2 +) + 1 + 2 J(&Y + 2° + JP).

We have also seen (Problem (1) above) that a? + a? + »® becomes a

square if only y =a + 1. Put then y =x + 1, when we have
P=2x+ 2x +z re f(5 +245 + 30% + 28 + 1);

that is, P=yq(+ x41)

It only remains to make a® + a + 1 a square. Equate this to (—x +4)
and we have

£—1 > Prt+a
| — ———— Cc + = —_——
% ad NEE ree 1) Blh1’
A elt)
h = x24 x + = .
whence g=13 J{af+ x +1) —

Therefore the roots of the required squares are

AL el eli otta
marr aan 27+ 1

 

Or, putting Z = (» — ¢)/2¢, the values become
J tegr=i2i yu 774 ogre ag suhag
Sw TTT aim
Yetg=1,7r=2 andwehave x=3 vy=4,2=1: or, if weput Z=2 in
7 ht =F T p
the values expressed in terms of #4 the values are x =%, y=§, z=14,

Xx

 

PROBLEM 17. 70 find two fourth powers A%, B* suck that their sum
is equal to the sum of two other fourth powers.

In other words, to solve the equation 4*+ B4= C*+ D* or (what is
the same thing) 4¢- D*= C¢- B*.

It is proved, says Euler, that the sum of two fourth powers cannot be
a fourth power, and it is confidently affirmed that the sum of #ree fourth
powers cannot be a fourth power. But the equation A+ Bi— Ci=D'is
not impossible.

First solution.
Suppose Ad=p+g, D=p—-g,C=r+s5, B=r—3;
thus the equation dA =Db= C'— B*
becomes pg (P+ PD =rs (4 8),
Put p = ax, ¢ = 4p, = 4x and s=y, and we obtain
ab (a’x* + 0%) = k (Bx? + 37) ;

L Novi Commentarii Acad. Petropol., 1772, Vol. XVIL (1773), pp. 64 sqq.= Commen-
tationes arithmeticae, 1. pp. 473-6; Mémoires de I’ Acad. Imp. de St Pétersbourg, X1. (1830),
PP: 49 sq.= Comment. arithm., 11. pp. 450-0.

24—3
378 SUPPLEMENT

2 gE
therefore Te Di which fraction has therefore to be made a square.
x alt fk
One obvious case is obtained by putting Z = ab, for then
Pen
2 wlli-1)
whence y=a,x=1, so that =a, g=ab, r= ab, s=a, and the result is
only the obvious case where p=s, g=7
Following up this case, however, let us put 2= ab (1 + 3).

We then have
Y_ AUP 1)+ 3+ 300 + BF 2 B-1+302+ 300+ 0
= ab (0* — 1 — 2) r BP—1—3 2

 

therefore, multiplying numerator and denominator by 4*— 1 —z and ex-
tracting the square root, we obtain

ya JIB = 12+ (302 —1) (BP — 1) 2+ 30° (0° — 2) 5° + 0% (0° — 4) 7° — b*2%}

Xx F—1—3

 

To make the expression under the radical a square, equate it to
{(6* = 1) + f3 + g3°),
and assume f; ¢ such that the terms in z, z* vanish.

In order that the term in z may vanish, f=1 (36*- 1), and, in order

that the term in z*> may disappear,
BE -2)=2(F-1)g+f=2(F~1)g +} (oF ~ 65 + 1),
Sst
whence g=toras

The equation to be solved is then reduced to
BB —4)- oe = (+0) 5,

SE -9 of

or =
OF + &

Now & can be chosen arbitrarily ; and, when we have chosen it and
thence determined z, we can put

s=Pot—s yma(F-r+fitg),
and accordingly

p=a(lP-1-23), r=ab (1 +2) (0*— 1-13),
g=ab(BP—1+fa +g), s=a(l®—1+/fz +g),
where we may also divide out by a.

If x, y have a common factor, we may suppose this eliminated before
7, 0, 7, 5 are determined.
SOLUTIONS BY EULER. PROBLEM 17 379

Ex. 1. Let 6=2 (for 4 cannot be 1, since then ¢ would be « ).

