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JOURNAL OF NUMBER THEQRY 26, 325-367 (1987) Solving Exponential Diophantine Equations Using Lattice Basis Reduction Algorithms B. M. M. DE WEGER* Mathematisch Insfiluuz, Rijks Uniuersireit Leiden, Postbus 9512, 2300 RA Leiden, The Nether1and.s Communicated by M. Waldschmidt Received June 5, 1986 Let S be the set of all positive integers with prime divisors from a fixed finite set of primes. Algorithms are given for solving the diophantine inequality 0 < x - y < y” in x, y E S for Iixed 6 E (0, 1 ), and for the diophantine equation x + y = z in x, y, 2 G S. The method is based on multi-dimensional diophantine approximation, in the real and p-adic case, respectively. The main computational tool is the L’-Basis Reduction Algorithm. Elaborate examples are presented. el 1987 Academic Press. Inc 1. INTRODUCTION In 1981, L. LovQsz invented an algorithm for computing a reduced (i.e., nearly orthogonal) basis of an arbitrary lattice in [w” from a known basis of the lattice. It has a surprisingly good theoretical complexity (polynomial time), and also performs very well in practice. This algorithm, together with an application to the factorization of polynomials, is described in Lenstra et al. [9]. It has several other interesting applications, such as in public-key cryptography (cf. Lagarias and Odlyzko [8]), and in the dis- proof of the Mertens conjecture (cf. Odlyzko and te Riele [13]). We shall refer to the algorithm as the “L3-Basis Reduction Algorithm,” (L3-BRA). The L3-BRA can also be used for solving multi-dimensional diophantine approximation problems, as Lenstra et al. already indicated [9, p. 5251. In the present paper it is shown that this enables us to find all solutions of certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems are associated to linear forms in logarithms of algebraic numbers (see, e.g., Baker [3, Chap. 41, Shorey and Tijdeman [ 181, Stroeker and Tijdeman [20, pp. 343-3531). Namely, for any large solution of the diophantine problem some linear form in logarithms is extremely close to zero. The Gelfond-Baker method provides effectively computable (and in many cases * Present address: Department of Applied Mathematics, University of Twente, P.0. Box 217, 7500 AE Enschede, The Netherlands. 325 0022-314X/87 $3.00 CopyrIght ,(‘ 1987 by Academx Press, Inc All nghta of reproductmn m any form reserved

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Page 1: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

JOURNAL OF NUMBER THEQRY 26, 325-367 (1987)

Solving Exponential Diophantine Equations Using Lattice Basis Reduction Algorithms

B. M. M. DE WEGER*

Mathematisch Insfiluuz, Rijks Uniuersireit Leiden, Postbus 9512, 2300 RA Leiden, The Nether1and.s

Communicated by M. Waldschmidt

Received June 5, 1986

Let S be the set of all positive integers with prime divisors from a fixed finite set of primes. Algorithms are given for solving the diophantine inequality 0 < x - y < y” in x, y E S for Iixed 6 E (0, 1 ), and for the diophantine equation x + y = z in x, y, 2 G S. The method is based on multi-dimensional diophantine approximation, in the real and p-adic case, respectively. The main computational tool is the L’-Basis Reduction Algorithm. Elaborate examples are presented. el 1987 Academic

Press. Inc

1. INTRODUCTION

In 1981, L. LovQsz invented an algorithm for computing a reduced (i.e., nearly orthogonal) basis of an arbitrary lattice in [w” from a known basis of the lattice. It has a surprisingly good theoretical complexity (polynomial time), and also performs very well in practice. This algorithm, together with an application to the factorization of polynomials, is described in Lenstra et al. [9]. It has several other interesting applications, such as in public-key cryptography (cf. Lagarias and Odlyzko [8]), and in the dis- proof of the Mertens conjecture (cf. Odlyzko and te Riele [13]). We shall refer to the algorithm as the “L3-Basis Reduction Algorithm,” (L3-BRA).

The L3-BRA can also be used for solving multi-dimensional diophantine approximation problems, as Lenstra et al. already indicated [9, p. 5251. In the present paper it is shown that this enables us to find all solutions of certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems are associated to linear forms in logarithms of algebraic numbers (see, e.g., Baker [3, Chap. 41, Shorey and Tijdeman [ 181, Stroeker and Tijdeman [20, pp. 343-3531). Namely, for any large solution of the diophantine problem some linear form in logarithms is extremely close to zero. The Gelfond-Baker method provides effectively computable (and in many cases

* Present address: Department of Applied Mathematics, University of Twente, P.0. Box 217, 7500 AE Enschede, The Netherlands.

325 0022-314X/87 $3.00

CopyrIght ,(‘ 1987 by Academx Press, Inc All nghta of reproductmn m any form reserved

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326 B. M. M. DE WEGER

explicitly computed) lower bounds for the absolute values of such linear forms. Thus, explicit upper bounds for the solutions of many diophantine problems can be obtained. The bounds that are found in this way are so large that enumeration of the remaining possibilities is practically impossible. However, it is generally assumed that the bounds are far from the actual largest solution. Therefore it is worthwile to search for methods to reduce the found upper bounds.

If the linear form in logarithms under consideration has only two terms, a simple method applies, based on continued fractions. For example, Cijsouw, Korlaar, and Tijdeman (Appendix to Stroeker and Tijdeman [20]) found in this way all solutions of the diophantine inequality

1 p v - q I 1 < p”-x- (1.1)

for all primes p, q with p < q < 20, and 6 = 4. In Section 4.B we extend this result for many more values of p, q, and 6 = 6.

A natural generalization of (1.1) is the following problem. Let S be the set of all positive integers composed of primes from a fixed finite set { pI ,..., p,}, where t > 2, and let 6 E (0, 1). Then lind all solutions of the diophantine inequality

0 < .Y - 1’ < y’ (1.2)

in x, y E S. Putting X/-V = p;’ . ‘p:‘, the corresponding linear form in logarithms is

/i=.u,logp,+ ..’ +x,logp,.

The continued fraction method applies only for t = 2. For f 2 3, multi- dimensional continued fraction algorithms are available (cf. Brentjes [S]), but they are not useful for our purposes. In Section 4.C we shall show that the L3-BRA leads to substantial improvements of the upper bounds for (1.2). Usually the new bound is of the size of the logarithm of the initial bound. For t = 6, {p, ,..., p6} = { 2, 3, 5, 7, 11, 13 >, 6 = 4, we show in detail how (1.2) can be solved completely with this method.

If the linear form is inhomogeneous of the form

/l=x,logcr, + .‘. +x,loga,+loga,+,,

it can of course be made homogeneous by introducing the variable x,+ i as coefficient of the last term. We may then solve this (n + 1)-dimensional approximation problem, and select all solutions with x,+ , = 1. There is, however, a different approach, which may ,be faster. See Baker and Daven- port [4] for n = 2, and Ellison [6 3 for n > 2. It is then needed to find good simultaneous approximations pi/q to log a,/log a, (i = l,..., n - 1). Lenstra

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DIOPHANTINE EQUATIONS 321

et al. [9, p. 5251 have indicated how the L3-BRA can be used to find such approximations. We do not work this out in the present paper.

Up to now we have only considered real linear forms in logarithms. There is a p-adic counterpart of the Gelfond-Baker theory, which provides lower bounds for the p-adic values of linear forms in p-adic logarithms of algebraic numbers. It is therefore a natural problem to devise p-adic analogues of the diophantine approximation methods sketched above. The simplest case is that of an inhomogeneous form with only one variable, such as

A =x log, a, + log, c(2.

Then it suffices to compute the p-adic expansion of log, aJog, a, far enough. See Wagstaff [21], Petho and de Weger [14], and de Weger [24].

In the case of a form with two terms, such as

/1=x,log,a,+x,log,cr,

a practical p-adic analogue of the real continued fraction algorithm is needed. Such an algorithm was first formulated by Mahler [ 11, Chap. 41. A similar algorithm has been studied by de Weger [23] in the context of p-adic approximation lattices. See Agrawal et al. [ 1 ] for an application to a Thue-Mahler equation. We shall show in Section 5.C how to solve

p-y + pi? = )$lp’;’ (1.3

for fixed pr , p2, pj, w using this algorithm. A natural generalization of (1.3 is the diophantine equation

x+y=z (1.4)

in x, y, z E S, with S as above. We may assume gcd(x, y, z) = 1. Put p = pr, and suppose p 1 z. Then p i xy. Put x/y = p-T’ *. .pf?j. Then we have the p-adic linear form in logarithms

n =x1 log/$, + ... S-X,-I 10$&p,-,.

The concept of approximation lattices of p-adic numbers, as introduced in [23], can be generalized to the multi-dimensional case, as we shall see in Section 5.B. Then we can apply the L3-BRA. In Section 5.D we show how this can be used to solve (1.4) explicitly. We give details for t = 6, {pl ,..., p6} = (2, 3, 5, 7, 11, 13). This generalizes the results of Alex [2], who gave a complete solution of (1.4) for t = 4, {pl ,..., p4} = (2, 3, 5, 7) by elementary arguments. The case where z has only one prime divisor was treated by Rumsey and Posner [16], also by elementary means.

Many diophantine equations, such as the Thue equation, the Thue-

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328 B.M.M.DE WISER

Mahler equation, the hyperelliptic equation and the Mordell equation, lead to linear forms in logarithms similar to those described above. These equations differ from our examples (1.2) and (1.4) in that the path from the equation to the linear form in logarithms is not as straightforward; it leads through some algebraic number theory. This clearly does not affect the applicability of our approximation methods for reducing upper bounds, since they are based only on the linear forms themselves.

2. BOUNDS FOR LINEAR FORMS IN LOGARITHMS

In this section we quote the results that we use from the theory of linear forms in logarithms. We do not quote the theorems in full generality, since we apply them only for logarithms of rational integers, and for rational coefficients. The results provide lower bounds for linear forms in logarithms in the real and p-adic cases. We chose results that give completely explicit constants and lead to convenient upper bounds for the solutions of the diophantine problems we want to solve. We stress that our methods for reducing these bounds are in principle independent of the size of the bounds.

Let p, ,..., pI1 (n > 2) be rational integers such that 2 <p, < .. <p,,, and [Q(p;l’,..., pf,“): Q] = 2”. Let b ,,..., h,,eZ’, and put B=max, SiG,l IhJ. In the real case we have the following result.

LEMMA 2.1. (Waldschmidt). Let

n =h, logp, + .” +h,,logp,,

he nonzero. Put

V, = max( 1, logpi) (i= l)..., n), Q= V, ... V,,,

c,=2 “I + 26n’7 + 4!2 log( e V,, , ), c, = c, log(e v,, ).

Then

IAl >exp(-(C, log B+C,)j..

