Number Theory

Number Theory An approach through history from Hammurapi to Legendre

AndrC Weil

Reprint of the 1984 Edition Birkhauser Boston Base1 Berlin

Andr6 Weil (deceased) Institute for Advanced Study Princeton, NJ 08540

U.S.A.

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Originally published

as a monograph

Cover design by Alex Gerasev. Mathematics Subject Classification (2000): 00A05, OlAxx, 11-03, 1I-XX

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Fermat and His Correspondents

105

abundant (and indeed superabundant) material collected by Billy in the Inventum Novum ( = Fe.III.325-398). In modern terms, the "ascent" is nothing else than a method for deriving new solutions, and in most cases infinitely many, from a few "visible" solutions for the equations of a curve of genus I. What was new here was of course not the principle of the method; it had been applied quite systematically by Diophantus (cf. Chap.1, §X), and, as such, is referred to, by Fermat (1oc.cit.) as well as by Billy, as "metho d vulgaris", ~ i.e., as the traditional method (cf. also Bachet's prolix commentary to Di0ph.VI.24~).T h e novelty consisted in the vastly extended use which Fermat made of it, giving him at least a partial equivalent of what we would obtain by the systematic use of the group-theoretical properties of the rational points on a standard cubic (cf. infra, Appendix 111). Obviously Fermat was quite proud of it; writing for himself in the margin of his Dwfdzantus(Fe.1.334,Obs.XLIII), he calls it "nostra inventio", and again, writing to Billy: "it has astonished the greatest experts" ("C'est une de mes inventions qui a quelquesfois estonne' les plus pan& maistres, et particulierement Monsieur Frpnicle . . ."; Fe.11.437). T h e record shows that Fermat treated in this manner the following types of problems, most of which occur already in Diophantus: (I) "Double equations" of the form AX('

+ BX + C = u2,

A'X~

+ B'x + C' = u2,

provided either A and A', or C and C', are squares; it may be assumed that the left-hand sides have no common zero on the projective straight line, since otherwise this would define a curve of genus 0 (for these, cf. 9XIV above). For our purposes, we will regard the curve as embedded in projective space, the homogeneous coordinates being (1, x, u, v). If A = a', A' = af2,we have four rational points at infinity

Similarly, if C = c', C' = cI2, we have four "trivial" solutions

Appendix I11 Fermat's "Double Equations" as Space Quartics

In our discussion of the "double equations" studied by Fermat and Euler and of the infinite descent as practised by them, we shall need some elementary facts about space quartics of genus 1. Let R be such a curve; as we have seen in Appendix 11, it can be defined by a pair of equations Q, = = 0, where @ = 0 and 9 = 0 are the equations of two quadrics in P3. For any 5,call Q5 the quadric given by Q, - @ = 0; these quadrics make up the pencil of quadrics through 0 (including of course the quadric 9 = 0, which may be written Q=). For any quadratic form 4

(a =

C .,xixi i,j= 1

we shall write det(@) = det(aV).A change of homogeneous coordinates with the determinant 6 multiplies det(Q,) with iS2; if Q, is multiplied with a constant A, det(@)is multiplied with h4; therefore det(@) is well defined up to a square factor when the quadric cD = 0 is given. For @, V as above we shall write F(5) = det(@-(9). For F(5) = 0, Qe is a cone; if 12 is a space quartic this cone cannot degenerate into a pair of planes. Clearly the "pencil" {Q6}must contain at least one cone; one may assume this to be Q,,; then 0 is a root of F; by taking coordinates so that @ = X: - 2X1X2, it is easily seen that 0 cannot be a double root of F unless

136 Number Theory

Chap.11, App.111

the quadric 9 = 0 goes through the vertex (0, 0, 0, 1) of the cone @ = 0, in which case R is of genus 0 and not 1. Thus F is of degree 4 or 3 and has four distinct zeros on the projective straight line; in other words, the pencil of quadrics through R contains exactly four cones. Now we will sketch a proof for the following result:

Proposition. There is a rational mapping o from R x R to the curue F(6) = v2 such that: (i) if M, M', N, N' are four points on $2, then o(M, M') = o(N, N') i f and only i f M + M' N + N'; (ii) if A is a rational point on $2, then the mapping M -,o(A, M) is an isomorphism of R onto the curve F(f1 = y2. For convenience we begin by recalling some well-known facts:

-

Lemma. A quadric @ = 0 (not a pair of planes) carries one family of straight lines or two according as det(@)is 0 or not; i f it contains a rational straight line, then det(@)is a square. If det(@)= 0, the quadric is a cone and the above assertions are obvious. If det(@) # 0, then, after suitably extending the groundfield, one can assume that the quadric contains a rational point A ; take coordinates so that A = (0,0, 0, 1) and that the tangent plane at A is X3 = 0; after possibly another rational change of coordinates, one can write, up to a constant factor: here det(@) = d # 0. Then A lies on the two straight lines XI = +x&, X3 = 0. In particular, if we had assumed that the quadric contains a rational straight line, we could have chosen for A a rational point on it and proceeded as above; then must be rational, and det(@)must be a square. In any case, the two straight lines through A belong respectively to the two families

a

Clearly each point of the quadric lies on one and only one

Space Quartics

137

line in each one of these families; this implies that all straight lines lying on the quadric belong to one or the other of these families. Now, going back to the quartic 0 defined by Q, = V = 0, take two points M, M' on R; call the straight line through M and M' (or the tangent to ll at M, if M = M'). If the quadric Q, - [V = 0 contains one point of AM,, beside M and M', it must contain AM,, ;thus Q,/Vhas a constant value on AMMt,for which we write ,$(MyMI); clearly this is a rational function of the coordinates of M and M', with coefficients in the groundfield K. In view of the lemma, F[t(M, M')] must be a square after the adjunction of the coordinates of M and M' to the groundfield; thus there is a rational function q(M, M') of those coordinates, such that AMMI