 

Therefore =r g=-23 5=58
As a does not enter into the calculation, we may write 1 for it;
therefore

 

 

 

 

6600 218% It 6600 25 /[66ooN®
r=3- = rz, JEsiis. -5.( )
2929 2929 2.2929 24 N2020
wy 38407 1300 3. 28804041
fi 2929’

But the ratio x: y is what we want, and

Y _3.28894941 28894941 _ 3210549 1070183

x 2187.2029 @ 2920.720 2920.31 27.2029]
so that we may put

 

x =700%3, y=1070133.
2.9529 19058

 

Further A=a(1+2)= = .
2029 2929
Therefore
2 =179083, 7 =2% 10053 = £14500,

gd =2.1070183 = 21403606, s= 1070133.
Consequently
A=p+g=2219449, C=7+5=1584749,
B=p—s=—555617, D=p-g=-20061233,
and A+ B= C+ DY
Ex. 2. Let b= 3; therefore f=13, g=3, 2=122;
3.360 1107 27.4%

 

 

A=3(1+2)= i TT
ip 20018. 744
further r=3 RE,

 

299 5 229) =s 2004: 2447 _ 8.89736
( 4 169 = 169 169 T1690 :
x . 6. . 16
Therefore ¥ 8.144.169 6.1 2
7 S.%9736 " 3730]

and we may put v=Tord, y= 3739:
Accordingly  p=u014, 7% EI 1014 = 6642,

g= 1121}, $= 3730,
and therefore A=n12231, C=10331,

B= 2903, D'=— 10203,
and again d+ B= C+ D*
380 SUPPLEMENT

Another solution in smaller numbers.

In the second of the papers quoted Euler says that, while investigating
quite different matters, he accidentally came across four much smaller
numbers satisfying the conditions, namely,

4 =542, B=103, C=359, D = 514,
which are such that 4% + B= C* + D4

He then develops two methods of analysis leading to this particular
solution ; but, while they illustrate the extraordinary ingenuity which he
brought to bear on such problems, they are perhaps of less general interest
than the above.
INDE,

[The references are to pages.]

| I. CREEPER.

ddvvaros, “impossible,” 53

( = “undescribed ” apparently),
Egyptian name for certain powers, 4I

adptoros, indeterminate: mwAHfos movddwy

abpiworor, an undetermined number of

dAoyos

units=the unknown, dpfués, i.e. x, 32,

115, 130; év 7) doploT, indeterminately,

or in terms of an unknown; 177
dpufunTicy distinguished from Aoyiorik), 4
apifués, number, used by Dioph. as techni-

cal term for unknown quantity (=x),
i 32, 115, 130; symbol for, 32-37, 130
apbposTéy (=1/x)-and sign for, 47, 130
dromwos, “absurd,” 53

Surf lobrys or SNonirys double-equation,
q. 2.

dvvaus, igpinre,} used for square of un-
known (=x2): distinguished from rerpd-
ywros, 37-38 ; sign for, 38, 129; TeTpamAy

_ dbvams, ‘‘quadruple-square,” Egyptian
name for eighth power, 41

duvapodvvamus, fourth power of wnkadwh
(==?) sign for, 38, 129 |

duvapuoduvapuoatéy, submultiple of durapmodv-
vaus (=1[/x*) and sign, 47, 130

duvapbkuBos, ‘‘square-cube” (=x), sign
.for, 38, 129

duvapuokvBooTov, submultiple of dvvapbdkvBos
(=1/+%) and sign, 47, 130

duvapooTéy, submultiple of durapucs (= 1%?)
and sign, 47, 130

»

€ldos, “species,” used for the different terms
in an algebraic equation, 7, 130, 131

Eewyns, “deficiency”: év é\heiyeal Twa
eldy, ‘‘any terms in deficiency,” i.e.
‘‘any negative terms,” 7, 131

