€¦ · ON THE DIOPHANTINE EQUATION xf + xf + x\ = y\ + y\ + y\ 705 Using suitable matrices Mx,...

mathematics of computationvolume 59, number 200october 1992, pages 703-715

ON THE DIOPHANTINE EQUATION xf + xf + xf = y\ + y\ + y¡

JEAN-JOËL DELORME

Abstract. In this paper, we develop an elementary method for producing para-metric solutions of the equation xf + xf + xf = yf + yf + yf by reducing theresolution of a system including it to that of the equation

(s2 + (sx + tx)2)(s¡ + (s2 + t2)2)(s] + (s3 + A3)2)

= [t\ + (SX + tx)2)(t\ + (S2 + t2)2)(t2 + (s3 + /3)2).

We give such solutions of degrees 4, 5, 7, 8, 9, and 11.

1. Introduction

We say that a solution (xi ; x2 ; X3 ; yx ; y2 ; y3) of the equation xf+xf+xf =y\ + y\ + y\ , or of any system including it, is trivial if and only if each of thetriples (x2 ; x2; x2) and (y2 ; y2;y2) may be deduced from the other by a per-mutation of its components. The first nontrivial solution, in positive integers,was given in 1934 by Subba-Rao [6], viz., 36 -I- 196 + 226 = 106 + 156 + 236.It is "the smallest" nontrivial solution, in that xf + x2 + xf is minimum. In1967, Lander, Parkin, and Selfridge [5] gave ten nontrivial solutions in integerswithout any common factor. All these solutions, except one, satisfy the addi-tional equality x2 + x\ + x\ = y\ + y\ + y\ . In 1970, Brudno [2] proved thatthe following system (B|) has an infinity of solutions:

' xf + xf + x\ = y\ + y\ + y\,(Bl) \x2 + x2 + x2^y2+y2 + y2,

y2 = x2 - X3 ,

. yi = x2+x3.

In 1974, Brudno and Kaplansky [3] obtained all the solutions of the precedingsystem. In 1976, Brudno [4] exhibited an infinity of solutions of the followingsystem (B2) :

r xf + xf + xf = yf + y\ + y\ ,(B2) < x2 + x22 + x2 = y2 + y\ + y\ ,

[ 3xx + x2 + Xi = 3yx + y2 + y3,

in the form of homogeneous polynomial functions of degree four in two pa-rameters. At the end of this paper, we mention important results obtained by

Received by the editor July 25, 1990 and, in revised form, June 28, 1991.1991 Mathematics Subject Classification. Primary 11D41.

©1992 American Mathematical Society0025-5718/92 $1.00+ $.25 per page

703

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704 JEAN-JOËL DELORME

A. Bremner in 1979. In the following, we shall exclusively consider solutions inrational numbers.

2. Preliminaries

We first reduce the resolution of a system including the equation xf + xf +xf = yf + y\ + y\ to that of a system of three quadratic equations.

Theorem 1. Consider the two systems:

(I)xf + xf + xf = y\ + y\ + y\,x\ + xf + x2 = y\ + y\ + y\,xiyi(x2 -y2x) + x2y2(x¡ -y\) + x3y3(x32 -y\) = 0,

' x¡ + xxyx -y\ + x\- x2y2 -y2 = 0,(II) I xf + x2y2 -y\+ x2 - x3y3 - y\ = 0,

. x2 + x3y3 - y] + x2 - xxyx - y\ = 0.

Then the following hold:(a) Every solution of (11) is a solution of (I).(b) If (xx ; x2 ; x3 ; yx ; y2 ; y3) ¿s a nontrivial solution of (I), then (xx ; x2 ;

x?, ; y\ ; yi ; y-i) or (xx ; x2 ; x3 ; -yx ; -y2 ; -y3) is a solution of (II).

Proof of Theorem 1. For 1 < / < 3, put xf+x¡yi-y2 = A¡ and -x2+x¡yi+y2 =B¡. It is easily verified that A¡ - B¡ = 2(xf -yf), A2 - B2 = 4x/y,(x2 - y2),and A] - B3 = 2(xf - xf), so that (I) is equivalent to the following system:

(AX+A2 + A3 = BX+B2 + B3,

A2 + A2 + A2 = B2 + B2 + B2 ,

A3 + A3 + A3 = B3 + B\ + B\ ,and (II) is equivalent to {Ax = Bx ; A2 = B2 ; A3 = B3}. So every solution of(II) is obviously a solution of (I).