F[8(M, M1)l = r)(M,MI)'. Write r for the curve F(6) = q2 and o for the mapping (M, M') -,(t(M, M'), q(M7 M')) of 0 x 0 into r ; we will show that this has the properties described above. We begin with (ii). As in Appendix 11, take a groundfield K such that Q, and V have their coefficients in it; assume that there is on fl a point A with coordinates in K. For any 8, take q such that q2 = F(e).Let A be one of the straight lines through A on Qe; as shown in the proof of the lemma, it is rational over the field K(8,r)). So is therefore the point P, other than A, where A meets some quadric other than Q p ,say the quadric Q, = 0, in the pencil (QZ}.As P is then on 0 , we have A = AAP,t(A, P ) = 5,and therefore q(A, P) = +r); replacing q by -7 if necessary, we may assume that q(A, P) = q, SO that o(A, P) = ( 6 , ~ )Since . at the same time P is rational over K(6, r ) ) , this proves that P -* @(A,P) is indeed a birational mapping, and therefore an isomorphism, between cR and r. In particular, this shows that w(A, P) = w(A, P') implies P = P', a conclusion which remains valid even if A is not rational, since we can always enlarge K so as to make it such. Now take M, M', P on 0 ; put o(M, M') = (t, 7). Call P'

138 Number Theory

Chap.11, App.111

the fourth point of intersection of fi with the plane ll through M, M', P; put u = M + M' + P + P'. As AM,* and App. lie in n, they intersect at a point R. The quadric Qc contains AMM,,hence R, hence Apy ; therefore we have [(P, P') = 6. Let N, N' be two more points on a;we have N N' M M' if and only if N N' + P P' a; as has been shown in Appendix 11, this is so if and only if N, N', P, P' lie in a plane; if so, the above proof shows that we have ((N, N') = [(P, P') = 6, and therefore q(N, N') = + q. Now, M and M' being kept fixed, take N variable; as q(N, N') is given rationally in terms of N, N', and as 17 is constant, q(N, N') must be independent of N. Taking N = M, we get N' = Mr. Therefore q(N, N') = 7,and o(N, N') = o(M, M'), as asserted. Conversely, assume that w(N, N') = w(M, MI), and take N" such that N + N" M + M'. Then we have

+

+

+ -

+

-

-

o(N, N") = o(M, M') = o(N, N'). As proved above, this implies N" = N', completing the proof of our proposition. Our result can be interpreted by saying that the equivalence classes of divisors of degree 2 on iR are parametrized by T. Our proposition can be applied for instance to the curves defined, in the language of Diophantus and of Fermat, by a "double equation" Since they consider only curves of that form which have in projective space at least one "visible" rational point, we may, for our present purposes, assume that this is so. Then we see that the curve is isomorphic to the one given by

Y2 = x[(b1X- b)2 - (a'X - a)(clX- c)]. Take for instance the problem for which Euler proved (cf. above, PXVI) that there are no solutions other than the obvious ones. We see that these

Space Quartics

equations define a curve isomorphic to

Y2

=

-X(X- l)(X-4).

On the other hand, take the problem, considered by Fermat (cf. 5XVI) of finding four squares in arithmetic progression. This amounts to 2y2 = x2

+ z2,

2z2 = y2

+ t2.

By the same rule as before, this curve is isomorphic to Y2 = X(X+2)(2X+ 1) which is changed into the previous equation by substituting - 4X, iY for X, Y. Thus, as Euler found out, the two problems are equivalent.

Appendix IV The Descent and Mordell's Theorem

Our purpose is now to provide some background for Fermat's "method of descent" in his treatment of elliptic curves (cf. above, 3XVI) in the light of modern work on the subject1'. As has been shown in Appendix 11, if an elliptic curve has at least one rational point, it can be exhibited as a cubic Y2 = f(x), where f is a polynomial of degree 3. The rational points on the cubic make up a group, and Fermat's method of descent has been refined by Mordell in 1922 so as to show that this group is "finitely generated"; this means that there are on the cubic finitely many rational points such that all others can be derived from these by the group operation defined in Appendix 11. The same is true if the groundfield, instead of being the field Q of rational numbers, is any algebraic number-field of finite degree over (1. Let r be the largest integer such that there are on the cubic r rational points which do not satisfy any linear relation with integral coefficients not all zero; then r is called the rank of the curve. Even now no systematic method is known for determining the rank of a given cubic, let alone a set of generators for its group of rational points. 11

Cf. e.g. J. W. S. Cassels, J . London Math. Soc. 41 (1966), pp. 193291, and A. Weil, Coll. Papers 1.47-57.