évurdpyovra, ‘‘existént,” used for positive
terms, 7, 130

émdvinua (‘““flower” or
Thymaridas, 114-116

‘“bloom ”’) of

{oos, equal, abbreviation for, 47-48

kvBékuBos, ‘‘cube-cube,”’ or sixth power
of unknown’ (=x9%), and sign. for, 3%
129

xvBoxvBooréy (=1/x5) and sign, 47, 130

Kk¥Bos, cube, and symbol for cube:of un--
known, 38, 129; kUBos éfehkTbs, Egyp-
tian term for ninth power (29), 41:

kvBooréy (=1/x%) and sign, 47, 130°

Nelmew, to be wanting : parts of verb used
to express subtraction, 44; Aelmovra fi
negative terms, I 30

Netyus, “wanting,” term for stot
or negation, 130; Aelye (dat.)=minus,
sign common to this and. parts of verb
Nelmew, 41-44

NoytoTik, the science of calealatids 115 5
distinguished from dptOunriksy, 4

pelfwy 7 év Noy, (32 n., 144 7.

pépos, “part,” =an aliquot part or sub-
multiple ; uépn, “parts,” used to describe
any other proper fraction, 191 :

pnhirys dplBubs (from pfidor, an’ apple), 4
113

words, ‘‘unit,” abbreviation for, 39, 130

MoptaoTikd, supposed work by ois dni
374

poplov, or év ponte expressing division or
a fraction, 46, 47 :

uupias mpd, devrépa, 47-48
382

opom\nd7 (etn), (powers of unknown)
“with the same coefficient,” 7

w, value of, calculated by Archimedes
and Apollonius, 122

waprbrys, wapiobrnTos dywyrn, approxima-
tion to limits, 95 sq., 207, 208, 209 7.,
211

mAaopuarikéy, ‘‘formativum?”’
meaning of, 140 7.

wAevpd, side, =square root, 65 7.

wAjfos, ‘‘number’ or ‘‘multitude,” used
for “coefficient,” 64 7.

mpbTagts, proposition, 9

(Tannery),

orepeds, solid, used of a number with
three factors, 183 7.

INDEX

ocwpela povddwr, ‘“heap” or collection of
units, 37
owpbs, “heap,” 37

TerpamAf)  Olwamus, ‘‘quadruple-square,”
Egyptian term for eighth power, 41

Umapkis, ‘‘ existence,” denoting a positive
term (contrasts with Aetyus), 41

Umdpxovta (eldn), ‘‘existent” or positive
(terms), 1307.

prallrys dpfuds (from ¢uddny, a bowl), 4,
113

akurbdkiov of Apollonius, 122
wpopéros, determinate, 115

1I. ENGLISH.

Abi’l-Faraj, 1

Abw’l-Wafa al-Biizjani, 6, 19

Achmim Papyrus, 45

Addition, expressed in Dioph. by juxta-
position, 42; Bombelli’s sign for, 22;
first appearance of +, 49 7.

Ahmes, 112

Alfraganus, 20

Algebra: three stages of development,
49-51

Algebraical notation : Diophantus, 32-39,
41-44; Bombelli, 22, 38; earlier Italian
-algebraists, 38; Xylander, 38, 48; Bachet
and Fermat, 38; Vieta, 38, 39, 50 7. ;
beginnings of modern signs, 49-30 7.

Aljabr, 64

al-Karkhi, 5, 417.

al-Khuwarazmi, Muhammad b. Misa,
34, 50

Almukabala, 64

Amthor, 122

Anatolius, 2, 18

Andreas Dudicius Sbardellatus, 17, 23

Angelus Vergetius, 16

Anthology, arithmetical epigrams in, r13-
114; on Diophantus, 3; indeterminate
~ equations in, 114

Apollonius of Perga, 5, 6, 12, 18, 122

Approximations: Diophantus, 95-98; Py-
thagoreans, 117-118, 278; Archimedes,
278-279

Arabian scale of powers of unknown
compared with that of Diophantus,
40, 41

Arabic versions and commentaries, 19

Archimedes, 11, 12, 33, 278, 279, 290;
Codex Paris. of, 48; Cattle-Problem,
121-124, 279; Arenarius, 35, 122

Arenarius of Archimedes, 35, 122

Arithmetica of Diophantus: different titles
by which known, 4-35; lost Books, 5-12;
division into Books, 5, 17-18; notation
in, 32-53; conspectus of problems in,
260-266