Further, if (xx ; x2 ; x3 ; yx ; y2 ; y3) is a nontrivial solution of (I), we eas-ily deduce from (Ibls) that (Ax ; A2; A3) is equal to one of the six triples(Bx; B2; B3), (Bx; 53 ; B2), (B2; Bx; Bt) , (B2; Ä3 ; Bx), (B3; Bx; B2), or(£3; R2', Bi) ■

Now suppose that for some integer i satisfying 1 < /' < 3, we have A¡ — B¡.For instance, assume that Ax = Bx. We then have x\ + xxyx - y\ = -x,2 +x\y\ + y\ > hence x\ = y\. From the first two equations of (I), we then deducethat xf + xf = y\ + y\ and x\ + x2 = y\ + y\ , so that (xf ; xf) = (y\ ; y\) or(xf ; xf) = (y\ ; y2). Thus the solution (xi ; x2 ; x3 ; yx ; y2 ; y3) is trivial.

Finally, (Ax ; A2 ; A3) is necessarily equal either to (B2 ; 53 ; Bx) or to (Bt, ;Bx ; B2), so that (xx ;x2;x3;yx;y2 ; y3) or (x, ; x2 ; x3 ; -yx ; -y2 ; -y3) is asolution of (II). D

3. Link matrices

Notation. Let (xi ; x2 ; x3 ; yx ; y2 ; y3) be a sextuple of rational numbers. For1 < i < 3 , put Vi = (yi). In the following, we shall write (xx; x2; xy, yx; y2; y3)or (Vx ; V2 ; V3) interchangeably.


ON THE DIOPHANTINE EQUATION xf + xf + x\ = y\ + y\ + y\ 705

Using suitable matrices Mx, M2, and M3 such that Mx Vx = V2, M2 V2 =V3, and M3 V3 = Vx, we reduce the resolution of system (II) to that of a singlefactorized equation.

Definition. 1. We call every 2 by 2 matrix of rational entries of the form(a~¿b a^_b) with determinant -1 a link matrix.

2. We call every triple of link matrices, (Mx ; M2; M3), such that the traceof the matrix M3M2MX is zero, a solving triple.

Theorem 2. (a) Let (Vx; V2; V3) be a nonzero solution of (II). There exists aunique solving triple such that MXVX = V2, M2V2 = V3, and M3V3 = Vx.

(b) Let (Mi ; M2; M3) be a solving triple. There exists a nonzero solutionof (II), (Vx; V2; V3), unique apart from a nonzero rational factor, such thatMx VX = VX, M2V2 = V3, and M3V3 = Vx.

Definition. We shall say that (Mx ;M2; M3) is the solving triple associated withthe nonzero solution (Vx; V2; V3) of (II), and that (Vx; V2; V3) is an antecedentsolution of the triple (M ; M2 ; M3).

Lemma. Let (xx ; x2 ; x3 ; yx ; y2 ; y3 ; ) be a nonzero solution of (II). For (i, j)G {(1; 2), (2; 3), (3; 1)} we have.

(a) (x;2 + x2)(x,2 + Xjy¡ - yf) = (x,y/ + y;x;)(x/x;' + x,y;- - y¡Xj),(b) (yf + y2)(xf + xtyt - yf) = (xtyj + ytx^ytyj + xtyj - y¡Xj),(c) (Xj ;y¡)¿ (0 ; 0), x2 + x,y(- - y2 ¿ 0, x(2 + x) # 0, and x¡yj + y¡Xj ¿ 0.

Proof, (a) For any rational numbers xx, yx , x2, and y2, we have the identity

(x2 + xf )(x2 + xxyx - y\) - (xxy2 + yix2)(xix2 + xxy2 - yxx2)= xf (xf + xivi - y\ + xf - x2y2 - y\).

If (xi ; x2 ; x3 ; yx ; y2 ; y3) is a solution of (II), we have x2 + xxyx -y\ + x\-X2y2 - y\ = 0, from which the result follows, for (i; j) - (1 ; 2). The othertwo cases are obtained by cyclic permutations of the indices 1,2, and 3.