The Descent

141

For simplicity we consider the case when the roots o f f are rational integers a , p, y, the cubic I' being given by Put x = X/Z, where Z is a natural integer and X an integer prime to 2. This gives As Z is prime to the other factors in the right-hand side it must be a square, Z = T2. Traditionally one observes that the g.c.d. of any two of those factors divides a fixed integer, so that one may write X - cuZ = AU2,where A is susceptible of only a finite number of values, therefore x - a = Au2 with u = U / T , and similarly x - P = B U ~x , - y = cw2, where u, v, w are rational and B, C can take only finitely many values. If one uses only the first one of these relations, one gets a rational point (u, z=y/u) of the quartic which is one of finitely many thus attached to T. On the other hand, after properly choosing the signs of u, v , w, one can write

Here ABC has to be a square; (2) defines a space quartic S2 in the (u, v , w)-space, and (3) defines a mapping of fl into r. The original problem is thus reduced to finding rational points on finitely many such quartics. Either one of these methods is typical of the traditional descent. An alternative procedure, equivalent to the one described above but more consonant with the modern theory would be as follows. Let (x, y) be a solution of (I); we can write, in one and only one way, x - a = Au2, x - P = BV*, x - y = Cw2, with rational u, v, w and squarefree integers A, B, C (positiveor negative); equations (2),(3) are satisfied, and ABC has to be a square.

142 Number Theory

Chap.11, App.IV

For any squarefree integers A, B, C (positive or negative) such that ABC is a square, call a(A, B, C) the space quartic defined by (2). T h e first step in the descent thus consists in reducing (1) to the finding of rational points on infinitely many quartics R(A, B, C). The next step will be to reject all but finitely many of these because they can have no rational points. In homogeneous coordinates, O(A, B, C) may be regarded as defined by equations Q(A,B, C) has a rational point if and only if (4)has a solution in integers U, V, W, T without a common divisor; a necessary condition for this is obviously that the same relations, regarded as congruences modulo some integer m, should have such a solution, whatever m may be. In the language of p-adic fields, this can also be expressed by saying that (2) should have a non-trivial solution in every p-adic field Qp; to this one can add the condition that it should have a nontrivial solution in the field R = Q, of real numbers. In particular, take for p a common prime divisor of B and C; as ABC is a square, and A, B, C are squarefree, p does not divide A. Take m' = p2, and suppose that p does not divide B , - y; then the congruences imply firstly that p divides T, then that it divides U, then that it divides V and W; R(A, B, C) can then have no rational point (not even in Q,,). Thus, if LR(A, B, C) has rational points, and if we write a for the g.c.d. of B and C, a must divide /3 - y. Write now B = ab, C = ac, with b prime to c; then we must have A = bc. As B and C are squarefree, a is prime to b and c, so that 161 is the g.c.d. of A and C, and IcI is the g.c.d. of A and B; therefore b divides y - a,and c divides a - P. This leaves us with finitely many curves LR(A, B, C), out of which further arguments and the consideration of the real field may lead to reject some more; it is even possible

T h e Descent

143

to discard all except those which have points in all fields Qp and in R. In contrast, however, with classical results, going back to Legendre (cf. Chap.IV, SVI), on curves of genus 0, not a11 such quartics have rational points, nor is there any known procedure for finding one if such points exist. In order to proceed further, it will now be assumed that, among all quartics R(A, B, C), the quartics Ri = R(Ai,Bi,C,) (1 G i d n) and no others have at least one rational point. Then, if M = (x, y) is any rational point on T,there is one and only one i such that (2) and (3) are satisfied for A = Ai, B = Bi, C = Ci and for some rational values for u, v, w; when that is so, we will say that M belongs to Ri and that it corresponds to the point N = (u, v, w ) on Ri. Let R = R(A, B, C) be one of the Ri; instead of defining it by (2), we write

where we have put then R may be regarded as defined by @ = 4'' = 0. With the notations of Appendix 111, we have

For any pair of points N, N' on R, let t(N, N'), q(N, N') be defined as in Appendix 111; put

so that oomaps R x R into r. Let M = (x, y) be a point on r, belonging to R and cor-

Chap.11, App.lV

144 Number Theory

responding to a point N = (u, v, w) of fZ, so that the relations (2), (3) are satisfied. The tangent ANNto In at N goes through the point with the homogeneous coordinates (u', v', w', 0) given by Auu' = Bvv' = Cww'. Therefore e(N, N) is given by @(uf,v', w ' , 0) [(N? N) = *(,,, ,I, ,I, 0). An easy calculation gives [(N, N) = x; then we have q(N, N)2 = F(x) = 6'~BCf(x) = 6*ABCy2. As 7,according to its definition in Appendix 111, is defined only up to a factor 2 1, we may, after changing its sign if necessary, assume that q ( N , N) = 6y, so that oo(N,N) = M. If we write A for the mapping N -,oo(N,N), this shows that A is no other than the mapping of R into r defined by (3). Now choose a rational point Moon I', belonging to cR and corresponding to a point Noon R; by definition, this means wo(No,N) that X(N,) = Mo. Write p for the mapping N of R into r; the proposition in Appendix I11 shows that this is an isomorphism of LR onto I?; write p - I for its inverse; we have p(No) = A(No) = Mo. Take any rational point M on r ; put P = @-'(M),so that M = p(P). By definition, M belongs to R and corresponds to a point N on SZ if and only if (3) maps N onto M, i.e. if and only if A(N) = M = p(P), i.e., in view of the definition of X and p, if and only if oo(N,N) = oo(No,P). Again by the proposition in Appendix 111, this is so if and only if 2N No + P on SZ. Put M' = p(N); as p is an isomorphism and maps N, No, P onto M', Mo,M respectively, the relation 2N No + P on R is equivalent to 2M' Mo M. This proves that M belongs to R if and only if there is a rational point M' on I' such that 2M' Mo M, in which case M corresponds to the point p- ' ( M ' ) on (n. For the proof of Mordell's theorem, one also needs a quantitative estimate for the height of M' in terms of the