Arithmetical progression, summation of,
248-249

Ars rei et census, 20

Aryabhata, 281

Auria, Joseph, 15, 18

Bachet, 12, 16, 17, 21, 22 22., 25, 26-29, 35,
38, 45, 48, 80-82, 87 »., ror, 107 22.,
109, IIO, 140 7., 173 #., 196-197 7.,
213 #., 220 72., 230 7., 232 7., 234-235 7.,
246, 271, 273, 287, 293

¢¢ Back-reckoning,” 56, 89, 93

Baillet, 45 7.

Bessarion, Cardinal, 17, 20

Bhiaskara, 281 3

Bianchini, 20

Billy, Jacques de, 28, 165 7., 166 7., 184 n.,
221 n., 267, 304, 308, 320, 321, 326
INDEX

Bodleian MSS. of Dioph., 15, 34, 35;
MS. of Euclid, 35

Bombelli, Rafael, 11, 27; Algebra of,
21, 22; symbols used by, 22, 38

Brahmagupta, 281

Brancker, Thomas, 286 7.

Brouncker, William, Viscount, 286, 288

Camerarius, Joachim, 21

Cantor, Moritz, 3 #., 6, 63 2.,
120 #., 125 »n., 281

Cardano, 21, 23, 40

Cattle-Problem of Archimedes, 11, 12,
121-124, 279

Cauchy, 188 7., 274

Censo, or Zensus, =square, 40, 41

Charmides, scholiast to, 111, 113, 121-
122

Chasles, 11

Cleonides, 16 7.

Coefficient, expressed by mA7fos, multitude,
39, 647.

Colebrooke, 6, 281 7.

Cosa, =the unknown, 22, 40

Coss,” 23

Cossali, 1,
2207.

Cracow MS. of Dioph., 5#., 14, 18

Cube: Vieta’s formulae for transforming
the sum of two cubes into a difference
of two cubes and wice wersd, 101-103;
Fermat’s extensions, 762d. ; a cube cannot
be the sum of two cubes, 144 7. ; Euler's
solution of problem of finding all sets of
three cubes having a cube for their sum,
329-334; sign for cube of unknown or
8%, 38, 120

¢¢ Cube-cube ” (=sixth power of unknown,
or x8), sign for, 38, 129

Cubic equation, simple case of, 66-67,
242

Cuttaca (‘‘pulveriser”), Indian method of,
283

112, 118 n.,

91 t.; 40, 41, 140 2,

Definitions of Diophantus, 32, 38, 39,
129-131

‘“ Denominator,” 137

Descartes, 271, 273; notation, 5072.

Determinate equations : of first and second
degrees, 58; pure, 58-59; mixed quad-
ratics, 59-65; simultaneous equations
leading to quadratics, 66; a particular
cubic, 66-67

‘‘ Diagonal-” numbers, rry, 118, 310

»

383

Dionysius, 2 7., 9, 129

Diophantus: spelling of name, 1; date,
1-2; epigram on, 3; works, 3-13; in
Arabia, 5-6, 19; *‘Pseudepigraphus,”
12, 31; MSS. of, 14-18; commentators
and editors, 18-31 ; notation of, 32-53 ;
methods of solution, 54-98; porisms
of, 3, 8-10, 99-101 ; other assumptions,
103 5qq. ; theorems in theory of numbers,
105-110; on numbers which are the
sum of two squares, 105-106; numbers
which are not the sum of two squares,
107-108; numbers not sum of three
squares, 108-109; numbers as sums of
four squares, 110;  Dioph. not inventor
of algebra, 111-116; nor of indeter-
minate analysis, 115-124; his work
a collection in best sense, 124; his ex-
tensions of theory of polygonal numbers,
127

Division, how represented by Dioph.,
44-47

Doppelmayer, 20 7.