(b) Similar proof, with the help of the identity

[y] + yl){x\ + xiyi - y\) - {xxy2 + yxx2)(yxy2 + xxy2 - yxx2)

= vf (xf + x,yx - y2 + xf - x2y2 - yf ).(c) (i) If, for instance, (xi ; yx ) = (0 ; 0), the first equation of (II) implies that

xf-x2y2-yf = 0, so that (x2 ; y2) = (0 ; 0), and the third equation implies thatx\ + x3y3 - yf = 0, so that (x3 ; y3) = (0 ; 0). Thus, (xx ; x2 ; x3 ; y, ; y2 ; y3)would be the zero solution, contrary to our hypothesis.

(ii) If, for instance, x\+xxyx -yf = 0, then (xi ; yx) = (0 ; 0), which bringsus back to (i).

(iii) If, for instance, xf+xf = 0, then xj = 0 and x2 = 0. The first equationof (II) implies that yf + yf = 0, so that yx = 0 and y2 = 0. Therefore, wehave (Xi ; yx) = (0; 0), which brings us back to (i).

(iv) By the identity proved in (a), since x2 + x2 ^ 0 and xf + x¡y¡ -yf ^0,we have x;y; -t- y¡x¡ ^ 0. D



Proof of Theorem 2. (a) Let (Vx; V2; V3) be a nonzero solution of (II). Let ussolve the system

fax -bx ax \ ixx\ _ ix2\ bx ax-bx) \yx) \y2

for ax and bx. It may be written as (xx+yx)ax-xxbx = x2, yxax + (xx-yx)bx =y2 . Its determinant, x2+xxyx -yf, is nonzero, by (c) of the lemma. Therefore,it has a unique solution, given by

xix2 + xiy2-yix2 yiy2 + Xiy2 -yix2ax —-=-■-2—' °i — —t~\-2—•x2x+xxyx-y2 xf + x^i-yf

By (a) and (b) of the lemma, we also have

*i + X2 ¿

■xiy2 + yi*2 '

Finally,

xf + xf _ y2i+y¡ai - ~-;—TT7" ' °i

a _b xf + xf -y\-y\ = x2y2 - xxyx' ' *iy2 + yi*2 xxy2+yxx2

from the first equation of (II). Thus,

M = 1 f-*2V2 - xxyx xf + xf1 xxy2+yxx2\ y\+y\ x2y2-xxyx

It then follows from Lagrange's identity that det(M) ) = -1. Analogous proofshold for M2 and M3, by cyclic permutations of the indices 1, 2, and 3. So thereexists a unique triple (Mx ; M2; M3) of link matrices such that MXVX = V2,M2V2 = F3,and M3V3 = Vx .

Now we prove that the trace of the matrix M3M2MX is zero. It is easilyverified that for any elements of a commutative ring, ax, a2, a3, ¿^ , fr? > and¿?3, we have

Trífa3-b3 a3 \(a2-b2 a2 \ Í ax - bx ax \\\\ b3 a3-b3)\ b2 a2-b2)\ bx ax - bx ) )

= 2(axa2a3 - bxb2b3).

It follows that Tr(M3M2Mx) = 2%, with

N = (x2 + xf )(xf + xf)(xf + xf) - (y2 + yf)(yf + y2)(yf + y2),D = (xiy2 + yix2)(x2y3 + y2x3)(x3y, + y3x, ).

N = i((xf + xf + xf)3 - (xf + xf + xf) - (yf + yf + yf)3 + (yf + yf + yf)).We may write

i3

Now every solution of (II) is, by Theorem 1, a solution of (I), so that xf + xf +xf = yf + yf + yf and xf + xf + xf = yf + yf + y] . It follows that N - 0, sothat Yr(M3M2Mx) = 0. Thus, (Mx ; M2; M3) is indeed a solving triple.