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THE HIGHER ARITHMETIC AN INTRODUCTION TO THE THEORY OF NUMBERS

Eighth edition

H. Davenport M.A., SC.D., F.R.S.

late Rouse Ball Professor of Mathematics in the University of Cambridge and Fellow of Trinity College Editing and additional material by

James H. Davenport

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521722360 © The estate of H. Davenport 2008 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2008

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paperback

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140

The Higher Arithmetic

Next consider the possibility that p and q are both odd. In this case, if we put p + q = 2P and p − q = 2Q, the numbers P and Q are relatively prime integers. One of them is even and one odd, since P + Q = p is odd. Substituting for p and q in terms of P and Q in (4), we obtain x 2P Q = 2 , z P + Q2

P 2 − Q2 y = 2 , z P + Q2

after cancelling a factor 2. The position is therefore the same as before, except that x and y are interchanged, and P and Q take the place of p and q. It follows that all solutions of x 2 + y 2 = z 2 in integers are given by the formulae (5), where m, p, q are integers, and p and q are relatively prime, and one of them is even and the other odd, apart from the possibility of interchanging x and y. These are the formulae of Euclid. The simplest solution (apart from trivial solutions with one of the unknowns zero) is x = 3, y = 4, z = 5, which arises by putting m = 1, p = 2, q = 1. The first few primitive solutions (that is, solutions with x, y, z relatively prime, and therefore m = 1) are (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (21, 20, 29), (9, 40, 41). Since the formula for z (taking m to be 1) is z = p 2 + q 2 , we can make z a perfect square by choosing p and q suitably, and so obtain a parametric solution for x 2 + y 2 = z 4 . Repetition of the process enables one to give solutions for x 2 + y 2 = z k , where k is any power of 2. Alternatively, the formulae for such an equation could be deduced from the formulae for x 2 + y 2 = z 2 by employing the identity (1) of Chapter V.

3. The equation ax 2 + by 2 = z 2 The method used above for the equation x 2 + y 2 = z 2 would also apply to the equation ax 2 + y 2 = z 2 , and would again lead to formulae for the general solution. As before, there are infinitely many primitive solutions. But the method will not apply to the more general equation ax 2 + by 2 = z 2 ,

(6)

where a and b are natural numbers, neither of which is a perfect square. Indeed, a moment’s consideration shows that such an equation may not be soluble (apart from the solution x = y = z = 0, which we shall exclude throughout). For example, the equation 2x 2 + 3y 2 = z 2

141

Some Diophantine Equations

is insoluble. For we can suppose that x, y, z have no common factor greater than 1, whence it follows in particular that neither x nor z is divisible by 3. But then the congruence 2x 2 ≡ z 2 (mod 3) is impossible, since 2 is a quadratic non-residue to the modulus 3. Similar considerations apply to the general equation (6), and give congruence conditions which must be satisfied if the equation is to be soluble. We can suppose that a and b are both square free, that is, not divisible by any square greater than 1; for the introduction of square factors into the coefficients a and b does not affect the solubility of the equation. If the equation (6) is soluble, we can divide out any common factor of x, y, z and so obtain a solution in which x, y, z have no common factor greater than 1. The equation implies the congruence ax 2 ≡ z 2 (mod b). Now x and b must be relatively prime; for if they had a prime factor in common, this prime would divide x and z, and therefore its square would divide by 2 , and since b is square free this would require the prime to divide y, which is impossible. Multiplying the congruence throughout by x 2 , where x x  ≡ 1 (mod b), we obtain a congruence of the form a ≡ α 2 (mod b),

(7)

b ≡ β 2 (mod a)

(8)

where α = x  z. Similarly

for some integer β. That is, a must be a quadratic residue (mod b), and b must be a quadratic residue (mod a). Here we are using the term quadratic residue in a more general sense than in Chapter III, since themoduli a and b are now not necessarily primes. If a and b have H.C.F. h > 1, there is another congruence besides (7) and (8) which must be soluble if the equation (6) is to be soluble. Put a = ha1 and b = hb1 , so that a1 , b1 , h are relatively prime in pairs. In any solution of (6), z must be divisible by h, so that a1 x 2 + b1 y 2 must be divisible by h. Multiplying throughout by b1 x 2 , we obtain a congruence of the form a1 b1 ≡ −γ 2 (mod h).