Double-equations (for making two ex-
pressions in x simultaneously squares),
11, 73-87, 91-92; two expressions of
first degree, 73-80, 80-82 7.; two ex-
pressions of second degree or one of first
and one of second, 81-87 ; general rule
for solving, 73, 146; double equations
for making one expression a square and
another a cube, 91-92

Dudicius Sbardellatus, Andreas, 17, 25

Egyptians: %au, sign for, 37; names for
successive powers, 41 ; beginnings of alge-
bra, Aau-calculations, 111-112; method
of writing fractions, 112

Eisenlohr, 112 7.

Enestrom, 63 7., 286 7.

Epanthema of Thymaridas, 114-116

Epigrams, arithmetical, in Anthology, 113—
114; on Diophantus, 3; one in Dio-
phantus (V. 30), 124

Equality: abbreviation for, 47-48; sign in
Xylander, 48 ; the sign = due to Recorde,
50 7.

Equations, see Determinate, Indeterminate,
Double, Triple, etc.

Eratosthenes, 121

Euclid, 8, r1, 12, 19, 63, 11%, 124, 132 7.,
144 7., 191

Eudoxus, 124

Euler, 56, 71-72 n., 83-85 n., 86 #n., go 7.,
384

1007., 102 #., 107, 110,. 145 #.; 151 7.,

. 160 n., 162 n.,. 158 #., 181-132 ».,
188 72.) 224 7.; 2306 #., 241 2., 242 12.,

208, 373, 274, 275, 286, 283-293,
294, 297, 299 7., Supplement, 329-379
passim ;

Euler, J. A., 360

Eutocius, 5, 6

Exponents, modern way of writing due
to Descartes, zo 7. :

Fakhri, 5, 41 n.
‘False supposition,” use of, in Egypt,

I12—-113 its
Fermat;, 28, 29, 30, 38, 73, go, 101,
: 102, 103, 106, 107, 108, 109, 110,

© 144-145 n., 163 n., 173 #., 179-180 7.,
- 182 ni, -183 n., 184 n., 188 n., 190—
1.10%: %., 197 #:, 202 R., 204 #., 205 7%.,
213-214 #., 218 m., 220 n., 223%.
“229 7.,.-230 My 231 7., 232 7.,.233 N.,
235 72., 236 #., 239 72., 240 7.5 241 7.,
242 7., 246, 254 n.,- Supplement, 267-
-- 328 passim; 364; ‘‘great theorem of
+ Fermat,” 144-145 7n.; Fermat on num-
bers which are, or are not, the sums of
. two, three, or four squares respectively,
106-110,-267-275 ; on numbers of form
~x?— ay? or. 2a%-3% 276-277, of form
x24 332% 275, and of form a2+ 532 276,
277; on equation x2 — 4y?=1, 285-287 ;
af —y4=2% cannot be solved in integers,
224, 293-297; problems on right-angled
triangles, 204-205 72. , 218-219 7., 220 72.,
229 72., 230 72., 231-233 7., 235 #., 2306 n.,
239-240 7.; 207-318 ; Fermat’s “ triple-
equations,” 321-328
Fractions : representation of, in Diophantus,
44-47; sign for 3, 45; for }, 45;
sign for submultiple, 45-47
Frénicle, 102 n.,. 276, 277, 235, 237,
295-297," 309, 3I0, 313, 314

=

Gardthausen, 3s, 36

Geminus, 4

‘Georg v. Peurbach, 20

Georgius Pachymeres, 18, 19, 31, 37

Girard, Albert, 30, 106 7.

Gnomons, 125 :

Gollob, 14, 18

Grammatens (Schreiber), Henricus, 49 7.,
50 71.

Greater and less, signs for, so 7.

_Giinther, 6, 2787., 279 7.

INDEX

Hankel, 6, 54-55, 108 72., 281, 283, 284,
286 7.

Harriot, 50 7.

Hau (=‘heap”), the Egyptian unknown
quantity, 37, 112

Heiberg, 35, 48 7., 118, 205 7.

Henry, C., 13, 23 7.

Hérigone, so 7.

Heron, 12, 13, 35 36, 43, 44: 45, 63,
129 72,

Hippocrates of Chios, 63, 124

Holzmann, Wilhelm, see Xylander

Houltseh, 27, 3, 4, 9, 10, IT, 12, 19%,
35,36, 37, 47 7., 63 7., 118 #., 122, 253 7.