(b) Let (Mx ; M2 ; M3) be a solving triple. We have det(M3M2Mx) = -1,because det(A//) = -1 for 1 < / < 3, and Yr(M3M2Mx) = 0. Hence, thematrix M3M2MX has the eigenvalues 1 and -1 . Let Vx be an eigenvector ofM3M2MX , with rational coordinates, associated with the eigenvalue 1. DefineV2 and V3 by MXVX = V2 and M2V2 = V3. We have M3V3 = M3M2V2 =


ON THE DIOPHANTINE EQUATION x\ + x\ + x\ = y\ + y\ + y\ 707

M3M2MXVX = Vx . We now prove that (Vx; V2; V3) is a solution of (II). Wehave

/x2\ = iax -bx ax \ (xx\\y2) \ bx ax-bx) \yx ) '

It follows that

x\ - x2y2 - yf = det(vW! )(xf + Xiy, - yf ).Now det(Mx) — -1 . We obtain xf+xiyj-yf+xf-x2y2-yf = 0, which is thefirst equality of (II). The other equalities are obtained by cyclic permutations ofthe indices 1, 2, and 3. Therefore, (Vx, V2, V3) is a solution of (II), nonzero,since Vx is nonzero. □

Remark. If (Mx ; M2 ; M3) is a solving triple, so is (Mx ; M2 ; -M3).We say that two nonzero solutions of (II) are conjugate if and only if, one

being an antecedent solution of a solving triple (Mx ; M2; M3), the other isan antecedent solution of the solving triple (Mx ; M2 ; -M3). If we considertwo such solutions, (Vx ; V2; V3) and (V(; V2; V3), then Vx and V[ are twoeigenvectors of the matrix M3M2MX associated with the eigenvalues 1 and -1,respectively. We shall use this remark in §4.

Theorem 3. The link matrices are matrices of the form ¿('vpr x-y) • w^tnX = s2 + (s + t)2, Y = t2 + (s + t)2, and Z = s2 + 3st + t2, where (s ; t) is anarbitrary nonzero pair of rational numbers.Proof. Every matrix M of the type given in Theorem 3 is of the form (a¿é f ¿),with a = y and b = £ ; it is easily verified that its determinant is -1. So Mis a link matrix.

Conversely, let M be a link matrix. We have M = (a~bb a^b), and thedeterminant of M is -1, so that a2 - 3ab + b2 = -1. Consider the followingsystem in 5 and t :

(a-b- l)s + (2a-b)t = 0,(a-2b)s + (a-b + l)t = 0.

Its determinant, equal to -a2 + 3ab - b2 - 1, is zero. Hence this systemhas nonzero solutions in rational numbers. Let (s ; t) be one of them. Putco = ( 1 + \/5)/2 and œ = ( 1 - \/5)/2. Then (to ;W) is a basis of the Q-vectorspace [\/5], so that the preceding system is equivalent to the equality

[(a-b-l)s + (2a - b)t]œ + [(a - 2b)s + (a-b+ l)t]W = 0.We may write this in the form (aw + bW)(tw - sœ) = seo - tW. Now (s; t) ^(0 ; 0), hence tco - sœ ^ 0, and sco - tcô ̂ 0. From this we deduce

, _ sco - tœ (sco - tlo)2aco + bto =-= = t- _..-—tco-sco (tco -sco)(sco - tco)

_ (2s2 + 2st + t2)co + (s2 + 2s? + 2t2)cos2 + 3st + t2 '

hences2 + (s + t)2 , t2 + (s + t)2 na = -^—i,--t and b = -¿—\-+=. Ds2 + 3st + t2 s2 + 3st + t2



Remark. We can give s and t explicitly in terms of a and b. Indeed, thenumbers a-b-l and a-b+\ cannot both be zero. If, for instance, a-b-l ^0, the system in s and t is equivalent to the equation (a - b - \)s+(2a - b)t= 0, whose nonzero solutions (s; t) are given by s = K(-2a + b) and t =K(a-b- 1), with K £Q*.

From Theorems 2 and 3, we deduce the following consequence. Let M¡,with 1 < / < 3, be three link matrices of the type given in Theorem 3,

WA--7, X,M' ZA r, x,-Yi

Then (Mx ; M2; M3) is a solving triple if and only if the trace of the matrixM3M2MX is zero, hence, from the proof of Theorem 2, if and only if XXX2X3 =Yx Y2 Y3 ; this may be written as

(sj + (sx + tx)2)(sj + (s2 + t2)2)(s¡ + (s3 + t3)2)

= (t\ + (SX + tx)2)(t2 + (S2 + t2)2)(t2 + (53 + t3)2).

We call this last equality the condition of closure.

4. Solutions of the Diophantine equation

Using the condition of closure, we now give solutions of system (II). First weneed matrices slightly simpler than link matrices.