(9)

The fact that the congruences (7), (8), (9) must be soluble imposes restrictions on a and b which are necessary for the solubility of the equation (6). It is by no means obvious that if the congruences are soluble then the equation is soluble. We shall now prove, following Legendre, that this is in fact the case, and so shall establish that the equation (6), where a and b are square free natural numbers, is soluble if and only if the congruences (7), (8), (9) are all soluble.

155

Some Diophantine Equations

create a new and extensive theory, that of ideals in algebraic number-fields. Wiles’s proof of Fermat’s Last Theorem is actually a major step forward in the theory of elliptic curves—see later in this section. In an elementary account such as this, we must content ourselves with proving the truth of Fermat’s conjecture for some particular value of n. The simplest case to treat is n = 4, where the insolubility of the equation was proved by Fermat himself. Fermat proved, more generally, that the equation x 4 + y4 = z2

(21)

has no solution in natural numbers, and his proof is an outstanding example of his technique of ‘infinite descent’, which is simply another form of the principle of proof by induction. From any one hypothetical solution of the equation in natural numbers, Fermat derived another with a smaller value of z. Repetition of this process leads eventually to a contradiction, since a decreasing sequence of natural numbers cannot continue indefinitely. The principle is the same as that underlying Legendre’s method, described in §3, except that here it is used to prove insolubility, whereas there it was used to prove solubility. Suppose x, y, z are natural numbers which satisfy (21). We can suppose that x and y have no common factor greater than 1, for the fourth power of such a common factor can be cancelled from the equation. The numbers x 2 , y 2 , z constitute a primitive solution of X 2 +Y 2 = Z 2 , and therefore, by the result proved in §2, they are expressible (possibly after interchanging x and y) as x 2 = p 2 − q 2 , y 2 = 2 pq, z = p2 + q 2 , where p and q are relatively prime natural numbers, one of which is even and the other odd. Looking at the first equation, and recalling that any square must be congruent to 0 or 1 (mod 4), we see that p must be odd and q even. Putting q = 2r, we have x 2 = p 2 − (2r )2 , ( 12 y)2 = pr. Since p and r are relatively prime and their product is a perfect square, each of them must be a perfect square. If we put p = v 2 and r = w 2 , the first equation becomes x 2 + (2w 2 )2 = v 4 . This equation is somewhat similar to (21) in its general form. When similar reasoning is applied again to the new equation, we obtain one exactly like (21). The last equation implies that x = P 2 − Q 2 , 2w 2 = 2P Q, v 2 = P 2 + Q 2 ,

156

The Higher Arithmetic

where P and Q are relatively prime integers, one of which is even and the other odd. Since P Q = w 2 , each of P and Q must be a perfect square. Putting P = X 2 , Q = Y 2 , the third equation becomes X 4 + Y 4 = v2 , which is of the same form as (21). In this equation X, Y, v are natural numbers and v2 = p
3, X n +Y n = 1 has only finitely many rational solutions, i.e. that for fixed n, (20) has only finitely many different (without common factors) solutions. Frey suggested, in 1985, that the existence of a non-trivial solution to u p + v p = w p would imply the existence of a non-modular elliptic curve, viz. y 2 = x(x +u p )(x −v p ), now known as the Frey curve. This suggestion was proved by Ribet in 1986. This curve is semi-stable (see the previous section), and in 1993 Wiles announced a proof (subsequently found to need another key ingredient, furnished by Wiles and Taylor) that every semi-stable elliptic curve is modular, the semi-stable case of the Taniyama– Shimura–Weil conjecture. Hence no non-trivial solutions to u p + v p = w p can exist.

50

DIOPHANTINE ANALYSIS

[CH

2,6

formulae (9) we obtain a Pythagorean triangle whose perimeter length is the number x+y+z = 2m(m+n) = [2t(2t-l)]s. For s = 2 we obtain the triangle (63, 16, 65), whose perimeter length is 12 2 • It is easy to find all the Pythagorean triangles whose areas are equal to their perimeter lengths (see de Comberousse [1], pp. 190-191). The sides x, y, z of such a triangle must satisfy the equations x 2 + y2 = Z2 and x + y + z = t xy. Eliminating z we obtain the equation (25)

(x-4)(y-4)

=

8.

This implies that x - 418. We cannot have x - 4 < 0, because in the case x - 4 = - 1 or x - 4 = - 2 we would have y - 4 = - 8 or y - 4 = - 2, respectively; this. in turn, would give y = - 4 or y = 0, which is obviously impossible. But if x - 4 = - 4, or x - 4 = - 8, then x ~ 0, which is also impossible. Thus we conclude that x - 4 > 0 and therefore, by x-418, we see that x-4 = 1,2,4 or 8. whence x = 5. 6, 8 or 12. Consequently, using (25), we obtain y = 12,8,6 or 5. This leads us to the conclusion that there are precisely two non-congruent relevant triangles, namely (5,12,13) and (6,8.10). The area and the length of the perimeter of the first is 30, of the other 24. It is easy to prove that there exist infinitely many Pythagorean triangles whose sides are rational and areas equal to the lengths of their perimeters. It can be proved that all such triangles (u, v, w) are given by the formulae

u=

2 (m +n)

n

4m

V=--,

m-n

w=

where m and n < m are natural numbers. 6. On squares whose sum and difference are squares Now we consider the problem of existence of natural numbers x, y, z, t satisfying the following system of equations (26)

In other words, we are going to answer the question whether there exist two natural numbers x and y such that the sum and the difference of their squares are squares. The answer is given by the following theorem of Fermat.