Hydruntinus, Ioannes, 16

Hypatia, 5, 6, 14, 18

Hypsicles, 2; on polygonal
125-126, 252, 253

numbers,

Tamblichus, 2, 3, 37 7., 49, 50, 115-116, 126

Ibn abi Usaibi‘a, 19

Ibn al-Haitham, 19

Identical formulae in Diophantus, 104, 105

Indeterminate equations: single, of second
degree, 67-73 ; of higher degrees, 87-

91 ; how to find fresh solutions when one
is known, 68-70; double-equations for
making two expressions simultaneously
squares, 11, (1) two expressions of first
degree, 73-80, 80-82 7., (2) two of second
degree, or one of second and one of first,
81-87; double-equations for making one
expression a square and another a cube,
91-92 ; rule for solving double-equations
in which two expressions are to be made
squares, 73, 146 ; indeterminate equations
in Anthology, 114; other Greek ex-
amples, 118-121; 2x2-p2= 1 solved
by Pythagoreans, 117-118, 278, 310

‘‘ Indian method,” 12-13, 217.

Indian solution of x%2- A4y?=1, 281-283,
200, 292 :

Inventum Novum of J. de Billy, 28,
165 #., 184 n., 198 n., 204 %., 205 n.,
221 #., 230 72.; 231 n., 239 n., ‘Supple-
ment, 265-328 passim

Toannes Hydruntinus, 16

Ishaq b. Yinis, 19

Italian scale of powers, 40, 41

Jacobi, 108 #., 288.

Ka'b6, Arabic term for cube of unknown,
40.72.
INDEX

al-Karkhi, 5, 41 7.
Kausler, 31
Kaye, G. R., 281
Kenyon, 45

Konen, 278 #., 279 #., 281 »., 2835 #.,
. 286 2., 288, 2927.

Kronecker, 288

Krumbiegel, 122

Kummer, 1457.

Lagrange, 72 n., 110, 188 n., 272, 273,
274, 275, 276, 277, 285, 287, 288,
2Q0, 292, 299, 300

Lato, “side,” 40n.

Legendre, 1077., 109 n., 1887., 273

Lehmann, 35

Lejeune-Dirichlet, r457., 288

Leon, 124

Leonardo of Pisa, 11, 41 2., 120

Less ‘and greater, signs for, 507.

Limits : method of, 57, 94, 95; approxi-
mation to, 95-98

Logistica speciosa and: Logistica numerosa
distinguished by Vieta, 49

Loria, 62 7., 157 7., 163 m., 175 #:;
176 n., 195 n., 197 n., 207 7., 240 7.,
241 7.

Lousada, Abigail, 31
Luca Paciuolo, 21, 40

Madrid MS., 14, 15, 16

Mal, Arabic term for square, 471 7.

Manuscripts of Diophantus, r4—18

Maximus Planudes, 13, 14, 19, 2I, 3I,

43, 44; 45> 46, 48 ;

Measurement of a circle (Archimedes),
122

Mendoza, 17

Metrica of Heron, 43, 44, 45, 63, 129 72. }
= MS. of, 113

Metrodorus, 5, 113 LRT

Minus, Diophantus’ sign for, 41-44,
130; same sign in Heron's Metrica,

43, 44; Bombelli’s abbreviation, 22;
modern sign for, 49 #.; Vieta’s
sign for difference between (=for ~),
5072.

Montchall, Carl v., 18

Montucla, 28

Moriastica of Diophantus, 3—4

Muhammad b. Misa al-Khuwarazmi, 34,
50

Multiplication, signs for, 50 7.

Murr, Ch. . Th. v., 20%.