This lemma was proved by Waldschmidt [22]. In the case n = 2 a sharper bound was given by Mignotte and Waldschmidt [ 121. In the p-adic case we have the following result:

LEMMA 2.2. (van der Poorten). Let p be a prime with p i pi (i = l,..., n). Let

A =h, log,p, + ... +h,log,p,,

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DIOPHANTINE EQUATIONS 329

be nonzero. Choose p, ti with 2/(n + 1) d p 6 2, 0 < K < p/2. Put

Vi = max(e, log pi) (i= l,..., n), Q= If,*.. v,,

G = P(P-1) P

b

if p=2,3

-1 if p>,5,

&=@--)/(I +K)(l +p)(n+l),

~=max{(~6n)‘~+““““+“, (@)(‘+V’)(“+‘), 161’“},

C,=4(n+ 1) cn+“k(l+~‘l(Gp/logp) Q.

Then B < I, OY

ord,( /i ) < C,(log B)‘.

This lemma follows from the proof of Theorem 2 of van der Poorten [IS]. Note that we omitted the factor n(2D2)“+ I, since D = 1; cf. van der Poor- ten [ 15, p. 351. To save computation time we may choose p, K as a function of n such that C3 is minimal, for van der Poorten’s estimate (16(n + l))“(“+‘) for 4(n + l)(ncl’k(‘+V) (with p = 2, K = +) is rather crude. Note that for n = 2 a sharper bound was by Schinzel [ 171. It is expected that the constant C3 of Lemma 2.2 can be sharpened considerably (van der Poorten, private communication).*

We also need the following simple lemma. For its proof, see Petho and de Weger [14, Lemma 2.23.

LEMMA 2.3. Let a> 0, h > 1, b > (e’/h)“, and let x E R satisfv .Y 6 a + b(log x)“. Then

x < (2a”” + 2b’lh log(hhb))h.

3. THE L3-B~srs REDUCTION ALGORITHM

In this section we describe how we use the L3-BRA. All lattices that appear in this paper are integral lattices, that is, sublattices of Z”. In the algorithm as stated in [9, Fig. 1, p. 5211, non-integral rationals may occur, even if the input consists of rational integers only. We now describe a variant of the L3-BRA in which only integers occur. This has the advantage of avoiding rounding-off errors.

Let Tc Z” be a lattice with basis vectors b, ,..., b,. Define b:, pu, di as in [9], (1.2), (1.3), (1.24), respectively. The di can be used as denominators

* Note added in proof. Recently, K. R. Yu has obtained such an improvement, to be published in the Proceedings of the 1986 Durham Conference. His results lead (in Section 5) to bounds less than the square root of the bounds we derived using Lemma 2.2.

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330 B. M. M. DE WEGER

for all numbers that appear in the original algorithm [9, p. 523). Thus, put for all relevant i, j,

(3.1 1

(3.2)

They are integral, by [9], (1.2X), (1.29). Note that, with Bi= lb,Fj2.

d, = di , Bi. (3.3)

We can now rewrite the L3-BRA in terms of ci, d,, i.,,j instead of b,?, Bi,

pi.,, thus eliminating all non-integral rationals. We give this variant of the algorithm in Table 1. All the lines in this variant are evident from applying

TABLE I

Variant of the LA-Basis Reduction Algorithm

(A)

d,:= I: c,:= b. A., : = ;i,, c,); c, := (d/c,-A,,c,)/d,-,

fOfj= I,..., i- 1; for i== I, .__, n;

4:= (%C;)ld,. , k:= 2; /

(1) perform (*) for /=&I; if 4dke2dk <3a;-, -4;1;,,_ ,, go to (2); perform (*) for I = k - 2,.... 1; if k = n, terminate; k:= k+I;

go to ( 1 L

(Cl 4-l := (dk-zdk+~:.lr-,)/d~.l; ifkz2, then k:= k-l;

go to (1); (*) if 2 I&,,( > d,, then

r := integer nearest to &Jd,; b, := be-rb,; uk := u,-rru,; A,, := &,-rA,,, for j= l,..., I- 1; A,,, : = &, - rd,.

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DIOPHANTINE EQUATIONS 331

(3.1), (3.2), and (3.3) to the corresponding lines in the original algorithm, except the lines (A), (B), and (C), which will be explained below.

We added a few lines to the algorithm, in order to compute the matrix of the transformation from the initial to the reduced basis. Let C be the matrix with column vectors b,,..., b,, (we say: the matrix associated to the basis b, ,..., b,), and let B be the matrix associated to the reduced basis computed by the algorithm. Then we define this transformation matrix Y by B = CV. More generally, let U be the matrix of a transformation from some C, to C, so C = C,,U. Denote the column vectors of U by u, ,..., II,,. All manipulations in the algorithm done on b, ,..., b, are also done on UI ,**-, u,. Then the algorithm gives as output matrices B and u’, such that B is associated to a reduced basis, B = CV, and u’ = UV. (Note that V is not computed explicitly). Hence B = CU-’ U’ = Co U’, so u’ is the matrix of the transformation from C, to B. In particular, if U = I, then C = Co and U’= v.

We now explain lines (A), (B), and (C).

(A) From [IS], (1.2) it follows that

Define for j=O, l,.,., i- 1,

ci(j)=djbi- .f $f& &k~k. k=l k-1 k

Then c,(O) = bi, and cj( i - 1) = ci. The ci( j) is exactly the vector computed in (A) at the jth step, since

tdjc,(j- l)FA,jcJ)ldj- I J-1

zdjb,- C A AikCk-A n;,.jCj=ci(j).

k=l dk-,dk ’ J-1 I

This explains the recursive formula in line (A). It remains to show that the occurring vectors q(j) are integral. This follows from

d, i -!- k=l h-,4

Ai,kCk=dj i P;,kbk*, k=l

which is integral by [9, p. 523, 1. 111.

(B), (C) Note that the third and fourth line, starting from label (2), in the original algorithm are independent of the first, second, and fifth line. Thus a permutation of these lines is allowed. We rewrite the first, second,

641126/3-l

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332 13. M. M. DE WEGER

and fifth line as follows, where we indicate variables that have been changed by a prime,

B; - 1 : = B, + ,L& ~, B, , ; (3.4)

B;:= B,m,B,/B;p,; (3.5)

/L;,~-, := p+,Bk-,IB;. -1: (3.6)

i4.k - I := &k-l~,.k-l+(l -pk.k I&k~I)/h,k;

1

(3.7) for i = k + l,..., n.

p1.k : = pLi,k - I - pk,k I pi.k (3.8)

The d, remain unchanged for i = 0, l,..., k - 2, and by (3.5) also for i = k. Now, (3.4) is equivalent to

d;-, dk A:,-, h-1 d, >=clfi+K d,mz’

(3.9)

which explains (C). From (3.6) we find .I lb,, I j.h.k I 4 , 4 2 A==--- 4 I 4 I dk 7 d;. , ’

hence &.& , remains unchanged. From (3.7) we obtain

whence, by multiplying by d,

dk , 3.i.k - , = &.A-

= &,k

Finally, from (3.8) we see

~, d;, _ , and using (3.9)

and (B) follows. In our applications we often have a lattice r, of which a basis is given

such that the associated matrix, A say, has the special form 1 0

Acoy , t 1 8, ... 8,

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DIOPHANTINE EQUATIONS 333

where the 13, are large integers (they may have several hundreds of digits). We can compute a reduced basis of this lattice directly, using the matrix A itself as input for the L3-BRA. But it may save time and space to split up the computation into several steps with increasing accuracy, as follows.

Let k be a natural number (the number of steps), and let 1 be a natural number such that the fIi have about kl (decimal) digits. For i= l,..., n, j = I,..., k, put

and define $I” by

fl(.i+l)= 1O’Q!jJ+$)jJ. I I

Thus, the +I” are blocks of 1 consecutive digits of Oi. Define for the relevant j,

i

1

E= 0’

Then it follows at once that

Ai+, = EA, + Yj.

Note that A, = A, since elk’ = Bi. Put UO = Z, C, = A,. For some j 2 1 let C, and U,- , be known matrices. Then we apply the L3-BRA to C= C,, Ii = U, _, . We thus find matrices B, and U, such that

Now put

Bj= CjU,:‘, U,.

c I+1 = EBj+ !P,Uj.

By induction the matrices B,, C,, and U, are well defined for j= l,..., k. Note that

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334 R. M. M. DE WEGER

so the C,U, I, satisfy the same recursive relation as the A,. Since C, U;’ = A,, we have C,U, !, = A, for all j. Hence

Bi= C,lJmml, U,= A,U,.

and it follows that B, and A, are associated to bases of the same lattice, which is ZY Moreover, since B, is output of the L3-BRA. it is associated to a reduced basis of f.

Let us now analyse the computation time. For a matrix A4 we denote by L(M) the maximal number of (decimal) digits of its entries. If the L3-BRA is applied to a matrix C, with as output a matrix B, then according to the experiences of Lenstra, Odlyzko (cf. Lenstra [ 10, p. 71) and ourselves, the computation time is proportional to L(C)3 in practice. Since B is associated to a reduced basis, we have

L(B) ‘v ‘Olog(det(r))/n.

In our situation, L(A,) 2: fj, L( Y,) N 1, and since det(Bj) = det(A,) = Bi”, we have L(B,) 2: lj/n. Put B,= (b$j’), U, = (u:,i’). Then by B,= A,U, and the special shape of A, we have b,,, ( j) = ~fjh) for i = l,..., n - 1, h = I,..., n, and

It follows that L( U,) N L(B,). So

Instead of applying the L3-BRA once with A as input, we apply it k times, with Cl,..., C, as input. Thus we reduce the computation time by a factor

For k between 2.5n and 3n this expression is maximal, about 0.4n2. So the reduction in computation time is considerable. The storage space that is required is also reduced, since the largest numbers that appear in the input have I( 1 + ((k - 1)/n)) digits.

We use the L3-BRA for finding a lower bound for the length of the non- zero vectors of a lattice f. Let 1.1 denote the euclidean length on [w”. Put

Then the following inequality holds (cf. [9, (l.ll)]).

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DIOPHANTINE EQUATIONS 335

LEMMA 3.1. (Lenstra et al.). Let b, ,..., b, be a reduced basis of the lattice IY Then

l(f)>,2- (n-l)/2 (b,l,

In some applications we want to compute all vectors in a lattice with length bounded by a given constant. To do this we employ a recent algorithm of Fincke and Pohst [7], in combination with the L3-BRA.

4. A DIOPHANTINE INEQUALITY

Let p, < . . . <pr be prime numbers, where t > 2. Let S be the set of all positive integers composed of these primes only, so

S= {p;1...p.$x,~Z, x,20 for i= l,..., t}.

Let 0 < 6 < 1 be a fixed real number. We study the diophantine inequality

O<x-y<yb (4.1)

in x, y E S. For a solution x, y of (4.1), the finitely many z E N for which zx, zy is also a solution of (4.1) can be found without any difficulty. Therefore we may assume that (x, y) = 1. Put

Tijdeman showed that there exists a computable number c, depending on p, only, such that for all x, y E S with x > y 2 3,

x - y ’ Yl(h2 Y 1’

(cf. Shorey and Tijdeman [ 18, Theorem 1.11). Thus, for any solution of (4.1) a bound for X can be computed, and we do so in Section 4.A. In Sections 4.B and 4.C we show how to reduce such an upper bound, in the cases t = 2 and t 2 3 respectively.