CH

2,6]

ON SQUARES WHOSE SUM AND DIFFERENCE ARE SQUARES

51

THEOREM 3. There are no two natural numbers such that the sum and the difference of their squares are squares. PROOF. Suppose that there exist natural numbers x and y such that x 2 + y2 = Z2 and x 2 - y2 = t 2, where z and t are natural numbers and, of course, z > t. Among all the pairs x, y there exists a pair for which the number x 2 + y2 is the least. Let x, y denote such a pair. We must have (x, y) = 1. For if dl x and d > y with d11, then, in virtue of x 2 + y2 = Z2, x 2 - y2 = t 2 , we would have d 2 1 Z2, d 2 1 t 2 , whence d I z and d I t, but this would imply that the equation can be divided throughout by d 2 , contrary to the assumption that x, y denote the solution for which the sum x 2 + y2 is the least. It follows from (26) that 2x 2 = Z2 + t 2 • Therefore the numbers z and t are both odd or both even. Hence the numbers z +t and z - t are both even and therefore -1(z + t) and t (z - t) are natural numbers. If dlt(z+t) and dlt(z-t) and d is greater than 1, then dlz, which in virtue of (27)

implies d 2 1 x 2 and so d I x. Consequently, since xi + y2 = dIy, which is clearly impossible since (x, y) = 1. Thus (28)

(

Z2,

we also have

z + t z-t)_ 1 2 '2 -.

-- --

From (28) and (27) we infer that the numbers t (z + t), t (z - t), x form a primitive solution of the Pythagorean equation, which by Theorem 1, implies that there exist relatively prime natural numbers m, n with m > n, one of them even and the other odd, for which either

t(z-t) = m 2 _ n 2 ,

t(z+t) = 2mn

t(z+t) = m 2 _ n 2 ,

t(z-t) = 2mn

or

hold. Since 2y 2 =

2

Z2 - t ,

in either case we have

2y 2 = 2 (m 2 - n 2 ) 4mn,

whence

y2

= (m 2 - n 2 ) 4mn.

As the number y is even, y = 2k, where k is a natural number. Using the formulae for y2 we obtain

(29)

(m 2 - n 2 ) mn

= k2 •

52

[CH

DIOPHANTINE ANALYSIS

2,6

Since (m, n) = 1, we have (m± n, m) = 1, whence (m 2 - n 2, m) = 1 and (m - n 2 , n) = 1. From (29) we infer that, according to the corollary of Theorem 8 of Chapter I, each of the numbers m 2 - n 2 , m, n is the square of a natural number, thus m = a 2 , n = b 2, m 2 - n 2 = c", where a, b, e are natural numbers. From (m, n) = 1 and from the fact that one of the numbers m, n is even and the other is odd we infer that (m + n, m - n) = 1. In fact, every common divisor of the odd numbers m + nand m - n is odd, but it is also a divisor of the numbers 2m and 2n, thus, since (m, n) = 1, it equals to 1. From the equalities (m + n, m - n) = 1 and (m + n) (m - n) = m 2 - n 2 = c 2 (by the already mentioned corollary) it follows that the numbers m + nand m - n are squares. Thus, since m = a 2, n = b 2, the numbers a 2 + b 2 and a 2 - b 2 are squares. But a 2 + b 2 = m + n < 2m ~ 2mn ~ t(z+t) < z ~ Z2 = X 2 + y 2, whence a 2+b 2 < X 2 + y 2, contrary to the assumption concerning the pair x, y. Thus the assumption that there exist natural numbers for which the sum and the difference of their squares are squares leads to a contradiction. This completes the proof of Theorem 3. 0 2

On the other hand, there exist infinitely many pairs of natural numbers x, y for which there exist natural numbers z and t such that x 2 + y2 = Z2 + 1 and x 2 - y2 = t 2 + 1. For instance, if q is even then for x = y = q3 we have X2+y2 = (q2+ q4j2)2+1,

X2_ y2

4

!!..-- + 1, 2

= (q4j2_ q2)2+1. for n = 1,2, ... There

We also have (2n 2)2 ± (2n)2 = (2n 2 ± 1)2-1 exist other pairs of natural numbers x, y such that for some natural numbers z, t we have X2+y2 = z2-1, X2_ y2 = t 2-1, e.g. 21 2+12 2 = 14 2-1, 21 2 - 12 2 = 10 2 - 1. It is not difficult to see that there exist pairs of natural numbers x, y for which we can find natural numbers z, t such that X2+y2 = z2+1 and X2_ y2 = t 2-1, e.g. 13 2+11 2 = 17 2+1, 13 2 _11 2 = 7 2 -lor 89 2 + 79 2 = 119 2 + 1, 89 2 - 79 2 = 41 2 - 1. It follows from Theorem 3 that the system of equations (*) x 2 + y2 = u 2, x 2 + 2y2 = v2 has no solutions in natural numbers x, y, U, v. In fact, if for some natural numbers x, y, U, v formulae (*) hold, then u 2 + y2 = v 2, u 2 _ y2 = x 2, contrary to Theorem 3. COROLLARY 1. There

are no natural numbers a, b, c such that a 4

-

b4

= c 2.