385

Negative quantities not recognised by
Diophantus as real, 52-53

Nesselmann, 6-10, 21 #., 25, 206n., 29,
33, 34 49-51, 55-58, 67, 87, 89, 93,
108 #., 140 72.y 173 7. 501 72 707 29
252 #,, 3290 7. -

Nicomachus, 2, 126, 127

Notation, algebraic: three stages, 49-51;
Diophantus’ notation, 32-49, 5I-52

Numbers which are the sum of two squares,
105-107, 268-271; numbers which are
not, ro7-1o8, 271-272; numbers which
are the sum of three squares, 272-273 ;
numbers which are not, ro8-rog, 273}
numbers not square are the sum of two,
three or four squares, 110, 273, 274;
‘corresponding theorem for triangles;
pentagons, etc., 188, 273

Numerus, numero, term for unknown quan-
tity, 38, 40

Nufiez, 23

Oughtred, 50 7.
Ozanam, 288

Pachymeres, Georgina, 13, 10, 31, 3

Paciuolo, Luca, 21, 40

Pappus, 11, 13. .

Papyrus Rhind,
6619, 112

Paris MSS. of Diophanius, 15, 106, 13".

Pazzi, A. M.,,

Pell, John, 31, i 288

‘“Pellian” equation, origin of this er-
roneous term, :286 2

Peurbach, G. von, 20 ; Ie in

Philippus of Opus on polygonal ph
125

Planudes, Maximus, 13, 14, I9, 2I, 3%,
43s 44> 45 40, 48

Plato, 4; 38»., 111, 113, 116, 125

Plus, signs for, 22, 49 7.; expressed in
Diophantus by i 3g

‘Plutarch, 127 ?

Polygonal Numbers, treatise on, 3, 11-12,
247-259; sketch of history of subject,
124-127 ; began with Pythagoreans, 124-
125; figured by arrangement of dots,
125; Hypsicles on, 125-126, 28% i308 5
Diophantus!. extensions, 127

Porisms of Diophantus, 3 8-10, 99-roI,
201, 202, 2014

Poselger, 30, 98 Ha

Powers of unknown quantity and signs

112; Berlin i
386

for, 37-39, 129; Italian and Arabian
scale (multiplicative) contrasted with Dio-
phantine (additive), 40-41; Egyptian
scale, 41 :

Proclus, 4 72., 113, 116, 117 72., 118, 242 7.

Psellus, 2, 14, 18, 41, III

Ptolemy, 18, 44

Pythagoreans: 3; on rational right-angled
triangles, 116, 2427.; on polygonal
numbers, 124-125; on indeterminate
equation 2x2-yp%= x1, 117-118, 278,
310

Quadratic equations in Diophantus, 7-8,
59-66; in Hippocrates of Chios, 63;
in Euclid, 63; in Heron, 63, 64

Quadratic inequalities in Diophantus,
60-63 ; limits to roots, 60-63, 65, 95

Qusta b. Liiqa, 19

Radice (=x), 40

Radix (=x), 38

Rahn, 507., 286 7.

Ramus, 16 7.

Rationality, Diophantus’ view of, 52-53

Recorde, Robert, 50 7.

Regiomontanus, 5, 17, 20, 23, 49

¢¢ Regula falsi” in Egypt, 112-113

Relato, Italian term for certain powers
of unknown, 41

Res, alternative for radix, in sense of
unknown quantity, 38

Rhind Papyrus, 112

Right-angled triangles in rational numbers
in Diophantus, 93-94, 105-106 ; method
of “forming,” 93-94; other methods of
forming attributed to Pythagoras, r16-
117, and to Plato, 116-117; Euclid’s for-
mula for, 117, 120; Pythagorean formula
once used by Diophantus, 242; Greek
indeterminate problems on, other than
those of Dioph., 119-121; Fermat's
theorems and problems on, 204-205 #.,
218-219 7., 220 7., 229 72., 230 72., 231—
232 7., 235 7., 236 72., 239-240 7., 293—
318, 364-371

Rodet, 34, 35

Rosen, 50

Rudio, 63 7.

Rudolff, Christoff, 23, 50 7.

Salmasius, Claudius, 17
Sand-reckoner of Archimedes, 122
Saunderson, N., 27 7.

INDEX

Schaewen, P. v., 327, 328

Schmeisser, 31

Schone, 43, 45, 118

Schreiber, H., see Grammateus

Schuler, Wolfgang, 24

Schulz, 9, 11, 18, 30, 31, 108 7., 1407.,
219 72.