Definition. We call every matrix L of the form L = (XyY XX_Y), with X -s2 + (s + t)2 , Y = t2 + (s + t)2, s and t rational numbers, (s ; t) ^ (0 ; 0) apre-link matrix. We shall adopt the abbreviated notation L = ( * x ).

1st solution. The simplest solution consists in looking for triples of pre-linkmatrices of the form

W.-C; ;)(; c.)(c \We may choose, for instance,

• (s + t)2 + S2\ t -( • (S + t)2 + t2\s2 + t2 . y) ' ^2 - {(s + t)2 + s2

Lx = ({s + t*)2 + t2 s2+ut2y (s,t)£Z2\{(0;0)}.

The matrices L¡, for 1 < /' < 3, are obviously pre-link matrices, and thecondition of closure is satisfied. From this we deduce a solving triple,

., 1 ( -2st-t2 s2 + t2Mi =

M2

s2-st + t2 \(s + t)2 + t2 -2st - t2

1 ( t2-s2 (s + t)2 + t2

M3 =

s2 + 3s( + t2 \(s + t)2 + s2 t2-s2

1 fs2 + 2st (s + t)2 + s2

We find thats2 - st - t2 \ s2 + t2 s2 + 2st

1 faxx aX2M3M2MXD \a2x a22


ON THE DIOPHANTINE EQUATION xf + x26 + xf = y\ + yf + y\ 709

withD =-s6 -3s5/ +2s4t2 +9s3t3 +2s2t4 -3st5 -t6,axx =-s6 -6s5t -6s4t2 +6s2t4 +6st5 +t6,ax2 = 3s6 +8s5/ +9s4t2 -4s3/3 -9s2/4 -2s/5,a2x = -2s$t -9s4t2 -4s3t3 +9s2t4 +8s/5 +3t6,a22 =-an-

The eigenvalue 1 of the matrix M3M2MX leads to the antecedent solution(s; t; -(s + t) : t; -(s + t) ; s), which is trivial and nonzero.

In the case of the eigenvalue -1, write / for the identity matrix and putP(s ; t) = 3s5 + 8s4/ + 9s312 - 4s2t3 - 9st4 - 2/5. We find that

M3M2Mx+I=±(sP{t>s) sP(s>lA3 2 1+ D\tP(t;s) tP(s;t)J-

Therefore, an eigenvector of M3M2MX associated with the eigenvalue -1 isVx = (*'), with xi = P(s; t) and yx = -P(t; s). The vectors V2 and V3 arecalculated by V2 = Mx Vx and V3 = M2V2. We then find

x, = 3s5 + 8s41 + 9s312 - 4s2t3 - 9st4 - 2t5,

x2 = -2si-s4t+\2s3t2 + l3s213 + 4st4 - t5,

x3 = -s5 - 9s41 - \3s3t2 - ls2t3 - 1st4 - 3t5,

yx = 2s5 + 9s4t + 4s3t2 - 9s2t3 - 8s/4 - 3/5,

y2 = -3s5 - ls4t - ls3t2 - 13s2/3 - 9s/4 - /5,

y3 = s5 + 4s4/ + 13s3/2 + 12s2/3 - st4 - 2/5.

Remarks. 1. This solution may be characterized by the fact that it is the conju-gate solution of the trivial solution (s ; / ; -(s + t) ; t ; -(s + t) ; s). This explainsin a simple way the remarkable property of this parametrization, noticed by J.H. Conway (see [1, p. 574]).

2. The 6 by 6 determinant of the sextuple (x( ; x2 ; x3 ; yx ; y2 ; y3) in thebasis (s5 ; s4/ ; s3/2 ; s2/3 ; st4 ; t5) is nonzero. So, contrary to the case of thesolutions of systems (Bx) and (B2), the polynomial functions xx , x2, x3,yx, y2, and y3 are linearly independent.

2nd solution. We look for triples of pre-link matrices of the form

w.-(; B.){c°){;DrThus the condition of closure is satisfied.

Put• s2 + (s + t)2

S(5;l)

andL3~ \t2 + (s + t)2

v2 + (u + v)2u2 + (u + v)2

We have

Xx = AC = (t2 + (s + t)2)(v2 + (u + v)2) = (su + sv + tu + 2tv)2 + (sv - tu)2


710

and

JEAN-JOEL DELORME

y, = BD = (s2 + (s + t)2)(u2 + (u + v)2) = (2sw + sv + tu + tv)2 + (sv - tu)2.