CH

2,6]

53

ON SQUARES WHOSE SUM AND DIFFERENCE ARE SQUARES

PROOF. If the numbers a, b, c could be found, then we might assume that (a, b) = 1; for, if(a, b) = d > 1, then putting a = da., b = db, we would have d4(at-bt) = c 2, whence d 2lc, so c = d 2c 1 and therefore at-bt = d, where (aI' b 1) = 1. Thus assuming (a, b) = I, we have (a 2, b 2) = 1, whence in virtue of the equality b" + c 2 = a 4 , the numbers b", c, a 2 form a primitive solution of the Pythagorean equation. Then from Theorem 1 we infer that there exist natural numbers m, n, m > n, such that a 2 = m 2 + n 2 and either b 2 = m 2 - n 2 or b 2 = 2mn. The first case is impossible, since it contradicts Theorem 3. In the second case we have a 2 + b 2 = (m + nf and a 2 - b 2 = (m - n)2, which also contradicts Theorem 3. This completes the proof of Corollary 1. 0 It follows that there are no natural numbers for which the sum and the difference of their squares are both the k-th multiples ofsquares of natural numbers, for otherwise we would have a 4 - b 4 = (kUV)2, contrary to Corollary 1. By Corollary 1 the difference of the fourth powers of natural numbers is not the square of a natural number; the product, however, of two different differences of this kind can be the square of a natural number; for instance (3 4 - 2 4)(11 4 - 2 4) = 975 2 , (2 4 -1 4)(23 4 - 7 4) = 2040 2 ,

(54 _ 4 4) (21 4 - 20 4 ) = 3567 2 ,

(9 4 - 74) (11 4 - 2 4) = 7800 2 •

COROLLARY 2. There are no natural numbers x, y, z satisfying the equation x" + y4 = Z4 (this is the Fermat Last Theorem for the exponent 4, cf.§ 18). PROOF. If the numbers x, y, z existed, then we would have Z4 _ y4 contrary to Corollary L 0

= (X 2)2,

Corollary 2 can also be expressed by saying that there is no Pythagorean triangle whose sides are squares. K. Zarankiewicz has asked whether there exists a Pythagorean triangle whose sides are triangular numbers (i.e. numbers t; = n (n + 1)/2). The answer to this question is obtained simply by checking that the numbers t 132 = 8778, t 143 = 10296, t 164 = 13530 form a Pythagorean triangle. We do not know whether there exist any other Pythagorean triangle with this property. However, there exist infinitely many Pythagorean triangles whose catheti are consecutive triangular numbers. As a matter of fact, in § 4 we have proved that the equation x 2 + (x

54

[CH 2,6

DIOPHANTINE ANALYSIS

+ 1)2 =

has infinitely many solutions in natural numbers x, z. For each such solution x, z, we easily check thatrj; +t~x+ I = [(2x + 1) Z]2. For example we have t~ + t~ = 35 2, ti~ + t~1 = (41 . 29)2. It is known that there exist infinitely many primitive Pythagorean triangles whose catheti are triangular numbers. To this class belongs the triangle (t 7 , t 9 , 53). If for some natural numbers a, b, c we have t~ + t~ = then, as can easily be verified, we also have (2a+l)2-1)2+(2b+l)2-1)2 = ((2c +IV-l)2. Thus the equation (x 2_1)2+(y2_1)2 = (Z2_1)2 has a solution in odd natural numbers x, y, z, namely x = 263, y = 287, Z = 329. The equation has also another solution in which not all numbers x, y, Z are odd, e.g. x = 10, y = 13, Z = 14. We do not know whether this equation has infinitely many solutions in natural numbers> 1. It is easy to prove that there is no primitive Pythagorean triangle such that adding 1 to its hypotenuse we obtain the square of a natural number. In fact, the hypotenuse of a primitive Pythagorean triangle is, by theorem 1, of the form m2 + n2 , where one of the numbers m, n is even and the other is odd; consequently, dividing the number m 2 + n 2 + 1 by 4, we obtain the remainder 2, whence we infer that m 2 + n 2 + 1 cannot be the square of a natural number. It is easy to prove that the equation Z2

t;,

2

(x 2_1)+(y2_1)2 = (z2+W has infinitely many solutions in natural numbers x, y, z. This follows immediately from the identity

((2n 2 + 2n)2- 1»2 +((2n + 1)2 - 1)2 = (2n 2 + 2n)2 + 1)2 for n = 1,2, ..., which, in particular, gives (4 2 _1)2 +(3 2 _1)2 = (4 2 + 1)2, (12 2-1)2+(5 2-1)2 =(12 2+1)2, (24 2-1)2+(72-1)2 =(24 2+1)2. We note that the numbers 2 n 2 + 2n and 2n + 1 can always be regarded as the catheti of a Pythagorean triangle, for

(2n 2 + 2n)2 +(2n + 1)2 = (2n 2 + 2n + 1)2

for n = 1,2, ...