4.A. Upper Bounds

THEOREM 4.1. In the above notation, put

C 4- -29”“t’14max(1, &) logp,~~~logp,log(elogp,~,)/(1-6),

C5 = 2 log 2/logp, + 2C, log(eC, logp,).

Then the solutions of (4.1) satisfy X< C,.

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336 H. M. M. IX WFGER

Ptmji If y 6 $Y, then ~1” > .Y -J> 3~: which contradicts j’ 3 1. So J > 4.~. Put n = log(.u/jl), then

O<A<X/J’- 1 <F ” ,$‘<($r) ” ‘)‘. (4.2)

By s = max( .q y ) 2 p.Fq we obtain

()<A <21- ?)-(I -iiM, (4.3 1

We apply Lemma 2.1 to A, with n = t, q = 2. Since p, 2 3 we have Vi = log pi for i 3 2. Thus

n>exp{-(logX+log(elogp,))C,(l-J)logp,}.

Combining this with (4.3) we find

X~C,log(elogp,)+log2/logp,+CC,logX.

The result now follows from Lemma 2.3, since C, > e2. 1

EXAMPLES. With t=2, 2<p,,<199 and S=$, we have C,<2.3Ox 10” and C,<1.97x 1019. With t=6, 2<p,<l3 and S=+ we find C, < 8.37 x 1O33 and C, < 1.35 x 1036.

4.B. The Case t = 2

In this section we work out the example t = 2, 2 <pi < 199 and 6 = &. We find all solutions of (4.1) with these parameters, thus extending a result of Cijsouw, Korlaar, and Tijdeman (Appendix to Stroeker and Tijdeman [20]). We write

n=Ix,logP,-.Glogp,l,

where x1, x2 are both positive integers. We assume that

p;> 1025, (4.4)

since it is easy to find the remaining solutions. Let logp,/logp, have the simple continued fraction expansion

logp,/logpz = 10; a,, az,...l,

and let the converge& ~,,/q,, be defined by

r -1 -,-- 3 ro - - 0, r”=a,r,-, +r,-,,

4-1 =o, qo= 1, qn=anqn-l+qn-2 (n = 1, 2,...).

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DIOPHANTINE EQUATIONS 337

It is well known that r/q is a convergent of a real number CI if

la - rlql < 1/2q2,

and that a11 convergents r,/q,, of a = [ao; a,, a,,...] satisfy

MatI + 1 + 2) qt, < )a - r,/qnl -c l/a,+ lqz. (4.5)

We may assume that (x,, x2) = i. We now have the following criteria.

LEMMA 4.2. Assume (4.4).

(a) If (4.3) holds for some x1,x2, then x2=rk, x,=qk for some k 6 92, and

1 l%P* a,+,+2>py”--. qk 2l”O

(b) If for some k

1 l%P, ak+, >pq’/” - - qk

2’/10 ’

then (4.3) holds for x2 = rk, xl = qk.

Proof: First we show that x, > x2, hence X= x, . Namely, if x, < x2, then

A =xzlogp,-x, logp,>X(logp,-logp,)~Xlog~>O.OIO1 x.

From (4.3) and (4.4) we infer

0.0101 d 0.0101 x< A < 2”‘010-5’2 < 0.0034,

which is contradictory. Next we prove that

p;y”O > 3.1 x. (4.6)

Namely, suppose the contrary. Then 2x”o< 3.1 X, and it follows that X < 80. This contradicts 3.1 X2 pF”O > 105’*. Now, (4.3) is equivalent to

(4.7)

Page 14: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

Pl

TABL

E II

(The

orem

4.

3a)

P;’

PZ

.x2

,” cc

P;:

delta

3 ti 31 2 13 2 1 19 2 2 5 13

17

6: : 29

23

11 5 :; :: 11

9 23

23

;: :: 11

12

:: 17

j2

125

128

128

256

343

512

2187

2187

21

97

6859

29

791

3276

8 62

7 48

517

1 12

589

9906

8 42

624

1 62

841

3597

9 10

449

2 21

331

4919

0 66

161

2 25

179

9813

6 85

248

2 25

179

9813

6 85

248

2 38

418

5791

0 15

625

3 93

737

6385

6 99

289

9 90

457

8032

9 05

937

11

3988

9 51

853

7314

3 11

69

414

6092

8 34

141

11

9209

2 89

550

7812

5 11

92

092

8955

0 78

125

12

2005

0 97

657

0582

9 21

91

462

4432

0 20

321

45

9497

2 98

635

7216

1 59

60

464

4775

3 90

625

177

9176

2 17

794

6041

3 35

3 81

478

3205

4 69

041

504

0363

6 19

364

6738

3 50

4 03

636

1936

4 67

383

505

4470

2 84

992

9377

1

3 5 3 11

11 : :z 13

t7’ 83

173

181 i; 149 83

19

83

157 89

19

3 19

7 19

7 :: 199 43

71

:: 1:; 59

16

3

9

: 2 2 2 2 2 4 9 1

1191

3 I

6304

3 2

2522

9 3

2133

1

25

27

121

121

125

243

361

529

2197

2209

22

09

6889

29

929

3276

1 42

241

0276

7 03

549

3904

1 66

161

2 25

229

2232

1 39

041

2 35

124

3277

5 37

493

3 93

658

8805

7 02

081

9 97

473

0326

0 05

057

I1

5149

9 04

768

9841

3 11

51

499

0476

8 98

413

11

6941

4 60

928

3414

1 12

20

050

9765

7 05

829

12

3586

6 42

791

6139

9 21

61

148

2313

2 84

249

0.40

194

0.32

293

P

"0%

~: z r

0.18

716

0.48

703

s 0.

8525

9 m

0.

8089

8 2

0.88

568

0.88

532

% % 0.

7615

9 0.

8794

2 0.

7628

2 0.

8656

0 0.

8759

4 0.

8874

3 0.

8934

3 0.

8986

2 0.

8826

8 0.

8865

6

45

8485

0 07

184

4903

1 0.

8405

9 58

87

158

6708

2 67

913

0.88

64'

174

8874

7 03

655

1304

9 0.

8978

5 35

0 35

640

3707

4 85

209

0.88

568

498

3114

1 43

181

2112

1 0.

8904

0 51

1 11

675

3300

6 41

401

0.89

536

498

3114

1 43

181

2112

1 0.

8958

0

0.000

00

0.21

534

0.48

832

0.28

906

0.40

575

0.22

754

0.46

694

0.49

512

0.45

416

0.29

941

Page 15: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

1’: 7 i; 2 2 13

2 ; 2 : 11

3 53 5 1;

2 2 2

41 3

1x : :: 2 2 :

4: : 13

17 :: ;::

60

10

11

17

2:

65

66

42

68 :; 30 f ii :: 15

:z

21

:: :‘4

16

84

84 :i 90

iJ

; 10

2 28

505

4470

2 84

992

9377

1 23

50

5 44

702

8499

2 93

771

59

558

5458

6 40

832

8400

7 41

799

0066

8 57

828

8412

1 79

9 00

668

5782

8 84

121

1152

92

150

4606

8 46

976

1822

83

780

4551

7 61

449

2472

15

921

5084

0 12

303

8650

41

591

9381

3 37

933

9223

37

203

6854

7 75

808

3647

2 99

637

7170

7 86

403

3689

3 48

814

7419

1 03

232

7378

6 91

629

4838

2 06

464

1 09

418

9891

3 15

123

5920

9 2

9514

7 90

517

9352

8 25

856

3 39

456

7389

9 22

223

1484

9 4

9125

8 90

425

6726

1 54

641

9 31

322

5746

1 54

785

1562

5 54

80

386

8577

8 48

021

8593

9 61

32

610

4156

8 09

986

4896

1 94

44

132

9657

3 92

904

2739

2 37

7 78

931

8629

5 71

617

0956

8 37

7 78

931

8629

5 71

617

0956

8

379

2922

7 19

491

5558

8 02

161

2392

99

329

2306

1 75

295

9008

3 24

70

6452

9 07

345

0392

7 04

413

1425

7 60

886

8461

7 89

454

4784

1 21

536

9396

3 07

555

7766

3 10

747

3219

9 05

755

8131

7 97

268

3760

7 98

497

3267

5 80

761

1094

7 11

841

1 21

375

1191

4 21

716

6362

2 74

241

1 93

428

1311

3 83

406

6795

2 98

816

1 93

428

1311

3 83

406

6795

2 98

816

1 93

832

4566

7 68

001

9896

7 96

723

2 25

393

4029

0 69

225

8087

8 63

249

123

7940

0 39

285

3802

7 48

991

2422

4 58

7 44

031

0636

0 42

001

8879

5 53

643

6338

2 53

001

1411

4 70

074

8351

6 02

688

5 07

060

2400

9 12

917

6059

8 68

128

2150

4 15

50

293

2802

6 62

396

2152

6 95

351

0552

1

1::

181

107

199

127 1:;

9:

101 29

191

199 4”:

151 I

181 41

181 17

1:;

163 :: 191

199 3

199

1:: 71

;: 89

1;

11

12 Fl

9 ; 11

2: 10

10

14 ; ;: 2 10

14

10

f; :: 22

14 :: 53

504

0363

6 19

364

6738

3 51

1 11

675

3300

6 41

401

550

3290

3 17

162

4844

1

787

6627

8 37

885

4976

1 80

2 35

917

8476

0 91

681

1151

93

665

7823

5 00

641

1838

45

921

2420

1 54

507

2459

37

419

1553

1 18

401

8594

75

474

8609

3 97

887

9269

03

592

9372

1 91

597

3619

1 31

987

9620

1 91

349

3725

2 90

298

4619

1 40

625

7314

2 41

268

9492

8 26

049

I 10

462

2125

4 11

204

5100

1 2

9755

8 23

267

5799

4 63

481

3 38

298

6815

5 95

733

1731

1 4

8941

5 46

411

9070

5 61

799

9 25

103

1023

1 50

136

2932

1 54

60

999

7061

2 05

831

7732

7 61

62

677

9503

3 67

185

1400

1 93

87

480

3376

4 77

543

0564

9 37

7 38

596

8469

5 57

044

9980

1 37

9 29

227

1949

1 55

588

0216

1

377

3859

6 84

695

5704

4 99

801

2390

72

435

6851

5 13

248

4715

3 24

69

9040

3 56

526

2140

3 03

521

1428

5 52

404

4631

8 60

195

2509

3 21

580

6066

2 62

396

0090

4 07

387

3211

8 38

877

9548

5 51

051

5736

9 98

768

3253

3 36

131

8095

1 12

441

1 23

414

7420

1 97

479

4188

8 22

591

1 93

813

4179

4 57

931

3317

8 02

199

1 93

832

4566

7 68

001

9896

7 96

723

1 93

813

4179

4 57

931

3317

8 02

199

2 25

501

1677

4 16

274

3178

6 82

911

123

6354

1 71

303

1158

3 51

179

8056

1 58

7 32

059

5938

5 49

335

3867

3 30

551

6332

5 11

891

3678

9 38

604

3275

9 54

593

5 07

282

0298

9 53

863

7524

7 83

563

9968

1 L5

49

673

1425

1 78

936

4350

9 93

211

3056

1

0.85

578

0.88

985

0.89

708

0.89

710

0.86

722

0.83

013

0.88

680

0.87

580

0.88

441

0.87

844

0.89

170

0.89

721

0.83

799

0.89

916

0.89

800

0.89

368

g

0.89

828

2 0.