For Lx to be a pre-link matrix, it is sufficient that

(su + sv + tu + 2tv) + (2su + sv + tu + tv) = (sv - tu).

This may be written as (3s + 3t)u = -(s + 3t)v . Therefore, we may chooseu = s + 3/ and v = -(3s + 3/). From this we deduce a solving triple,

M3 = ~z3

M2 = ~¿2

with Z3

(s + 3/)2 + (2s)2'

(s2 + 3s/ + t2)

Mi = J-

S(4;4;a)

• s2 + (s + t)2't2 + (s + t)2 •

(s + 3/)2 + (2s)2\(3s + 3/)2 + (2s)2 • ) '

with Z2 = -s2 + 12s/+ 9/2,(s2-s/)2 + (3s2+4s/+3/2)2\

(2s2 + 5s/+3/2)2 + (3s2 + 4s/+3/2)2 • /with Zx = 1 Is4 + 27s3/ + 32s2/2 + 21s/3 + 9/4.

From the eigenvalue 1, we obtain the parametric solution

x, = 3s4 + 9s3/+18s2/2 + 21s/3 + 9/\x2 = 2s4 + 4s3/ - 5s2/2 - 12s/3 - 9/4,

x3 = -s4- 10s3/-17s2/2- 12s/3,

yx=s4- 3s3/ - 14s2/2 - 15s/3 - 9/4,

y2 = 3s4 + 8s3/ + 9s2/2,y3 = 2s4 + 12s3/ + 19s2/2 + 18s/3 + 9/4.

Apart from exchanging xi and x2, this is Brudno's solution [4] of (B2). Fromthe eigenvalue -1, we obtain

Xi = 67s4 + 165s3/ + 178s2/2 + 81s/3 + 9/4,x2 = 36s4 + 140s3/ + 189s2/2 + 108s/3 + 27/4,x3 = 37s4 + 42s3/ - 13s2/2 - 48s/3 - 18/4,y, = 15s4 - 9s3/ - 18s2/2 + 3s/3 + 9/4,y2 = 65s4 + 152s3/ + 169s2/2 + 96s/3 + 18/4,y3 = -52s4 - 152s3/ - 167s2/2 - 102s/3 - 27/4.

Apart from multiplicative coefficients, permutations, and changes of signs, thissolution is deduced from the preceding one by the transformation s' = 2s + 3/,/' = -3s - 2/, which realizes a one-to-one mapping from Q2 onto Q2 .

Remark. Every solution of the system

S(4;4;b)

( xf + xf + xf = yf + yf + yf,xf + xf + xf = yf + yf + y\,

_ x, + 3x2 + x3 = 3yi +y2 + y3


ON THE DIOPHANTINE EQUATION xf + xf + xf = yf + yf + yf 711

(that is, (B2), apart from exchanging xi and x2) is a solution of (II), therefore,a fortiori, of (I). In particular, it satisfies the additional equality

*iVi(xf - yf) + x2y2(xf - yf ) + x3y3(xf - yf ) = 0.3rd and 4th solutions. We look for pre-link matrices of the form

^

712 JEAN-JOEL DELORME

In this last case, we write [an ; ... ; 09] to denote the polynomial, homogeneousin s and /, ao^9 H-1- ¿M9 ■

5th and 6th solutions. As in the preceding case, we look for triples of pre-linkmatrices of the form

Put, this time,

L3L2LX =

L3 =

. BA •

u2 + v2

• CBX •

AX

(u + v)2 + V2

and X = s2 + t2. We have

BX = ((w + v2) + v2)(s2 + t2) = (su + sv + tv)2 + (sv -tu- tv)2.

Put su + sv + tv = bx and sv - tu - tv = b2. We have BX = b\ + b\.For L2 to be a pre-link matrix, it is sufficient that C = (bx+ b2)2 + b\ , that is,C = ((s - t)u + 2sv)2 + (-tu + (s - t)v)2. In the same way,

AX = (u2 + v2)(s2 + t2) = (su + tv)2 + (sv - tu)2.

Put su + tv — ax and sv - tu — a2. We have AX a\ + a\. For Lxto be a pre-link matrix, it is sufficient that C = (ax + a2)2 + a\ , that is, C((s - t)u + (s + t)v)2 + (-tu + sv)2.