Sebastian Theodoric, 24

Serenus, 12

“Side ” =square root, 63 7.

‘¢Side-” and * diagonal-” numbers, Py-
thagorean solution of 2x%-j3%?==1 by
means of, 117-118, 278, 310

Simon Simonius Lucensis, 25

Simplicius, 63 7.

Sirmondus, J., 27

Smith, H. J. S., 292

“ Species” (etdn) of algebraical quantities,
7, 130, 131

Speusippus on polygonal numbers, 125

% Square-cube” (=a), sign for, 38,
129

Square root, sign A/ for, 507.; =mhevpd
(side), 65 7.

‘¢ Square-square ” (=x%), sign for, 38,
129

Squares : numbers as sum of two, three,
or four, 110, 273, 274; of two, 105-107,
268-271; not of two, 107-108, 271-272;
of three, 272-273; not of three, 108,
109, 273

Stevin, Simon, 29, 30 7.

Stifel, M., 23, 497., 507.

Submultiples, sign for, 45-47; decom-
position of fractions into, 46, 112;
submultiples of unknown and powers,
47

Subtraction, symbol for, 41-44

Suidas, 1, 18, 22

Surdesolides, sursolida ox supersolida, 41

Surds, 23-24

Suter, H., 19.

Tannery, P., 2 2., 3, 5, 6, 8, 10-12, 14-19,
25, 28 72. 31, 32-37, 43-44 45, 108 72.,
yIL, 113 #., v25%., 138 nu, 138 2,
144 1., 148 n., 150 n., 156 n., 160 2.,
198 n., 219 n., 234 n., 256, 278, 279,
280, 281, 290, 308

Zanto, unknown quantity, in Bombelli, 22

Tartaglia, 21, 40

Theaetetus, 124

Theon of Alexandria, 2, 18

Theon of Smyrna, 2, 36, 117, 126, 310
INDEX

Theudius, 124

Thompson, D’Arcy W., 37

Thymaridas, Zpanthema of, 114-116

‘ Triple-equations”’ of Fermat, 163 7.,
179#., 182 n., 202:., 223 7., 224 2.,
246, 321-328

“Units ” (uorddes) =absolute term, 39-40;
abbreviation for, 39, 130

Unknown quantity (=x), called in Dio-
phantus dptfués, ‘‘ number’: definition

. of, 32, 115, 130; symbol for, 32-37,
130; signs for powers of, 38, 129;
signs for submultiples of unknown and
powers, 47, 130; Italian-Arabian and
Diophantine scales of powers, 40, 4I;
Egyptian scale, 41; x first used by
Descartes, 50 #7.; other signs for, 1 used
by Bombelli, 22, 38, V (for Numerus)
by Xylander, Bachet, Fermat and others,
38, R (Radix or Res), 38, Radice, Lato,
Cosa, 40 n.

Vacca, G., 1067.

Valla, Georgius, 48

Vatican MSS. of Diophantus, 5 7., 15, 16,
17

Vergetius, Angelus, 16

387

Vieta, 27, 38-39, 49, 507.
214 n., 285, 329, 331
Vossius, 31

101, 102,

Wallis, 40 7., 286, 287, 288, 289

Weber, Heinrich, 3 7.

Weber and Wellstein, 107 7., 145 7.

Wertheim, 30, 110 22., 137 #., 138 ».,
143 #., 151 n., 101 n., 200 n., 211 77.,
212 1., 216 %., 217 n., 254 %., 250, 257,
286 7., 294, 295

Westermann, 125 72.

Widman, 49 7.

Wieferich, 143 7.

Woepcke, 5 7.

“ Wurm’s problem,” 123

x for unknown quantity, originated with
Descartes, 50 7. .

Xylander, 17, 22-26, 27, 28, 29, 35, 38,
107-108 7., 1407. ; Xylander’s MS. of
Diophantus, 17, 25, 30

Zensus (= Censo), term for square of un-
known quantity, 38

Zetetica of Vieta, 2%, 101, 285

Zeuthen, 118-121, 205 7., 278,
200, 204-295

281 7.,

 

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