Also the equation

(x 2 _1)2 +(y2)2

= (Z2 _1)2

has infinitely many solutions in natural numbers. This follows from the identity

((8n4-1)2-1)2+((2n)6)2

=

((8n 4+1)2-1)2

Thus, in particular, (72 _1)2 +(8 2)2

= (9 2 _1)2.

for

n = 1,2, ...

2,6]

CH

55

ON SQUARES WHOSE SUM AND DIFFERENCE ARE SQUARES

However, there is no Pythagorean triangle for which by subtracting 1 from each of its catheti we would obtain the squares of natural numbers. The reason is that, as we know, in each Pythagorean triangle at least one of the catheti is divisible by 4. It c~n be proved that for each Pythagorean triangle (a, b, e) and for each natural number n there exists a triangle similar to the triangle (a, b, e) and such that each of its sides is the mth power of a natural number with m ~ n. To construct this triangle it is sufficient to multiply each of the sides of the triangle (a, b, e) by a 2 (4 n 2- 1) b 4 n(n - 1)( 2 n + 1)c 4 n2(2n-1). Using the fact that a 2 + b 2 = c2 , one easily sees that

(a 2 nb (n -

1 )(2n + 1 )e n(2n -1))2n)2

+( (a 2 n + 1 b 2 n 2- 1 C 2 n 2)2n -

= (a2n-1 b2(n Thus in particular for n

= 2, if a 2 + b 2 = e 2 ,

(a 4 b 5 c 6 )4 ) 2 +(a 5b 7 C8)3)2

=

1)2

l)n C2n 2- 2n+ 1

fn+ 1)2.

then

(a3b4e5)5)2.

It is not known whether there exist natural numbers x, y, Z, t such that + y4 + Z4 = t 4 • It is known that the equation has no solutions in natural numbers x, y, z, t with t less than 220000 (Lander, Parkin and Selfridge [IJ). It is interesting to know that 30 4 + 120 4 + 274 4 + 315 4 = 353 4 (Norrie, 1911) and 133 4 + 134 4 = 59 4 + 158 4 (Euler, 1778). We do not know whether the equation x 4 + y4 + Z4 + (4 == u 4 has infinitely many solutions in natural numbers x, y, z, t, U such that (x, y, z, t) = 1. Apart from the solution mentioned above there are precisely 81 other solutions of this equation with u ~ 20469 and (x, y, Z, t) = 1 (Rose and Brudno [IJ), e.g. 240 4+3404+4304+5994 = 651 4 (J. O. Patterson 1942). On the other hand, there exist infinitely many quadruples x, y, z, t such that (x, y, Z, t) = 1 and X 4+y4 = Z4+ t4 (cr. Lander and Parkin [IJ, Lander, Parkin and Selfridge [IJ, Zajta [IJ). We also have x4

24+24+34+44+44 = 54, 44+64+84+94+144 = 154, )4+8 4+124+324+644 = 65 4. Turning back to Corollary 1 we note that the equation x 4 - y4 = Z3 has solutions in natural numbers. In fact, for a natural number k we have

56

DIOPHANTINE ANALYSIS

[CH

2,6

Thus, in particular, for k = 2, 450 4 - 225 4 = (15 3 ) 3 . E. Swift [1] has proved that the equation x 4 - y4 = Z3 has no solutions in natural numbers x, y, z such that (x, y) = 1. COROLLARY 3. There are no three squares forming an arithmetical progression whose difference is a square. PROOF. If for natural numbers x, y, z, t the equalities y2 - x 2 = t 2 and Z2 _ y2 = t 2 were valid, then y2 _ t 2 = x 2, y2 + t 2 = Z2, contrary to Theorem 3. 0 COROLLARY 4 (Theorem of Fermat). There is no Pythagorean triangle whose area is the square of a natural number e). PROOF. Suppose, to the contrary, that such a triangle (a, b, c) exists. Then a 2 +b 2 = c 2 and ab = 2d 2, where d and c are natural numbers. Without loss of generality we may assume that a > b, since the case a = b could not possibly occur because 2a 2 = c 2 is impossible. Hence c 2 +(2d)2 = (a +b)2, c 2 -(2df = (a - b)2, contrary to Theorem 3. 0 We lea ve to the reader an easy proof of the fact that there are no two rationals, each different from zero, such that the sum and the difference of their squares are the squares of rational numbers. Similarly, it is not difficult to prove that there are no rational numbers a, b, c, all different from zero, such that a 4 - b" = c 2 • To see this we suppose, on the contrary, that such numbers a, b, c exist. We may of course assume that they are all positive. So a = 11m, b = rls, e = ulo, where I, m, r, s, u, v are natural numbers. Since a4 - b4 = e 2 , we see that (lvs)4 - (rvm)4 = (uvm 2s2)2, contrary to Corollary 1. It can easily be proved that there are no squares of rational numbers, all different from zero, which form an arithmetical progression in which the difference is the square of a rational number. It follows that there is no rational number x for which each of the numbers x, x + 1, x + 2 is the square of a rational number.

el

C. M. Walsh devoted a long paper to this theorem [IJ. The paper contains detailed historical references as well as many remarks by the author himself.