8707

1 2

0.84

941

0.88

788

2 0.

8893

3 m

0.

8939

0 0.

8975

5 0.

8607

8 0.

8931

9 0.

8940

2

0.84

151

0.86

903

0.89

326

0.86

709

0.89

791

0.89

060

0.89

106

: .-

Page 16: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

TABL

E III

(T

heor

em

4.3b

)

PI

x 1

P;’

P2

.x2

P;’

delta

: : :: : 8 3 9 10 4 1: 7 1 :“o

11

:: 20

15 ;: 23

54

;: 21

55

2;

32

32

1%

128

128

216

256

343

512

1024

12

96

1728

20

48

2187

21

87

2197

50

625

2 19

936

I 12

589

9906

8 42

624

1 12

589

9906

8 42

624

I 52

168

1143

1 69

024

1 94

619

5068

3 59

315

2 25

179

9813

6 85

248

3 65

615

8440

0 62

976

4 11

724

8169

4 15

651

8 29

350

9467

4 71

872

1oOO

OOOO

oOOo

ooOO

O 11

92

092

8955

0 78

125

18

0143

9 85

094

8198

4 21

91

462

4432

0 20

321

21

9369

5 06

403

7785

6 21

93

695

0640

3 77

856

36

0287

9 70

189

6396

8

9 25

27

15:

121

125

225

243

361

529

tE

1764

20

25

2197

22

09

2209

50

653

2 79

841

I 11

913

0473

1 02

767

0.85

259

1 15

683

1381

4 26

176

0.89

628

1 53

157

8985

2 64

449

0.85

597

1 95

312

5000

0 00

000

0.83

986

2 21

331

4919

0 66

161

0.88

532

3 67

034

4486

9 87

176

0.84

507

4 09

600

0000

0 00

000

0.89

095

8 14

040

6085

1 91

601

0.89

154

9 90

457

8032

9 05

937

0.87

396

12

2005

0 97

657

0582

9 0.

8986

2

17

7147

0 oo

oo0

oooo

o 0.

8909

6 21

61

148

2313

2 84

249

0.88

656

21

9146

2 44

320

2032

1 0.

8169

0 21

61

148

2313

2 84

249

0.88

845

36

5203

4 74

360

5657

6 0.

8873

5

.ooo

oo

0.2

1534

0.

4883

2 0.4

oOOo

0.

2890

6 0.

4057

5 0.

2275

4 0.

4087

6 0.

4669

4 0.

4951

2

0.45

416

0.46

007

0.49

607

0.48

070

0.41

184

0.29

941

0.40

194

0.32

293

0.30

762

0.36

309

Page 17: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

19 3 52

35

:: 15

12

11

3: 1:

3:

19 ! 20

14

17

;i 17

:: 49

20

:; 3”:

;:

23

20

: 3’

: 15

25

42

0529

8 34

622

5705

9 50

03

154

5098

9 99

707

51

1858

9 30

140

9075

7 95

42

895

6661

6 82

176

96

5491

5 73

730

4687

5

155

5680

9 55

578

1222

4 50

5 44

702

8499

2 93

771

558

5458

6 40

832

8400

7 78

9 73

022

3053

6 02

816

789

7302

2 30

536

0281

6 79

9 00

668

5782

8 84

121

2481

15

287

3203

7 36

576

6502

11

142

2497

9 47

648

6568

40

835

5712

8 90

625

6568

40

835

5712

8 90

625

3689

3 48

814

7419

1 03

232

2 43

569

2242

1 60

813

0539

7 2

9514

7 90

517

9352

8 25

856

6 72

749

9949

3 25

600

0920

1 9

3132

2 57

461

5478

5 15

625

41

3954

5 12

236

9384

7 65

625

54

8038

6 85

778

4802

1 85

939

61

4094

2 21

446

4815

4 97

216

94

4473

2 96

573

9290

4 27

392

131

0720

0 00

000

0000

0 00

000

188

8946

5 93

147

8580

8 54

784

377

7893

1 86

295

7161

7 09

568

2392

99

329

2306

1 75

295

9008

3 33

25

2567

3 00

796

5087

8 90

625

1978

4 19

655

6603

1 35

891

2397

9 32

199

0575

5 81

317

9726

8 37

607

1 93

428

1311

3 83

406

6795

2 98

816

2 25

393

4029

0 69

225

8087

8 63

249

10

8347

0 59

433

8837

2 20

418

3025

1 22

95

856

9288

6 98

149

5482

2 20

544

171

6155

8 31

334

5863

4 29

238

9520

1 17

1 90

707

9974

8 42

259

1028

6 58

176

171

9070

7 99

748

4225

9 10

286

5817

6 25

251

1682

9 40

423

4886

1 69

433

5937

5

10

42

4207

4 74

827

7657

6

:: ::,

50

5421

0 65

117

2681

7 50

54

210

6513

7 26

817

96

5491

5 73

730

4687

5 97

65

625

WOO

0 00

000

13

154

4723

7 77

391

1946

1 13

50

4 03

636

1936

4 67

383

,iO

3290

3 17

162

4844

1 78

7 66

278

3788

5 49

761

799

0066

8 57

828

8412

1 78

7 66

278

3788

5 49

761

:: 24

72

1592

1 50

840

1230

3 65

82

9520

0 58

400

3528

1 13

65

02

1114

2 24

979

4764

8 12

65

82

9520

0 58

400

3528

1

28

:i

3725

2 90

298

4619

1 40

625

2 44

140

6250

0 00

000

0000

0 2

9755

8 23

267

5799

4 63

481

6 71

088

6400

0 OO

OOO

OOOO

O 9

2510

3 10

231

5013

6 29

321

41

2906

5 87

69.8

35

408

0153

6 54

60

990

7061

2 05

831

7732

7 61

32

610

4156

8 09

986

4896

1 93

87

480

3376

4 77

543

0564

9 13

0 90

925

5398

6 67

734

3846

4

188

3234

9 19

413

1742

6 09

041

379

2922

7 19

491

3558

8 02

161

2390

72

435

6851

5 13

248

4715

3 33

34

4626

7 95

181

5307

0 88

493

1977

9 85

201

4625

5 88

779

3408

1 32

118

3887

7 95

485

5105

1 57

369

1 93

832

4566

7 68

001

9896

7 96

723

:: 2

2550

1 16

774

1627

4 31

786

8291

1 10

84

280

3560

5 96

593

2354

2 07

744

21

22

9468

2 51

895

1294

0 71

398

7276

8

::,

171

7986

9 18

400

0000

0 00

000

0000

0 17

1 61

558

3133

4 58

634

2923

8 95

201

17

171

7986

9 18

400

OOOO

O OO

OOO

000@

3 18

25

259

9333

5 73

498

0608

1 18

208

0664

9

0.87

619

0.88

076

0.88

656

0.88

631

0.88

575

0.87

497

0.85

578

0.89

708

0.85

579

0.89

216

0.89

710

0.86

739

0.89

872

0.89

414

0.85

892

~1

0.89

721

3 0.

8710

1 2

0.89

800

0.87

486

0.89

638

8

0.87

993

3 0.

8873

0 m

g;

;;;

s 5 0:

8686

3 E;

0.88

695

3 0.

8936

8 0.

8707

1 0.

8912

6 0.

8494

3 0.

8939

0 0.

8940

2 0.

8690

3 0.

8799

1 0.

8751

6

0.89

088

0.89

829

0.88

250

0.88

234

&

Page 18: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

342 B. M. M. DE WEGER

It follows from (4.6) that

hence .X,/X is a convergent of log p,/log p2, say .x2 = rk, X = qk. Since qk is at least the (k + 1)th Fibonacci number, and by XC 1.97 x lOI (from the examples at the end of Sect. 4.A), we obtain k d 92. The lemma now follows from (4.5) and (4.7). 1

To solve (4.1) we computed the continued fraction expansions and the convergents of logp,/logpz exactly, up to the index n such that q, ~ , < 1.97 x lOI < q,!. Lemma 4.2 guarantees that n 6 93. Doing so, we obtained the result,

THEOREM 4.3. (a) The diophantine inequality

(4.8 1

with p,, p2 primes such that p, <p: < 200, and

x ,, .x,EZ, x,22, .x,>2, andeither S=J

or 6 = 6 and min(p-;I, p-z’) > lOI (4.9)

has only the 77 solutions listed in Table II.

(b) The diophantine inequality (4.8) with p,, pz non-powers such that 2 <p, <p2 < 50 and conditions (4.9), has on1.v the 74 solutions listed in Table III.

In Tables II and III, the column “delta” gives the real number with 1~;’ -p’;21 = min(p.;l, ~;z)~~“~. Note that in Theorem 4.3 we do not demand (xi, x2) = 1. The numerous solutions of (4.8) with 6 = & and min(p.;‘, p?) < lOI can be found without much effort. The computations for the proof of the theorem took 35 sec. We computed approximations of log pi by writing it as a suitable linear combination of numbers of the form log( 1 + x) for small x, and evaluating log( 1 + X) by a Taylor series, taking care to avoid mistakes by rounding-off procedures. Thus we computed explicit rational numbers 8,, 0, with

8, <logp,/logp,<e,<e, +E

for a small enough E. Then as far as the partial quotients of the continued fraction expansions of 8, and l32 coincide, they coincide with the partial quotients aj of log p, /logp,. It appeared to be sufficient to take E = lOPso.

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DIOPHANTINE EQUATIONS 343

Note that Lemma 4.2. does not yield a decision if

ak + 1 < pft”’ 1 hP2 --<ak+,+2.

qk 2l”O

Since this gap is relatively small, this situation is unlikely to occur. We met only one such a coincidence, namely for p, = 15, p2 = 23. Here, log15/log23=[0;1,6,2,1,51,... 1, so that a,=51, r,=19, q4=22, and 1522”o & log 19/2 “‘O = 51.4... E [Sl, 53). We have further /1 = 0.002714... < 0002771- = 2’/‘015-22/‘o so (4.3) holds. But (4.1) does not hold, since ldg(1522~2319)/10g(2319j=0.9008... . This example illustrates that (4.3) is weaker than (4.1). Therefore all found solutions of (4.3) have been checked for (4.1) as well.

