The Diophantine equation (axk−1)(byk−1)=abzk−1

Journal of Number Theory 136 (2014) 252–260

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Journal of Number Theory www.elsevier.com/locate/jnt

The Diophantine equation (axk − 1)(by k − 1) = abz k − 1 ✩ Zhongfeng Zhang School of Mathematics and Information Science, Zhaoqing University, Zhaoqing 526061, PR China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 28 July 2013 Received in revised form 13 October 2013 Accepted 14 October 2013 Available online 7 December 2013 Communicated by David Goss

Let a, b be positive integers. In this paper, we prove that the equation (axk − 1)(by k − 1) = abz k − 1 has no solutions in integers x, y, z and k with |x| > 1, |y| > 1 and k  4. © 2013 Elsevier Inc. All rights reserved.

MSC: 11D41 11D61 Keywords: Diophantine equation Rational approximation Continued fraction

1. Introduction In [4], Bugeaud considered the problem of finding positive integers a and b such that the set {a + 1, b + 1, ab + 1} is comprised entirely of perfect kth powers, for k  3. These correspond to the Diophantine equation  k   x − 1 y k − 1 = z k − 1.

(1.1)

✩ This research was supported by the Guangdong Provincial Natural Science Foundation (No. S2012040007653) and NSF of China (No. 11271142). E-mail address: [email protected].

0022-314X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jnt.2013.10.013

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Bugeaud showed that there are no solutions to Eq. (1.1) with z  2, provided k  75, or if k  5 and the smaller of x and y is sufficiently large. Utilizing a variety of techniques, including the hypergeometric method of Thue and Siegel, as well as an assortment of gap principles, Bennett [3] proved that Eq. (1.1) has only the solutions (x, y, z, k) = (−1, 4, −5, 3) and (4, −1, −5, 3) in integers x, y, z and k with |z|  2 and k  3. Following the techniques developed by Bennett [3], we consider the more general equation (axk − 1)(by k − 1) = abz k − 1 and prove the following theorem. Theorem 1.1. Let a, b be positive integers, the equation    k ax − 1 by k − 1 = abz k − 1 has no solutions in integers x, y, z and k with |x| > 1, |y| > 1 and k  4. 2. A lemma Defining μn =



p1/(p−1) .

p|n

To prove Theorem 1.1, we need a result of Bennett [1, Theorem 1.3], that is Lemma 2.1. Let k and u are positive integers with k  3 and √ √ ( u + u + 1 )2(k−2) > (kμk )k ,

(2.1)

then    k  1 + 1 − p  > (8kμk u)−1 q −λ  u q with λ=1+

√ √ log(kμk ( u + u + 1 )2 ) √ . √ log( kμ1 k ( u + u + 1 )2 )

(2.2)

3. Proof of the theorem Proof of Theorem 1.1. We begin by noting that proving Theorem 1.1 reduces to consideration of equation  k   ax − 1 by k − 1 = abz k − 1 with k = 4, 6, 9 or k  5 a prime, or to one of the equations

(3.1)

Z. Zhang / Journal of Number Theory 136 (2014) 252–260

254

 k   ax + 1 by k + 1 = abz k − 1

(3.2)

 k   ax + 1 by k − 1 = abz k + 1,

(3.3)

or

where k = 9 or k  5 a prime, and, in all cases, x, y and z may be taken to be positive integers. Now we discuss in detail how to treat Eq. (3.1), for positive integers x, y, z. Let us suppose, in the following discussion, that we have a solution to Eq. (3.1) in positive integers x, y, z and k with x > 1, y > 1 and k  4. Let us write u + 1 = axk ,

v + 1 = by k ,

uv + 1 = abz k .

(3.4)

We may suppose, without loss of generality, that vu>1 and either k = 4, 6, 9 or k  5 is prime. Since abxk y k = (u + 1)(v + 1) > uv + 1 = abz k , we can write xy = z + t, t  1, that is 1/k  k uv + 1 (u + 1)(v + 1) = ab +t . ab Expanding this, we have (k−1)/k  (k−2)/k  uv + 1 uv + 1 k(k − 1) ab u + v = kab t+ t2 + · · · + abtk . ab 2 ab Since k  4 and v  u we get (k−2)/k  uv + 1 ab t2 > u, ab and then (k−1)/k   (k−1)/k uv + 1 uv v > kab t > kab t, ab ab which yields v > kk abuk−1 tk .

