A NOTE ON THE DIOPHANTINE EQUATION a x b y =c

MATH. SCAND. 107 010, 161 173 A NOTE ON THE DIOPHANTINE EQUATION a x b y =c BO HE, ALAIN TOGBÉ and SHICHUN YANG Abstract Let a,b, and c be positive integers. We show that if a, b = N k 1,N, where N,k, then there is at most one positive integer solution x, y to the exponential Diophantine equation a x b y =c, unless N, k =,. Combining this with results of Bennett [3] and the first author [6], we stated all cases for which the equation N k ± 1 x N y =c has more than one positive integer solutions x, y. 1. Introduction Let a,b,x, and y be positive integers and c an integer. The Diophantine equation 1 a x b y = c has a very rich history. It has been studied by many authors see for examples [], [3], [5], [6], [7], [9], [10], [13], [14], [15], [16], [17], [19], [0]. This Diophantine equation has some connections with Group Theory [1] and with Hugh Edgar s problem i.e., the number of solutions m, n of p m q n = h [4]. In 1936, Herschfeld [7] proved that equation 1 has at most one solution in positive integers x,y if a, b = 3, and c is sufficiently large. The same year, Pillai [13], see also [14] extended Herschfeld s result to any a,b with gcda, b = 1, a>b, and c >c 0 a, b, where c 0 a, b is a computational constant depending on a and b. Moreover, Pillai has conjectured that if a = 3 and b = then c 0 3, = 13. In 198, this conjecture was proved by Stroeker and Tijdeman [19]. For more information about the history of this Diophantine equation, one can see for example [], [15], [16], [19]. In this paper, we consider the exponential Diophantine equation a x b y =c. There are infinitely many pairs a, b such that the equation has at least two solutions. For example, let r and s be positive integers with r = s and Received 4 December 008, in final form 6 Februar010.

16 bo he, alain togbé and shichun yang max{r, s} > 1. If a = b r + b s / and c = a b r, then x, y = 1,rand 1,sboth satisfy equation. In 003, Bennett [3] proved the following result. Theorem 1.1. If N and c are positive integers with N, then the Diophantine equation N + 1 x N y =c has at most one solution in positive integers x and y, unless N, c {, 1,, 5,, 7,, 13,, 3, 3, 13}. In the first two of these cases, there are precisely 3 solutions, while the last four cases have solutions apiece. Very recently, the first author [6] extended Theorem 1.1 to obtain: Theorem 1.. If a, b = N k + 1,Nwith min{n,k}, then equation has at most one solution, except N,k,c {,, 3,,, 13,,t, t 1} t 3. In the first case, there are precisely 3 solutions, while the last two cases have solutions. The aim of this paper is to study the number of solution of the equation 3 N k 1 x N y =c and to prove the following result: Theorem 1.3. If a, b = N k 1,Nwith min{n,k} and N, k =,, then equation has at most one positive integer solution x, y. Naturally, from Theorems 1.1 1.3 we state Corollary 1.4. If a, b = N k ± 1,N with N, then equation has at most one solution, unless a,b,cor b,a,c {, 3, 1,, 3, 5,, 3, 7,, 3, 13,, 3, 3, 3, 4, 13,, 5, 3,, 5, 13,, t + 1, t 1 t 3}. These cases having more than one solution are listed here: 3 = 3 = 3 3 = 1 3 3 = 3 = 5 3 3 = 5 5 = 3 5 = 7 5 3 = 3

a note on the diophantine equation a x b y =c 163 and 3 = 4 3 = 7 4 3 = 8 3 5 = 13 3 3 = 5 3 = 3 5 3 = 7 5 = 13 t + 1 = t+1 t + 1 = t 1. The organization of this paper is as follows. In Section, we prove some useful results and recall a result due to Mignotte [11]. The proof of Theorem 1.3 will be given in Section 3 by the means of lower bounds for linear forms in two logarithms.. Preliminary work Let p be a prime and let ord p n denote the highest exponent of p in the prime factorization of an integer n. We define the number ν p n by ν p n = p ordpn. This corresponds to n p defined on pages 00 01 of [1]. Moreover log p 1 + n denotes the p-adic logarithm of n. The p-adic logarithm satisfies the identity log p 1 + n = nr r=1 1r+1 r. We have the following result. Lemma.1. Let x,y,n, and k be positive integers with N. If 4 N y 1 mod N k 1 x, then d y, where 5 { kn k 1 x 1, if N or x = 1 or N k 1 mod 4, d = 1 ord N k +1 kn k 1 x 1, otherwise. Proof. As N k 1 N y 1, we get k y. So there exists a positive integer z such that y = kz. The congruence 4 gives N y 1 mod N k 1 x. Let p be a divisor of N k 1. If p =, then we have N kz 1 mod 4 and if p is odd, then we have N kz 1 mod p. Also we know that ν p N kz 1 = ν p log p N kz, see Mordell [1]. Then we obtain ν p N k 1 x ν p N kz 1 = ν p z log p N k = ν p zν p log p N k = ν p zν p N k 1 = ν p zν p N k 1ν p N k + 1.