By equating the two expressions of C, we obtain uv(2s2 - 4s/ + 4/2) =t>2(-3s2 + 4s/). If v - 0, we obtain

L3 =

hence M3 = (° ¿), yielding a trivial solution. If v / 0, we may chooseu = -3s2 + 4st and v = 2s2 - 4s/ + 4/2. From this we deduce in successionLi , L2 , L3 , Mi , M2 , M3, then the following two parametric solutions:

x, =[15; -96; 295; -628; 920; -880; 608; -320; 64],x2 = [-25; 163; -483; 896; -1220; 1200; -784; 384; -128],x3 = [-5; 22; -84; 284; -620; 880; -816; 512; -192],

y, = [-25; 162; -480; 908; -1220; 1200; -816; 320; -64],y2 = [15; -109; 309; -544; 800; -960; 832; -512; 192],y3 = [-5; -12; 143; -376; 580; -720; 656; -384; 128]

S(8;8;a)

and

S(8;8;b)

xi = [13; -28; 31; 68; -160; 336; -256; 192; -64],x2 = [-21; 55; -151; 240; -380; 368; -240; 128; 0],x3 = [l; -58; 166; -308; 460; -368; 304; -128; 64],y, =[-21; 54; -154; 268; -300; 368; -272; 192; -64],y2 = [13; -39; -1; 48; -200; 256; -288; 128; -64],y3 = [-l; -64; 161; -320; 420; -432; 272; -128; 0].



7th and 8th solutions. We look for triples of pre-link matrices of the form

«*-U ")(¿ ")(¿ f )■The condition of closure is obviously satisfied. Put

j _( • ((s + t)2 + t2)(a2 + b2)\Li~ \(s2 + t2)(c2 + d2) . )'

(s2 + (s + t)2)(c2 + d2)L2 =

L, =

(t2 + (s + t)2)(e2 + f2)(s2 + t2)(e2 + f2)

((s + t)2 + s2)(a2 + b2) •

We may write

r -( * Xi + Yi\ th X3 = -as-at+bt, Y3 = -bs-bt-at.*~\Z2 + T2 . )' Wlt : Z3 = cs+dt, T3 = ds-ct,

• X2 + Y22\ ., X2 = cs - ds -dt, Y2 = ds + cs + ct,' Z2 + T2 . J' Wlth: Z2 = et-fs-ft, T2 = ft + es + et,

r-( • X2 + Y¡\ with. *i=et+fs, Yx=ft-es,l~\Z2 + T2 • J' • Zx = -as-at+bs, Tx=-bs-bt-as.

For every matrix L¡ to be a pre-link matrix, it is sufficient that Y, = T¡ andX¡ + Z, + T,■ = 0 (for 1 < /' < 3).

Now if Yi = T¡ for 1 < i < 3, we have Yx + Y2 + Y3 = Tx + T2 + T3.After cancelling, this equality is reduced to s(a + c - 2e) = t(a - 2c + e).A sufficient condition is that a + c - 2e = 0 and a - 2c + e = 0, that is,a = c — e. By replacing c and e by a in every equality Y¡ = T¡, we obtain theequalities -b(s + t) = ds = ft. A simple sufficient condition is that b = -2s/,d = 2/(s + /), and / = 2s(s + /).

Then, if we replace c and e by a, b by -2s/, d by 2/(s + /), and /by 2s(s + /) in the equalities X¡ + Z, + T¡■ = 0 for 1 < /' < 3, we obtaint(a - s2 - st - t2) = 0, (s + t)(a - s2 - st - t2) = 0, and s (a - s2 - st - t2) = 0.

A simple sufficient condition is that a - s2 + st + t2. There follow theparametric solutions

x,=[2;7; 15; 18 ; 9 ; 2 ; 1; 1],x2 = [l; 1; -4; -17; -22; -15; -7; -2],x3 = [-l;-6;-17;-21;-17;-ll;-6;-l],y1=[-l;-l;-2;-9;-18;-15;-7;-2],y2 = [l;6; 11; 17; 21 ; 17 ; 6 ; 1],y3 = [2;7; 15 ; 22 ; 17; 4; -1 ; -1]

S(7; IT



S(1I; 7)

andx, = [8; 43; 167; 547; 1296; 1863; 1631 ; 927; 352;

63;-17;-5],x2 = [-l; 39; 213; 610; 1005; 939; 587; 412; 341;

177; 45; 8],x3 = [5; 38; 42; -155; -769; -1554; -1744; -1217;

-613; -232;-50;-1],

Vl = [-5; -17; 63; 352;927; 1631;1863;1296;547;167;43; 8],

y2 = [l; 50; 232; 613; 1217; 1744; 1554; 769; 155;-42;-38;-5],

y3 = [-8; -45; -177; -341 ; -412; -587; -939;- 1005; -610; -213; -39; 1].