4.C. The Case t > 3

In this section we show how the L3-BRA can be used to reduce an upper bound for the solutions of (4.1) in the multi-dimensional case. This will enable us to find all solutions of (4.1) for given t >, 3, pl,..., pr and 6.

Let s, v be a solution of (4.1). Put xi = ord,(x/y) (i = l,..., t), and X=max , <, <, IxJ. Let C be an upper bound for X, for example, C = C, (cf. Theorem 4.1). Choose a positive constants y E Z, Co E R, and put

ei = CYCO log Pi1 (i = l,..., t). (4.10)

Consider the lattice Tc Z’, generated by the column vectors of the matrix

Y 0 A= 0 '..

i i

Y 8, ... 8,

Put Ib=x,e, + ... +x,0,. Then

With this notation we have the following useful lemma.

LEMMA 4.4. Suppose that for a solution of (4.1)

14 ’ c lx,1 i=l

(4.11)

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344 U.M.M.DEWEGER

COROLLARY 4.5. Let X0 he u positiw number such that

I(f)3(4r’+(t- l);s’)‘?X,,. (4.13)

Then (4.1) has no solutions with for i = l,..., t,

10g(2’-“yC,/t&)/(1-6)10gp,~ Ix,/ <X”. (4.14)

1 4.4. Put A = log(x/y) = XI= 1 x, log pi. Then

I = i .~,(CK, log P,l - YC” l%P;) 6 .i I-y,/9 ,= I ,=I

Proof of Lemma

/A-yC,A

whence, by (4.11),

IAl 2 /;I/ - i l-K,\ ( YC,)>O. r-1

By (4.2) we infer

XC2 JAI- “{’ 6) < (21 fiyco/( 11, -i, ,uK,l))‘i” --n’.

Now (4.12) follows, since pj’gl 6 max(x, y) =x. 1

Proof of Corollary 4.5. By x # y we have y # 0. Suppose that Ix,/ d X0 for all i. Then

l(lJ26 Iy12=y2 ‘x’ ~2 + i,’ < (t - 1) ?‘Xi + A’. i= I

By (4.3) it follows that

A’>l(f)‘-(t- 1)y2i@4t2X&

and we infer

IA/- f, /xi~~2tXo-tX*=tXo. i= 1

Now apply Lemma 4.4, and the result follows at once. 1

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DIOPHANTINE EQUATIONS 345

We use the corollary to reduce the upper bound C for X as follows. Choose C,, somewhat larger than (tC)‘. The parameter y is used to keep the “rounding-off error” lyCO log pi - oil relatively small. (If C,, is large, then this error is already so small compared to C, that it is safe to take y = 1.) The ei are integers, and are computed exactly. By the L3-BRA we can com- pute a lower bound for l(r) (cf. Lemma 3.1). We may expect that this bound is of size (det(T)) ‘I’, which is about ytC. Thus we may expect that (4.13) holds with X0= C. Otherwise we may try some larger C,. If (4.13) holds, then (4.14) gives bounds for Jxil, and thus for X, of size log(C,/C), which is of size log C. Hence the reduction of the upper bound is con- siderable indeed. Lemma 4.4 is more precise than its corollary, and therefore more suitable for reducing a small bound C.

We now proceed with an elaborate example. Let t = 6, p, ,..., p6 = 2 ,..., 13, and 6 = I. By the example at the end of Section 4.A, we know that X< C for C = 1.35 x 1036. We take C, = 10Z4’, y = 1. The values of the Bi were computed exactly. We applied the L3-BRA to the corresponding lattice I-,, and found a reduced basis c, ,..., c6 with (c, 1 > 9.40 x 1039. So Lemma 3.1 yields

I( r, ) > 2 ~ 5’2 x 9.40 x 1o39 > 1.66 x 1039.

This is larger than ,,/% C= 1.64... x 1037, so (4.13) holds with X0 = C. Hence, by Corollary 4.5,

X< log(2”2 x 10240/6 x 1.35 x lO36)/$ log 2 < 1350.4,

so X< 1350. Next we choose Co = 1032, y = 1, and C = 1350. The reduced basis of the corresponding lattice T2 was computed, and we found (c, 1 > 2.71 x 105. Hence I( r,) > 4.79 x 104, which is larger than ,,/% C = 1.64... x 104. So (4.13) holds for X0 = C, and Corollary 4.5 yields

lx,1 < 187, 1x21 < 118, 1x31 6 80,

1x41 G 66, 1x51 < 54, IXJ Q 50. (4.15)

Next we choose Co = 1012, y = 104. We use Lemma 4.4 as follows. If IA1 > lo6 then (4.11) holds by (4.15) and (4.12) yields

1x11 d67, ix21 G 42, 1x31 6 29,

ix41 d 24, 1x51 < 19, 1x61 d 18. (4.16)

All vectors in r3 satisfying (4.15) and IL1 < lo6 can be computed with the algorithm of Fincke and Pohst [7] (we omit the details of the com-

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346 R. M. M. DE WEGER

putations). We found that there exist only two such vectors, but they do not correspond to solutions of (4.1). Hence all solutions of (4.1) satisfy (4.16). Next we choose C, = lo*, 7 = 104. If /iI > 5 x lo’, then (4.12) yields

TABLE IV (Theorem 4.6)

-1 0

21 1

19 6

-2 11

1 -22

13 1 3

-26 3 8

25 -6

8 1

-4 -4

16 -8 -5

-25 2

- 14 -24

-5

2 18

7 - 10

-11 -1 0 6 0 4 5 I -6 0

-2 -2 -1-3 0 13 -1 -3 -1 -2 0 0 -8 1 0 2 -I 1-6 3

15 -1 -2 -4 0 -15 0 2 I 1

8 -1 -8 0 3 5 1 -I 1 3

1 3 -1 I -6 2 9 -4 -4 0 3 0 4 2 -7 10 530

-13 10 -2 0 0 -2 -10 4 1 1

1 -4 o-5 0 1 -2 -6 0 7

-13 0 3-2 3 -13 -3 7 2 0

-I -4 l-4 I 2 -II 2 6 0

-3 5 1 -1 -6 8 0 -8 3 2

-2 -5 11 0 -3 I 1 o-2 5 0 13 -9 -2 0

19 -2 -4 1 -1 -1 -2 12-l 0

5 10 0 1 -8

-4 -9 3 7 -2 7 0 -13 0 2

-5 3 -9-3 8

17 71561 17 71470 91 17 71875 17 71561 314 20 97152 20 96325 827 31 88646 31 88185 461 51 67168 57 64801 2367 88 58304 88 57805 499

143 48907 143 48180 727 143 50336 143 48907 1429 288 29034 288 24005 5029 293 62905 293 60128 2777

337 92000 337 87663 4337 351 56250 351 53041 3209 627 52536 627 48517 4019 671 10351 671 08864 1487 781 25000 781 21827 3173 878 95808 878 90625 5183

1006 63296 1006 56875 6421 1882 45551 1882 38400 7151 1929 14176 1929 13083 1093 1992 97406 1992 90375 7031

4392 39619 4392 3OOKI 9619 7812 58401 7812 5oooO 8401

14336 00000 14335 62273 37727 14758 24779 14757 89056 35723 19773 26743 19773 oooo0 26743 40600 88955 40600 86272 2683 48828 12500 48827 86447 26053

1 27848 76137 1 27848 44800 31337 1 38412 87201 1 38412 03200 84001 2 61035 15625 2 61033 83072 I 32553

2 67363 98612 9 68892 08832

1305 16915 36COO

2 67363 28125 70487 9 68890 10407 1 98425

1305 16881 72831 33 63169 2834 49760 00000 41 04623 10 -6 5 -6 4 2834 49801 04623

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DIOPHANTINE EQUATIONS 347

There are 143 vectors in r4 satisfying (4.16) and 121 < 5 x 105. Of them, 2 correspond to solutions of (4.1), namely the vectors with

(x ,,..., x6) = (7, -5, 3, -9, -3, 8) ,I = 257674,

(x , ,..., x6) = (- 10, 10, -6, 5, -6, 4) A= 144817.

Both also satisfy (4.17). Hence all solutions of (4.1) satisfy (4.17). At this point it seems inefficient to choose appropriate parameters Co, y

to repeat the procedure with. But the bounds of (4.17) are small enough to admit enumeration. Doing so, we found 605 solutions of (4.1). We cannot list them all here. Instead we give the following result, from which the reader should be able to find all solutions without much effort.

THEOREM 4.6. The diophantine inequality

o<x-y<y’J’

in x, yES={2.“... 13”?xi~Z2, xi20 (i= l,..., 6)) with (x,y)= 1 has exactly 605 solutions. Among those, 571 satisfy

ord,(xy) < 19, ord,(xy) d 12, ord,(xy) < 8,

ord,(xy) d 7, ord,,(xy) < 5, ord,,(xy) d 5.

The remaining 34 solutions are listed in Table IV.

The computation of the reduced basis of rI took 113 set, where we applied the L3-BRA as we described it in Section 3, in 12 steps. The direct search for the solutions of (4.17) took 228 sec. The remaining computations (computation of the log pi up to 250 decimal digits, of the reduced basis of r2, and of the short vectors in r3 and r,) took 8 sec. Hence in total we used 349 sec. The numerical details can be obtained from the author.

5. A DIOPHANTINE EQUATION

Let p1 < ... <pr be prime numbers, where t > 3, and let S be the set of all positive rational integers composed of those primes only. In this section we study the exponential diophantine equation

x+y=z (5.1)

in x, y, z E S. Without loss of generality we may assume that x, y, z are relatively prime. For any a E S we define

m(u) = ,yy, or-d,(a). .,

64112613-S

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348 R.M. M. DE WEGER

It was proved by Mahler that m(xyz) is bounded for the solutions of (5.1). An explicit bound can be computed (cf. Shorey and Tijdeman [ 18, Corollary 1.21). We do so in Section 5.A. In Section 5.B we introduce multi-dimensional p-adic approximation lattices, and in Sections 5.C and 5.D we show how to reduce the found upper bound, and to solve (5.1) completely, in the cases t = 3 and t 3 4, respectively. We conclude with some remarks on a conjecture of Oesterlt and Masser, which is related to Eq. (5.1) in Section 5.E.

5.A. Upper Bounds

THEOREM 5.1. In the above notation, put

s = [2t/3], p=p, “‘P,,

V, = max(e, log pi) (i’ I,..., t), Q= V,p,r+,.‘. v,,

G,=2, G,=6, GDx=p,- 1 ifp,>5, G= ,~~~tGp,llOg~i, .-.

C, = 29” + 26ss+ 4Q log(e V, , ).

Choose p, K with 2/(s + 1) < p 6 2, 0 < K < p/2, and put

E=(P--K)/(l +K)(l +PL)(J+ I),

k=max{(l6s) (I+l/X)(S+-I), (8/E)(1+Pc)(S+i), 161/C},

C,=4(s+ I)(s+‘)kcl+p’GfZ, ~,=4(~,+~~~~~(~/P,))fl~~Pl~

c, = Cdl% c, J2, C,. = max( C,, C,(log C,)2).