164 bo he, alain togbé and shichun yang Thus we have 6 ord p N k 1 x 1 ord p z + ord p N k + 1. When p is odd, as p N k 1wegetp N k + 1, i.e., ord p N k + 1 = 0. If N, then p and any divisor of N k + 1 are both odd. By inequality 6, we get the first case. If N, then we need to consider p =, then from 6 we have 7 1 + ord N k 1 x 1 ord z + ord N k + 1. We put z = α z, z and N k 1 = β μ, μ, then applying inequality 6 we get μ x 1 z. Similarly applying inequality 7 with α and β,wehave 1+βx 1 <α+ord N k +1. Thus we obtain 1 ord N k +1 N k 1 x 1 z, so the remaining cases are proved. We can prove the following lemma using a similar argument. If Lemma.. Let x,y,n, and k be positive integers with N, y k. 8 N k 1 x 1 mod N y, { 1, if N is even, then τn y k x, where τ =, if N is odd. Proof. Let p be a divisor of N. It is easy to see that x. Ifp =, then we have N k 1 x 1 mod 4. Otherwise, if p is odd, thus N k 1 x 1 mod p. We know ν p N k 1 x 1 = ν p log p N k 1 x. This and condition 8 imply Thus we obtain ν p N y ν p x/ν p N k ν p N k. ord p N y k ord p x/ + ord p N k. In the case N, we don t need to consider p =. We immediately get the result. If N, since k, this implies N k 0 mod 4. Then we have ord p N k = 1. So we obtain ord p N y k ord p x. Now we recall the following result on linear forms in two logarithms due to Mignotte see [11], Corollary of Theorem, page 110. For any non-zero

a note on the diophantine equation a x b y =c 165 algebraic number γ of degree d over Q, whose minimal polynomial over Z is a d j=1 X γ j, we denote by hγ = 1 d its absolute logarithmic height. log a + d j=1 Lemma.3. Consider the linear form log max1, γ j = b 1 log α 1 b log α, where b 1 and b are positive integers. Suppose that α 1,α are multiplicatively independent. Put D = [Qα 1,α : Q]/[Rα 1,α : R] and let ρ, λ, a 1 and a be positive real numbers with ρ 4, λ = log ρ, a i max{1,ρ 1 log α i +Dhα i }, i = 1, and a 1 a max{0, 4λ }. Further suppose h is a real number with { b1 h max 3.5, 1.5λ, D log + b } + log λ + 1.377 + 0.03, a a 1 χ = h/λ, υ = 4χ + 4 + 1/χ. Then we have the lower bound 9 log C 0 + 0.06λ + h a 1 a, where C 0 = 1 λ 3 { + 1 χχ + 1 1 3 + 1 9 + 4λ 1 + 1 + 3 } 1 + χ 3/ 3v a 1 a 3v. a 1 a

166 bo he, alain togbé and shichun yang 3. Proof of Theorem 1.3 Suppose that the equation N k 1 x N y =c>0 has two solutions x i,y i i= 1, with 1 x 1 satisfying the condition 10 N, k and N, k =,. Proposition 3.1. The equation 11 N k 1 x 1 + N k 1 = N y 1 + N has no solution x 1,,y 1, with the condition 10. Proof. We rewrite equation 11 into the form N k 1 x 1 N k 1 x 1 + 1 = N min{y 1, } N y 1 + 1. Since gcdn k 1,N= 1, we have N y 1 +1 0 mod N k 1. Therefore, there exist positive integers p, q such that y 1 =pk + q, for 0 q<k. Then we obtain 1 N y 1 N pk+q = N k p N q N q mod N k 1. Thus we get N q + 1 0 mod N k 1. This implies N k 1 N q + 1. But as q<k, we get N k 1 N k 1 + 1. It follows that N k 1 N 1. This is impossible when N, k =,. So Proposition 3.1 is proved. Let us consider the equation 1 N k 1 x 1 N y 1 = N k 1 N =±c, c > 0, with x 1 < and y 1 <. Taking equation 1 modulo N, wehave If N>, it follows that 1 x 1 1 mod N. 13 x 1 mod. We rewrite equation 1 into the form 14 N k 1 x 1 N k 1 x 1 1 = N y 1 N y 1 1. Since x 1 is even, so N k N k 1 x 1 1. Thus N k divides the right side of equation 14. As gcdn y 1 1,N= 1, we have y 1 k.