Remarks. 1. Consider system (Bx) that Brudno and Kaplansky [3] solved, byhaving recourse to the rational points on a cubic. It is easily verified that, forevery solution of (Bi),

if xix2x3yi < 0, then (xx ; x2 ; x3 ; yi ; y2 ; -y3) is a solution of (II)

and

if XiX2x3y[ > 0, then (xi ; x2 ; x3 ; -yx ; y2 ; -y3) is a solution of (II).

So it is natural to consider, for instance, the following system (Bi-II), deducedfrom (Bi) by replacing y3 by -y3, and by adding an inequality to it:

(BpII)

JÍ \ ~T~ Jv'y ~\ J^'l yf+yf + yf,xf + x2¿+x|=yf + y| + y3Sy2=x2-x3,-y3 = x2 + x3,xix2x3yi < 0.

From the beginning of this remark, every solution of this system is a solutionof (II). More precisely, a nonzero solution of (II), with associated solving triple(Mx ; M2 ; M3), is a solution of (Bi-II) if and only if M2 = (_'2 "j1 ) .

2. The seventh solution of Lander, Parkin, and Selfridge [5], after exchangingyx and y2, that is, put in the form (51; 113; 136; 125; 40; 129), is a solu-tion of (II). However, an examination of its solving triple shows that it can beobtained neither as one of Brudno's solutions of systems (Bi) or (B2), nor byone of our parametric solutions.

3. Permutations and changes of signs apart, all the known solutions of thesystem

A) I JC~) ~\~ JC'i yf + y\ + y\._2 _ „2I xf + X2Z + x3z =yf+y$+y$

are solutions of (II). It is still an unsolved problem to find a counterexample.4. In 1979, Bremner [1] studied the surface V defined by (II), by using

techniques of algebraic geometry. He determined a basis for the Néron-Severi



groups of V over the fields Q(i) and Q, and he exhibited a system of thirteengenerators over Q. This allowed him to determine all parametric solutions of(II) of a given degree thanks to an algorithm whose results corroborate our own,sometimes, however, with difficulty.

It would be interesting to determine how our results could be interpreted interms of the framework of [1], but this does not seem straightforward.

Acknowledgments

The author thanks Mr. Simon J. Agou, professor at the Université du Maineand at Claude Bernard University Lyon I, for his help and guidance. He alsowishes to thank the referee for his useful remarks and suggestions.

Bibliography

1. Andrew Bremner, A geometric approach to equal sums of sixth powers, Proc. London Math.Soc. 43(1981), 544-581.

2. Simcha Brudno, On generating infinitely many solutions of the Diophantine equation A6 +B6 + C6 = D6 + E6 + F6 , Math. Comp. 24 (1970), 453-454.

3. Simcha Brudno and Irving Kaplansky, Equal sums of sixth powers, J. Number Theory 6(1974), 401-403.

4. Simcha Brudno, Triples of sixth powers with equal sums, Math. Comp. 30 (1976), 646-648.5. L. J. Lander, T. R. Parkin, and J. L. Selfridge, A survey of equal sums of like powers, Math.

Comp. 21 (1967), 446-459.6. K. Subba-Rao, On sums of sixth powers, J. London Math. Soc. 9 (1934), 172-173.

Jean-Joël Delorme, c/o Professor Simon J. Agou, Department of Mathematics andComputer Sciences, Université du Maine, BP 535, 72017 Le Mans, France


€¦ · ON THE DIOPHANTINE EQUATION xf + xf + x\ = y\ + y\ + y\ 705 Using suitable matrices Mx,...

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Transcript of €¦ · ON THE DIOPHANTINE EQUATION xf + xf + x\ = y\ + y\ + y\ 705 Using suitable matrices Mx,...