Then all solutions of (5.1) satisfy m(xyz) < Cl,,.

ProoJ If we consider instead of (5.1) the equivalent equation

x f y = 2, (5.2)

then we may assume that xy has at most s prime divisors. Suppose that m(xy) < 7. Then

hence

py(‘) d z < 2 max(x, y) < 2(P/p, I’,

m(z) G ww/P,)‘)/l% PI 3

and it follows that m(xyz) < C,,. Now let m(xy) 2 8, and suppose m(z) > 2. Then for some p =pi,

m(z) = ord,(z) = ord,( +x/y - 1) = ord,(log,(x/y)).

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DIOPHANTINE EQUATIONS 349

Put x/y = pi, +*..,>. Then m(xy) =max((xi,l,..., IxJ). On applying Lemma 2.2 we obtain

m(z) < C,(log m(xy))2. (5.3)

Obviously (5.3) is also true if m(z) < 2. If in (5.2) the plus sign holds, then

(P/p,)““‘> z > max(x, y) >pyC”?‘).

By (5.3) it then follows that

MXY 1-c G l”z;E1) (log m(xy))! (5.4)

If in (5.2) the minus sign holds, then we apply Lemma 2.1 as follows. Suppose that

4XY) l%P, 3 (Cc + c7 logwP,)wg WY))‘. (5.5)

Then it follows that

C,~log~(xY))210g(~/P,)6~(xY)logP,-log2.

Note that by (5.3)

so that /y/x- II<+. Hence

I 1 WA) C7hw(xv)Y

llog(Ylx)l G 2 log 2 5-l 62log2 mkv) Pl

On the other hand, Lemma 2.1 yields

lb3(ylx)l >expi - G(log I + log(e~,))}.

Thus we obtain

WY) l%P, < wag wY))2 1%wP,) +log(2 log 2)

+ G(log m(v) + log(eV,)).

Obviously,

G(log m(w))* > log(2 log 2) + G(log WY) + log(eV,)),

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350 B.M. M. DE WISER

and we have a contradiction with (5.5). So from (5.4) or from the negation of (5.5 ) we infer

and from Lemma 2.3 we obtain rn(-~y) < C,. Now the result follows from (5.3). I

EXAMPLES. With t = 3, p, = 2, pz = 3, p3 = 5 we find a minimal value for k i+lc on taking p= 1, K-A, namely k’+jf=2’08. Then C,,<6.75 x 104’. With t=6, p ,,..., pb=2 ,..., 13 we take p=l, ti=+, and we find c <337x 107’. 10 .

5.B. Approximation Lattices

In [23] the concept of (2-dimensional) approximation lattices of a p-adic number was introduced. In this subsection we generalize this notion to multi-dimensional approximation lattices of a linear form of p-adic num- bers. We confine ourselves to the particular lattices that we use for solving Eq. (5.2), and indicate how a basis of such a lattice can be computed explicitly.

Let p be any of the primes p, ,..., pt. We may assume that p [ ~JY Rename the other primes as po,...,p,~2, such that ord,(log,(p,)) is mmimal. For i = l,..., t-2 and mEN, put

% m-l

@;= -lo~p(P~)llo~p(PO) = 1 ui.l P’, l9j’“’ = c u,.,p’, /= I I= I

where u,,, E (0, l,..., p - 13. Then 0, is a p-adic integer by the choice of po, and 6!“’ is the unique rational integer satisfying ord,((!Ij - Sj”)) > m and 0 d 0:‘) <pm. The Ojmr can be computed for the desired m by using the Taylor series for the p-adic logarithm,

log,(~)=$log,(I+(%~-l))=~ i (-1)“’ (Xk - 1 ,‘/I, /= I

where k is the smallest positive integer such that ord,(Xk - 1) 2 1. Consider the lattice r, c Zr-’ generated by the column vectors

b ,,..., b,_*, b, of the matrix

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DIOPHANTINE EQUATIONS 351

Put m, = ord,(log,(p,)). Then

rt?l= {(x I)...) X,&Z, x0) E Z’- l: 1x,8,+ .‘. +X,_*e,-*-X&<p-m}

= {c%...,xt-2r x,)EZ’-‘: (log,(p~...p:L-:)l,~~--‘“+“o’}.

We call such a lattice an approximation lattice of the p-adic linear form x, 8, + . . . + X, _ 2 6, ~ 2. For t = 3 we have exactly the approximation lattice of 0, in the sense of [23]. (Note that there is a different matrix notation there). Further, consider also the set

It is clear that rz is a sublattice of r,,,. In general, the two are not equal, since (x ,,..., x,+~, x0) E r,,, only implies p$. . . p::; = i (mod pm +“‘O) for some (p - 1)th root of unity 5, not necessarily + 1. (Recall that log,([) = 0 if and only if [ is a root of unity). For p = 2, 3 the only roots of unity in Q, are +I, so then rz=r,,,.

For p > 3 we show how a basis b:,..., b:- *, b,* of rz can be computed from a known basis b, ,..., b, ~ 2, b, of r,,,. Let < be a primitive (p - 1)th root of unity in Qep. For any x = (x, ,..., x, ~ 2, x0) E r, we define k(x) E: Z by

P?.. p:L-; E jW (mod pm + "o), O<k(x)<p-2.

Then k( x ) is (mod( p - 1)) a linear function on r,, and x E f 2 if and only if $(p - 1) ( k(x). Put

k = gcW(bo),..., k(b, ~ 2)h

and compute (by the euclidean algorithm) a basis b;,..., b;- 2r bb of r, such that k(bb) = k. Put for i = l,..., t - 2,

y,r k(bj)/k (mod(p - 1)/2), IYil 6 (P- I)/43

b* = b/ - y,bb.

Then k(bT) z k(bj) - yik(bb) = 0 (mod(p - 1)/2) (i = l,..., t - 2). Put

y. = lcm(k (P - I)/2 )lk,

which is the smallest positive integer such that y,k = 0 (mod(p - 1)/2). Every x E r, can be written as

x=y,b:+ ... +y,~zbT_2+yob;, Y,E&

since bf ,..., b:_ 2, bb is a basis of I-,,,. Now,

k(x) = y,k (mod(p - 1)/2).

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352 B. M. M. DE WEGER

So x E fz if and only if ‘J(, ( yo, Hence put

then it follows that b:,..., bT_ 2, b,* is a basis of r,$ In practice it may occur that p0 can be chosen such that it is a primitive root (modp). Then choose i 5 p0 (mod p), and it follows from k(b,) = 1 that b: = b, (i = O,..., t - 2). If pi sp; (mod p), then

m-1 y,ra,+8]“‘=ai+ 1 u,,, (mod(p- 1)/2) (i= l,..., t-2).

/=O

Yo = (P - 1 J/2.

So in this special case it is simple to find a basis of L’,$ For any solution x, y, z of (5.2) put x/y =p$ . . p;‘~:;, and consider the

point x = (x, ,..., xrM2, xO)e Z’- ‘. Suppose that ord,(z)82. Then

ord,(z) = ord,(x/y k 1) = ord,(log,(x/~v))

=ord,(x,8,+ ... +~,~~~,~~-x~)+m~.

Hence x E rz for m < ord,(z) - m,. With this notation we have the follow- ing useful lemma:

LEMMA 5.2. Let me N and X,>O be constants such that

l(l-2)> (t- l)“Z x0.

Then (5.2) has no solutions x, y, z with

(5.6)

m + m, 6 ord,( z) 6 m(xyz) < X0. (5.7)

ProoJ Suppose (5.7) holds. By ord,(z) B m + m, we have x E I% x # 0. Further, ixil < m(xyz) < X0 (i = O,..., t - 2). Hence

r-2

l(r;)2dIx12= c xfd(t-1)X& r=O

which contradicts (5.6). 1

Suppose that we know that m(xyz) < X0. We may expect that l(rz) is of size (det(rz))“‘‘-*‘, which is about p *A- ’ ), Thus we may expect that it

will suffice to take m somewhat larger than (t - 1) log(m X,)/log p. If (5.6) does not hold then, we may try some larger m. If (5.6) holds, then (5.7) yields ord,(z) <m + mo. We repeat this procedure for p =p, ,..., p,. Since (5.2) is invariant under permutation of x, y, z we find a new upper bound for m(xyz), which is of size m N log X0.

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DIOPHANTINE EQUATIONS 353

5.C. The Case t = 3

We illustrate the use of the p-adic analogue of the one-dimensional con- tinued fraction algorithm by solving the equation

xfy=wz, (5.8)

wherex,y,zE(pk:p=2,3,5,kEiZ,k~0},andwEZ, lw1<106,(w,z)=1. Put X = maxp = 2,3,5 ord,(xyz). The example at the end of Section 5.A shows that in the case 1 WI = 1 we have. X-C 6.75 x 104’. It can be checked without difficulties that the effect of the w with IwJ d lo6 can be neglected, so that for the solutions of (5.8) also X < 6.75 x 104’ holds. Put for p = 2, 3, or 5,

such that

0 = -log, PI /log, PO

is a p-adic integer. Then define the lattices r,,, and rz as in Section 5.B, so r,,, is generated by

b,=(D:“,). bo=(,“:)

For p = 2, 3, we have r; = r,, and for p = 5 a basis of r;? is

b:=b,-yb,,, b,* =2b,,

where y = 0 if t? (m) is odd, y = 1 if 8 ‘m) is even. Using the algorithm of [23, Sect. 31, we can compute a basis c, , c2 of rz that is reduced in the sense that

Cl.1 c, = ( ) Cl.2

has minimal norm @(c,)=max(lc,,,l, Jc~.~/) in r:\(O). We choose p, po, p,, and m as in the foliowing table, where m is chosen so that pm is somewhat larger than (6.75 x 1041)‘:

P p. pi m. m y O(q)> u< w 1x01 d 1x11 G

2 3 5 2 297 2 148 298 1O6x2298 222 152 3 2 5 1 189 3 94 189 lo6 x 3’89 354 152 5 2 3 1 135 0 567 135 lo6 x 513s 370 233

We give the values of eCrn) in Table V, and the reduced bases of the rz in Table VI. From this table we find the lower bounds for @(c,) given above.

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TABL

E V

(Sec

tion

5C)

(p-a

dic

nota

tion:

O

.abc

...

= a

+ b.p

+

c.p2

+ .

)

thet

a =

-log,

5/

lag,

3

= 0.

1010

1 11

101

cm01

11

110

llooo

10

101

OOOC

Q 01

001

1110

1 00

010

1oOG

.I 10

011

1011

0 10

000

0101

1 11

100

OOOo

l 11

010

@Jo

00

0000

1 00

010

1110

0 11

100

1011

1 01

001

0110

1 11

m

0101

0 01

110

0101

0 11

110

OwoO

00

101

0011

0 00

100

0001

1 01

111

0111

0 01

010

1110

1 10

010

0100

1 00

001

1010

0 00

111

0000

1 11

111

0111

1 00

011

1011

0 OO

OOO

0010

1 01

101

0110

0 00

010

0111

0 11

100

1110

1 11

011

0101

1 10

1 11

o.