a note on the diophantine equation a x b y =c 167 From Lemma.1, we have k y 1 k. It is easy to show that the special case k y 1 or k can be solved by Theorem 1.1. In fact, if k y i i = 1, then there exist positive integers t 1 and t such that y 1 = t 1 k and = t k. Let us put M = N k 1, thus the equation M + 1 X M Y =c have the solutions X, Y = x 1,t 1 and,t. From Theorem 1.1, we have M 3. Thus we get N k 1 3 which contradicts the condition 10. Therefore, using equation 14, we will consider 15 y 1 >k and k y i i = 1,. Assume N =. Considering equation 1 modulo k gives 1 x 1 1 mod k. Using condition 10, we get k 3. This leads to x 1 and. Proposition 3.. If the equation 16 N k 1 x 1 N y 1 = N k 1 N = c>0 has solutions x 1,,y 1, with the condition 10, then N k 1 < 4379. Proof. Either y 1 >kor x 1 implies x 1. We set Then we have = logn. 0 < <e 1 = c N < N k 1 x 1. N On the other hand, using equation 14 we get N y 1 1 mod N k 1 x 1. Then from Lemma.1 with x 1 and N k 1 3 1 >.8,wehave N k x1 1 1 N k 0.5x1 1 y 1 k k >kn k 1 0.3x 1. Thus we obtain < y 1 /k 3.15 < 3.15. N N We know that < y 1 /k 3.15 /N / 3.15 /. The function y/ 3.15 / y is a maximum when y is between 4 and 5, so <0.548. Now we apply Lemma.3 to. We take 17 D = 1, α 1 = N k 1, α = N, b 1 =, b =

168 bo he, alain togbé and shichun yang and 18 a 1 = ρ + 1, a = ρ + 1. Since N 4 with k = orn with k 3, we choose ρ = 4.8. It satisfies a 1 a max{0, 4λ }. The fact >0 implies We take >. { } x h = max 8.56, log + 0.8. First we suppose x h = log + 0.8, then 99. We obtain C 0 < 0.67, then we have x log > 3.1 log +.389. We have = + < + 0.407. Combining this and bounds of, wehave < 0.407 + 3.15 log + 3.1 x log < 1.698 +.317 log x + 3.1 log x We get < 415. Next we suppose h = 8.56, then we have also <e8.56 0.8 99 < 415. +.389 +.389.

a note on the diophantine equation a x b y =c 169 Since / < /, thus 19 < 415. Using 15, 19, and Lemma.1, we obtain N 0 N k k x1 1 1 1 <k + y 1 < < 415. This implies N k 1 < 4397. Proposition 3.3. If the equation 1 N y 1 N k 1 x 1 = N N k 1 = c>0 has solutions x 1,,y 1, with the condition 10, then N k 1 < 4455. Proof. We will use a similar method to that of Proposition 3.. We set again = logn. Then we obtain 0 < <e c N y 1 1 = <. N k 1 N k 1 The fact that the left side of equation 1 is positive implies y 1 >k. From equation 14, we get N k 1 x 1 1 mod N y 1. So Lemma. gives x 1 N y1 k. Therefore, as N k 8, then we obtain < N k N k 1 N y 1 k N k 1 < 1.15 x 1 1 N k 1 1.15 1 1 N k 1. 1 From congruence 13, we have 1 x 1 and 1 /3. Then we obtain 0.77 3 < N k 1. /3 Again, by 3 and N k 8weget <0.05. Thus we have 4 x < < + 0.05 logn < + 0.038. Now we apply Lemma.3 to. We take the same parameters as those in 17, 18 and we choose ρ = 4.1. Here we have { h = max 9.10, log } + 0.81.

170 bo he, alain togbé and shichun yang First we suppose h = log + 0.81, then 5 > 3983. We have C 0 < 0.859 and thus log > 3.91 log On the other hand, by inequality 3 we get +.. log < 0.7 + log 3. The upper and lower bounds imply < 1.5 log x 0.405 + 35.87 log Using this and the middle terms of 4, we get < 1.1 log + 35.87 log It results < 3969. This contradicts inequality 5. Next, we suppose h = 9.10. Then we have <e9.10 0.81 < 3984. Since / < /, thus 6 < 3984. By 15, 6 and Lemma., we get +.. +.. 7 N y 1 k x 1 < < 3984 3984 logn y 1 k.

a note on the diophantine equation a x b y =c 171 This implies N y 1 k < 4455. If y 1 k k, wehaven k < 4455. Otherwise, suppose that y 1 k k 1. From equation 1 we have N k 1 x 1 <N y 1. Then we obtain N k 1 x 1 <N k 1. If x 1, then we have N k N k <N k 1. This implies that N k 1 N 1 <, which is impossible. It remains x 1 = 1. Now from and y 1 k 1, we have 8 < 1 N k 1. Using 3 and N k = 8, we get N k 1 x >. Thus we obtain x < 1 x. Thus / is a convergent in the simple continued fraction expansion to /. It is known that see [8], if p r /q r is the r th such convergent, then p r q > 1, r a r+1 + qr where a r+1 is the r + 1st partial quotient to /. In the continued fraction expansion = [k 1, 1,a,...], by direct computation, one gets q = a + 1 and N k 1 1 <a <N k 1. Let / = p r /q r for some nonnegative integer r. From inequality 6 we have q r < 3984. IfN k 1 > 3984, then q 1 = a >N k 1 1 3894 1 >q r 1. This implies r<. But q 0 = q 1 = 1 such that = 1, which is impossible. Then we have N k 1 3984. This completes the proof of Proposition 3.3. Finally, running a Maple scripts by Scott and Styer [18], we found all solutions of the equation a x b y = c in the range 1 <a,b<53000, which are listed in [17]. This helps us to check the remaining cases stated in Propositions 3. and 3.3. We found no solution

17 bo he, alain togbé and shichun yang x, y satisfying a, b = N k 1,N with condition 10. Combining this with Proposition 3.1 completes the proof of Theorem 1.3. Acknowledgements. The authors are grateful to the anonymous referee and Prof. Bjørn Dundas for insightful and valuable comments that help to improve the manuscript. The first and third authors were supported by the Applied Basic Research Foundation of Sichuan Provincial Science and Technology Department No. 009JY0091. The second author was partially supported by Purdue University North Central. REFERENCES 1. Alex, L. J., Diophantine equations related to finite groups, Comm. Algebra 4 1976, 77 100.. Bennett, M. A., On some exponential equations of S.S. Pillai, Canad. J. Math. 53 001, 897 9. 3. Bennett, M. A., Pillai s conjecture revisited, J. Number Theory 98 003, 8 35. 4. Cao, Z. F., On the equation + m = y n and Hugh Edgar s Problem in Chinese, Kexue Tongbao 31 1986, 555 556. 5. Cassels, J. W. S., On the equation a x b y = 1, Amer. J. Math. 75 1953, 159 16. 6. He, B., The exponential Diophantine equation a x b y =c in Chinese, Adv. Math. China 38 009, 403 408. 7. Herschfeld, A., The equation x 3 y = d, Bull. Amer. Math. Soc. 4 1936, 31 34. 8. Khinchin, A. Ya., Continued Fractions, 3rd ed., Noordhoff, Groningen 1964. 9. Le, M. H., A note on the Diophantine equation ax m by n = k, Indag. Math. N.S 3 199, 185 191. 10. LeVeque, W. J., On the equation a x b y = 1, Amer. J. Math. 74 195, 35 331. 11. Mignotte, M., A corollary to a theorem of Laurent-Mignotte-Nesterenko, ActaArith. 86 1998, 101 111. 1. Mordell, L. J., Diophantine Equations, Pure Appl. Math. 30, Academic Press, London 1969. 13. Pillai, S. S., On a x b y = c, J. Indian Math. Soc. 1936, 119 1 and 15. 14. Pillai, S. S., On the equation x 3 y = X + 3 Y, Bull. Calcutta Math. Soc. 37 1945, 15 0. 15. Scott, R., On the equations p x b y = c and a x + b y = c z, J. Number Theory 44 1993, 153 165. 16. Scott, R., and Styer, R., On p x q y = c and related three term exponential Diophantine equations with prime bases, J. Number Theory 115 004, 1 34. 17. Scott, R., and Styer, R., On the generalized Pillai equation ±a x ± b y = c, J. Number Theory 118 006, 36 65. 18. http://www41.homepage.villanova.edu/robert.styer/reesescott/naivenew6jan006.mws 19. Stroeker, R. J., Tijdeman, R., Diophantine equations, pp. 31 369 in: Computational Methods in Number Theory, Math. Centre Tract 155, Math. Centrum, Amsterdam 198.

a note on the diophantine equation a x b y =c 173 0. Terai, N., Applications of a lower bound for linear forms in two logarithms to exponential Diophantine equations, Acta Arith. 90 1999, 17 35. DEPARTMENT OF MATHEMATICS ABA TEACHER S COLLEGE WENCHUAN, SICHUAN, 63000 P. R. CHINA E-mail: bhe@live.cn ysc100@sina.com MATHEMATICS DEPARTMENT PURDUE UNIVERSITY NORTH CENTRAL 1401 S, U.S. 41 WESTVILLE IN 46391 USA E-mail: atogbe@pnc.edu