...

thet

a =

-log,

S/

log,

2

= 0.

1102

2 12

121

2200

1 12

010

2110

2 10

210

1002

2 20

210

2001

0 10

112

2220

1 21

021

2102

2 10

000

2202

0 12

012

0202

2 21

001

0001

2 02

020

2121

0 12

202

1220

0 00

000

1012

0 00

211

1202

1 10

120

0210

0 10

222

2212

2 01

201

2111

1 11

121

1100

1 20

222

1000

0 20

121

2?22

1....

thet

a =

-log,

3/

lag,

2

=

0.33

002

0200

3 04

411

2312

0 44

012

0101

1 00

044

4320

4 30

340

0002

3 14

333

1241

3 43

420

4030

2 10

202

4410

4 32

433

-144

32

030-

71

1231

1 34

044

4023

1 04

112

3323

0 00

242

1423

2 14

400

3110

....

Page 31: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

TABL

E VI

(S

ectio

n 5C

)

p =

2(ba

se-2

no

tatio

n)

(

-101

11

001

0001

1 11

100

1010

1 01

010

0111

0 00

101

1111

0 10

000

1100

0 11

100

1001

1 11

001

0111

0 10

000

1000

1 00

011

1111

0 01

100

1001

1 01

101

0100

0 11

110

0100

1

b,=

1011

1 10

010

1100

1 10

111

-101

0 11

101

0101

1 00

011

0000

0 00

111

0100

1 11

001

1001

0 10

110

0011

1 00

001

0110

0 11

101

0001

1 01

001

0010

0 10

101

1101

1 10

011

ml1

01

011

0101

1 00

110

1111

1 11

110

1010

0 01

010

0110

1 00

011

1011

lo

o01

1010

1 01

101

1011

1 10

101

0111

1 10

111

0000

1 01

010

1101

0 10

001

1001

1 11

111

0110

1 10

110

0110

0 11

001

0010

1 lo

000

0001

1 00

100

0101

0 00

101

0101

1 10

000

0011

1 01

100

1010

1 10

110

bz =

-1

010

0110

1 01

110

1111

0 10

000

OOOO

O 10

001

llooo

10

111

0010

0 01

010

1100

0 11

110

1110

1 10

101

1111

0 00

011

1010

1 01

000

1101

0 01

001

0110

0 10

111

0011

0 10

101

0111

0 11

100

0001

0 11

110

1001

0

p =

3 (b

ase-

3 no

tatio

n)

3

b,

= 11

10

0010

2 10

211

2201

0 20

100

1002

2 10

001

2022

0 12

202

0110

2 11

202

1111

0 20

112

2210

1 20

022

2211

2 10

020

1002

2 12

000

%

2000

11

110

2111

1 10

112

1010

2 10

210

2210

1 01

021

1021

2 10

020

1100

2 12

222

2001

2 11

101

0122

2 11

211

0120

1 22

201

02oo

o

bz=

(-120

01 21

02

2111

2 02

001

2210

1 01

111

0022

2 12

000

2002

2 02

011

1220

0 21

221

0020

0 00

120

2021

0 s

0110

0 12

216

1022

2 00

022

2020

2 =!

10

002

2101

0 00

220

1020

0 02

122

0022

1 22

120

2210

0 10

120

1202

0 20

120

2201

2 10

011

1002

2 01

122

2222

2 22

212

0200

1 iF

.4

p =

5 (b

ase-

5 no

tatio

n)

m

b, =

-2

13

2104

1 20

044

2101

1 03

000

0042

0 40

302

1314

4 33

303

4214

3 22

021

3123

3 42

233

4231

4 -1

40

0122

1 40

144

1032

3 41

221

1011

3 13

410

4414

4 41

032

2113

1 43

034

4032

2 11

323

4302

2

b2 =

-2

00

2001

2 43

403

1323

2 12

424

4410

2 00

032

4232

1 20

012

1413

4 22

130

2010

3 00

020

1330

1 23

3 01

424

4201

3 24

004

4324

4 32

120

3023

0 23

141

2234

0 40

304

3111

3 30

442

3344

3 20

012

Page 32: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

356 B. M. M. DE WEGER

They are all larger than 6.75 x 10“‘. Hence for the solutions of (5.8) we have u <m + m,, and Iu’I z 6 W, as shown in the table above. We now find the new upper bounds for IxOI, lx,/ as follows. If in (5.8) the minus sign holds, then, on supposing that min(x, y) > W”19, we infer

Ix-),1 = IkV 16 W<rnin(,~,4:)~“~.

By Theorem 4.3a and Table II, the inequality Ix - yl < min(x, Y)~“’ has no solutions with min(x, y) > W, since W> 10’“. Hence min(x, v) < W10i9, and we infer

max(x, y) < min(x, y) + 1~11 z < W’O” + W.

If in (5.8) the plus sign holds, this inequality follows at once. So now the bounds for 1x01, lx11 follow from

lxij log pi < log max(x, y) 6 log( W1o’9 + W).

We repeat the procedure with m as in the following table:

P m Y @(Cl)> UG W 1x01 6 I-~‘1 6

2 17 260 18 106x2’* 31 21 3 13 531 13 106x 3” 49 21 5 8 1 818 8 lo6 x 58 49 31

The numbers are now so small that the computations can be performed by hand. For example, for p = 5 the lattice f,* is generated by

b: = (-3:,,,7)9 b,* = i7xp250)~

and a reduced basis is

c’=( ,:k), c2=(y$ Now, in all three cases, W”” < 1015. On supposing min(x, y) > 1015 we infer

lx-y1 = Iwl 26 W< 1015X9~10<min(x,y)9~10.

By Theorem 4.3a and Table II we see that the inequality Ix - yl < min(x, Y)~“’ has only two solutions: (x, y) = (265, 5**), (284, 353). However, both have Ix -yl > 10” x9’*o. So we infer min(x, y) < 1015, hence by max(x, y) d 1015 + W we obtain the bounds for 1x01, (~‘1 as given above.

Page 33: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

DIOPHANTINE EQUATIONS

TABLE VII (Theorem 5.3)

p=2,p,=3,p,=5

357

XI -

P;’

2 9 10 9165625 10 59049 10 9765625 4 81 12 244140625 6 729 10 9765625 2 9 8 390625 6 729 8 390625

10 59049 8 390625 14 4182969 10 9165625 4 81 8 390625 0 1 8 390625

8 6561 8 390625 0 1 6 15625 4 81 6 15625 8 6561 6 15625 6 129 6 15625 2 9 4 625 2 9 6 15625 0 1 4 625 4 81 4 625 0 1 2 25

2 9 1 3 1 3 2 9 3 27 4 81 4 81 6 729 6 729 3 27

5 243 5 243 I 2187 6 129 7 2187

11 177147 3 27 8 6561 I 2187 8 6561

2 1 3 0 1 0 2 2 4 3

3 1 5 0 1 1 5 0 3 4

25 5

125 1 5

25 25

625 125

125 5

3125 1 5 5

3125 1

125 625

sign u w

-1 4 -610351 -1 4 -606661 -1 9 -416831

-1 5 -305153 -1 3 -48827 -1 3 -48737 -1 3 -41447 -1 7 - 38927 -1 4 - 24409 -1 5 - 12207

-1 6 -6001 -1 3 - 1953 -1 3 - 1943 -1 3 -1133 -1 4 -931 -1 3 -17 -1 8 -61 -1 4 -39 -1 5 -17 -1 3 -3

-1 4 -1 1 3 1 1 I 1

-1 3 1 1 5 1

-1 4 5 -1 3 1 -1 6 11 -1 3 13

1 3 19

1 4 23 1 3 31 1 6 83

-1 3 91 1 4 137 1 10 173 1 4 197

-1 5 205 1 3 289

-1 4 371

Table continued

Page 34: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

358 B. M. M. DE WEGER

TABLE VII (Theorem S.3)pCon~inud

I 3 5 243 9 19683 8 6561

10 59049 5 243 9 19683 9 19683

IO 59049 I2 531441

I 3 9 19683

13 1594323 IO 59049 10 59049 12 531441

3 27 I 2187

11 177147 3 27

II 177147 11 177147 12 531441 12 531441 12 531441 II 177147 13 1594323 1 2187

14 4782969 13 1594323

13 1594323 1 3 5 243 9 19683

14 4782969 13 1594323 15 14348907 14 4782969 14 4782969 14 4782969

16 43046721 9 19683

15 14348907

3125 3125

125 25

15625 78125

-1 -1

1

5 2 4

4 0

9

9 2

8 4 0

0 11

3125 25

625

78125 78125 78125

625

390625 78125 78125

3125 1953125

125 78125

-I

-I -I

-I 15625

25 1953125

125 1953125

25

3125 1953125 1953125 1953125

I5625 1953125

3125 390625

625

-1

-I

48828125

3 391 3 421 5 619 3 817 5 I357 5 2449 3 2461 3 2851 4 3689 7 4147

4 4883 4 6113 8 6533 3 7303 3 7381 4 8801 3 9769 3 10039 4 11267 7 15259

3 22159 3 31909 4 33215 3 64477 3 66427 5 66571 4 99653 4 122207 5 149467 3 199291

3 199681 3 244141 3 244171 3 24660 1 4 297959 3 443431 5 448501 3 549043 3 597793 3 597871

6 672605 6 763247 4 896807

Table continued

Page 35: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

TABLE VII (Theorem 5.3+Continued

p=3,po=2,p,=5

14 16384 10 9 512 9 4 16 8

12 4096 6 7 128 5 2 4 4 1 2 2 5 32 1 6 64 3

11 2048 4

9765625 1953125 390625

15625 3125

625 25

c

12; 625

9 512 10 1024

3 8 15 32768 14 16384 17 131072 16 65536

8 256 19 524288 18 262144

1 25

15625 125

5 78125

3125 78125

25 1

23 8388608 1 5 13 8192 8 390625 22 4194304 8 390625 10 1024 11 48828125 18 262144 9 1953125 20 1048576 4 625

0 1 9 1953125 21 2097152 6 15625

5 32 10 9765625 24 16717216 3 125

23 8388608 10 9765625 26 67108864 7 78125

-1 4 - 120361 -1 3 -72319 -1 3 - 14467 -1 3 - 427 -1 4 -37 -1 3 -23

1 3 1 -1 3 1

1 3 7 1 5 11

1 3 19 -1 3 37

1 4 193 -1 4 403

1 3 607 -1 3 1961

1 3 2543 1 3 2903 1 4 6473

-1 3 9709

-1 6 11507 1 3 14771

-1 5 15653 1 7 22327 1 4 27349

-1 3 38813 1 3 72338 1 3 78251 1 3 361691 1 3 621383

1 3 672379 1 4 829469

p=5,po=2,p,=3

12 4096 16 43046721 -1 3 -344341 5 32 15 14348907 -1 3 -114791 7 128 1 3 -1 3 1 6 64 8 6561 1 3 53

14 16384 2 9 -1 3 131 13 8192 9 19683 1 3 223 20 1048576 10 59049 1 3 8861 21 2097 152 3 27 -1 3 16777

Page 36: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

360 B. M.M. DE WEGER

Those bounds are small enough to admit enumeration of the remaining cases. Thus we obtain the following result.

THEOREM 5.3. The diophantine equation

where x = p$, y = p ;I, = = p”, (P, po, P[) = (2, 3, 5), (3,2, 5), or (5,2,3), -x0, x, , u are nonnegative integers, M’ E L, 1 w( < 106, and p l w has exactly 29 1 solutions for p = 2, 412 solutions .for p = 3, and 570 solutions for p = 5. In Table VII all solutions with u > 3 are given.

The computer calculations for the proof of this theorem took 3 sec.

5.D. The Case t 3 4

In this section we present an elaborate example of the use of the L3-BRA for solving an equation of type (5.2) in the multi-dimensional case. Let S be the set of positive integers composed of the primes 2, 3, 5, 7, 11, 13 only. In the example at the end of Section 5.A. we have seen that the solutions x, y, ZE S of (5.2) satisfy m(xyz)< 3.37 x 1073. We show how to reduce this bound, and thus we are able to find all solutions. With the notation of Section 5.B we choose the following -parameters:

P PO PI Pz P3 P.% m. m 70 YI Y1 Y3 Y4

2 3 5 7 11 13 2 1320 3 2 5 7 11 13 1 840 5 2 3 7 11 13 1 600 2 1 0 0 1 7 3 2 5 11 13 1 480 3 0 O-l 1

11 2 3 5 7 13 1 360 5 -2 -1 2 0 13 2 3 5 7 11 1 360 6 3 1 -2 1

We computed the six values of the 0j”’ (i= 1, 2, 3,4), and the reduced bases of the six lattices r,$,. Thus we obtained

P l(r, *)> lc,l/4> ord,(xyz)d

2 6.34 x 107’ 1321 3 2.50 x 1O79 840 5 2.02 x 1os3 600 7 2.39 x 1oso 480

11 2.28 x 1O74 360 13 4.23 x 107’ 360

Page 37: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

DIOPHANTINE EQUATIONS 361

These lower bounds for l(r2) are all larger than ,,,$ x 3.37 x 10”. So we may apply Lemma 5.2 with X0 = 3.37 x 1073, which is the theoretical upper bound for m(xyz). For every p we thus find ord,(z) <m+m,. Since Eq. (5.2) is invariant under permutations of x, y, z, we even have ord,(xyz) < m + m,, as shown in the above table. Hence m(xyz) < 1321.

We repeated the procedure with X0 = 1321 and m as in the following table. After computing the reduced bases of the six lattices rz we found the following data (Note that in all cases l(Z-2) >fix 1321.)

P m YO YI Y2 Y3 y4 /(Cl) > ord,(xyz) <

2 77 8342 78 3 49 9026 49 5 35 2 0 1 1 0 22325 35 7 28 3 0 -1 1 0 14403 28

11 21 5 1 1 1 -2 5162 21 13 21 6 0 0 1 2 14779 21

Hence m(xyz) < 78. Next, we repeated the procedure with X0 = 78, and m as in the following table. We found

P m y. yI y2 Y3 Y4 Z(rz) > ord,(xyz) 6

2 55 364 56 3 35 301 35 5 25 2 1 1 1 0 622 25 7 20 3 1 -1 1 0 693 20

11 15 5 1 2 -2 -2 192 15 13 15 6 1 0 3 2 658 15

Hence m(xyz) < 56. To find the solutions of (5.2) with ord,(xyz) below the bounds given in

the above table, we followed the following procedure. Suppose that we are at a certain moment interested in finding the solutions with ord,(xyz) <f(p), where f(p) is given for p = 2,..., 13. Choose a p and an m<f(p) - mo, and consider the lattice l$. If a solution x, y, z of (5.2) exists with ord,(z) 2 m + m,, then the vector

Xl

xc : (9 x4

x0

Page 38: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

362 B. M. M. DE WEGER

TABLE VIII (Section 5D)

P m n P m n

2 3 5 I

II 13

2 3 5 I

11 13 2 3 5 7

11 13

2 2

2 2 2 2 2 2 2 2 2 3

3 3 3 3 3 3 5 5

44 28 20 16 12 12 33 21 15 12

- 7 7

11 11 11

9 9

22 14 10

8 6 6

21 20

13 13 13

- 2 2 2 2 2 2 2

19 18 17 16 15 14 13 12 11 13

- 3 1 3

2 5 2 5

5

12 .- 5 11 7

10 1 7

9 1 7 8 1 11 I 6 11

9 - 13 8 13

7 6 5 7 6 5 4 5 4 3

5 4 3

10 9 8 7 6 5 4

3 6 5 4 3 2 1 4 3 2

1 3 2 I 2 I 2 1

6

1 4 -.

1 4

1 2 3 6

15 16 26 31

44 5 8

16 35 54 87

1 5

18

36 -

6 18

1 8 -

4

(n = number of solutions found.)

Page 39: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

DIOPHANTINE EQUATIONS 363

with x, = ord,,(x/y) (i = O,..., 4) is in the lattice. Its length is bounded by (f(Po)*+ ... +f(p4) 1 . * “* All vectors in rz with length below this bound can be computed by the algorithm of Fincke and Pohst [7] (we omit details). Then all solutions of (5.2) corresponding to lattice points can be selected. Then we replace f(p) by m + m, - 1, and we may repeat the procedure for newly chosen p, m.

We performed this procedure, starting with the bounds for ord,(xyz) given in the above table for f(p), and with p, m as in Table VIII. At the end we have f(2) = 4, f(p) = 1 for p = 3,..., 13. The remaining solutions can be found by hand. Thus we obtained the following result.

THEOREM 5.4. The diophantine equation

x+y=z

in x, y, ZES= {2”‘... 13”6: x~EZ, xi>0 (i= l,..., 6)) with (x,y)= 1 and x 6 y, has exact1.v 545 solutions. Of them, 514 satisfy

ord,(xyz) 6 12, ord,(xyz) 6 7, ord,(xyz) 6 5,

ord,(xyz) < 4, ord,,(xyz) < 3, ord,,(xyz) < 3

The remaining 31 solutions are given in Table IX.

The computer calculations for the proof of this theorem took 2856 set, of which 2830 set were used for the first reduction step. In this first step, we applied the L3-BRA in 12 steps (cf. Sect. 3) which costed on average about 400 sec. The remaining 430 set were mainly used for the computation of the 24 8j”)‘s. Full numerical details can be obtained from the author.

5.E. Examples Related to the Oesterk-Masser Conjecture

Let x, y, z be positive integers. Put

G= n p. p I x.vz

p prime

For all x, y, z with (x, y) = 1 and

x+y=z

we define

c(x, y, z) = log z/log G.

Page 40: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

364 B. M. M. DE WEGER

00C40N000c-l- 0000 -00000 rrmooorl-oom -

3000000~~-0 o--oomo--0 OOOdONOONO 0

b--0-0000- NC-IO- OONPlOO 0000 -oocfoo -

0m0m000000 o-00m000miD 0-60000-00 0

000N006000 oowoo-o+oo -000000000 0

30000606’~ 000m000mb0 00*0000000 0

Page 41: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

DIOPHANTINE EQUATIONS 365

TABLE X (Section 5E)

x Y z 44 Y, z)

121=112 48234315 = 3256?3 48234496 = 2=‘23 1.62599 1 4314 = 2 37 4375 = 547 1.56789

343 = 13 59049 = 3 ‘O 59392 = 2”29 1.54708 198425 = 5=7931 96889010407 = 713 96889208832 = 2’*3’13’ 1.49762

121=11’ 255879 = 3913 256000 = 2”53 1.48887 37 32768 = 215 32805 = 3*5 1.48291

3200 =2’S 4823609 = 7641 4826809 = 136 1.46192 1 2400=2s3 52 2401 = I4 1.45567

702021632 = 21913 103 1977326743 = 7” 2679348375 = 3”531 1” 1.45261 1 512000 = 2125’ 512001 = 357243 1.4433 1 1 19140624 = 2437547 19140625 = 5V 1.43906

7168 = 2r07 78125 = 5’ 85293 = 3*13 1.43501 3 125=5’ 128=27 1.42657 5 177147=3” 177152 =2”‘173 1.41268

Recently, Oesterle posed the problem to decide whether there exists an absolute constant C such that c(x, y, z) < C for all x, y, z. Masser conjec- tured the stronger assertion that c(x, y, z) < 1 + E, when z exceeds some bound depending on E only. For a survey of related results and conjectures see Stewart and Tijdeman [19].

It might be interesting to have some empirical results on c(x, y, z), and to search for x, y, z for which it is large. From the preceding sections it may be clear that such x, y, z correspond to relatively short vectors in appropriate approximation lattices. As a byproduct of the proofs of Theorems 4.6 and 5.4 we computed the value of c(x, y, z), corresponding to many short vectors that we came across in performing the algorithm of Fincke and Pohst. All examples that we found with c(x, y, z) > 1.4 are listed in Table X. Our search was rather unsystematic, so we do not guarantee that this list is complete in any sense. The largest value for c(x, y, z) that occurred is 1.626, which was reached by

x= 112, y=32x56x73, ~=2~‘x23.

These results do not seem to yield any heuristical evidence for the truth or falsity of the above mentioned conjecture.

ACKNOWLEDGMENTS

The author wishes to thank Professor R. Tijdeman and Dr. F. Beukers for their helpful remarks. He was supported by the Netherlands Foundation for Mathematics (SMC) with

Page 42: Solving Exponential Diophantine Equations Using …certain exponential diophantine equations and inequalities in a routine manner. As is well known, many types of diophantine problems

366 B. M. M. DE WEGER

linancial aid from the Netherlands Organization for the Advancement of Pure Research (ZWO). All machine computations were performed on the IBM-3083 computer at the Centraal Reken Instituut of the University of Leiden.

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2. L. J. ALEX, Diophantine equations related to finite groups. Comm. Algebra 4 (1976), 77-100.

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coefficients, Math. Ann. 261 (1982), 515-534. 10. A. K. LENSTRA. “Polynomial-Time Algorithms for the Factorization of Polynomials,”

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Theorem,” Univ. of Notre Dame Press, Notre Dame, 1961. 12. M. M~GNO~TE AND M. WALDSCHMIDT. Linear forms in two logarithms and Schneider’s

method, Math. Ann. 231 (1978), 241-267. 13. A. M. ODLYZKO AND H. J. J. TE RIELE, Disproof of the Mertens conjecture, J. Reine

Anger. Math. 357 (1985). 138-160. 14. A. PETHO AND B. M. M. DE WEGER, Products of prime powers in binary recurrence

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