(3.5)

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Note that from (3.4), we have 

xy z

k −

u2 − 1 u u+1 = < , u u(uv + 1) abz k

whence    k  1 + 1 − xy  < u .  u z  kabz k

(3.6)

From the fact that u = axk − 1, x > 1 and k  4, it is not hard to show √

u+

√

√ u + 1 > 1.96 u + 1.

Then one has √   √ k 2(k−2) 3 k ( u + u + 1 )2(k−2) > 1.96 × 2 2 > 2k− 4 > k2k > (kμk )k when k  6 and √  √ k+1 2(k−2) ( u + u + 1 )2(k−2) > 1.96 × 2 2 > (kμk )k when k = 4, 5 and a  2. Therefore, (2.1) is satisfied except for

(x, k, a) ∈ (2, 4, 1), (2, 5, 1) .

(3.7)

For the time being, we will suppose that (x, k, a) is outside this set (so that λ < k). Then (3.6) together with Lemma 2.1 leads to 8μk 2 u , ab

(3.8)

8k μkk k+λ 8k μkk k+λ u  u  8k μkk uk+λ . λ (ab) (ab)2

(3.9)

z k−λ < therefore v k−λ <

Case 1. k = 6, 9 or k  5 prime. If k  7 is prime, then μk = k1/(k−1) . Since u = axk − 1, from Eq. (2.2), it is not hard to see that λ is monotone decreasing in x  2, k  7 and a  1, whereby λ < 3.15. Hence, (3.9) implies that v < 80u2.7 , contradicting (3.5). Similarly, if k = 9, then λ < 3.13, and so v < 57u2.1 , a contradiction to (3.5). If k = 6 and x  2, a  2 or x  3, a = 1, then λ < 3.9, and we have v < 15 517u4.72 . Combining this with (3.5), a contradiction. If, however, k = 6, x = 2 and a = 1, then we have

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63y 6 − z 6 =

62 = 1, 2, 31, 62. b

The case b = 1 is a special case of Bennett [3, Theorem 1.1], and the case b = 62 is also impossible from the result of Bennett [2] on the number of solutions to the equation |axn − by n | = 1. The equation modulo 9 can exclude the remaining two cases. If k = 5, x  3 and a  1, then λ < 2.83, and so v < 305u3.61 . Combining this with (3.5) leads to a contradiction. If k = 5, x = 2 and a  4, then λ < 2.96, and so v < 439u3.91 . Again, this contradicts (3.5). If k = 5, x = 2 and a = 3, then 3 < λ < 3.03, one has v 5−3.03 <

85 · 55/4 5+3.03 u . 33

That is v < 103u4.08 , a contradiction to (3.5). If k = 5, x = 2 and a = 2, then 3 < λ < 3.15, one has v 5−3.15 <

85 · 55/4 5+3.15 u , 23

that is v < 267u4.41 . Again a contradiction to (3.5). The remaining case is k = 5, x = 2 and a = 1. Then 31y 5 − z 5 =

30 = 1, 2, 3, 5, 6, 10, 15, 30. b

Similar to the case k = 6, x = 2 and a = 1, we can exclude b = 1, 30 by the results of Bennett and the remaining cases by congruences modulo 11 and 31. Case 2. k = 4. Now we introduce a “gap principle” that sharpens inequality (3.5) in this case. From (3.5), we get the inequality y > 3ax3 t.

(3.10)

v > 16bu4 .

(3.11)

We proceed to show that

From z = xy − t, t  1, we have 

  axk − 1 by k − 1 = ab(xy − t)k − 1,

that is 4abx3 y 3 t + 4abxyt3 + 2 = ax4 + by 4 + 6abx2 y 2 t2 + abt4 , therefore

(3.12)

Z. Zhang / Journal of Number Theory 136 (2014) 252–260

4abxyt3 + 2 ≡ ax4 + 6abx2 y 2 t2 + abt4

mod by 3 .

257

(3.13)

We now use these relations to show that t > x/2. If we have 4abxyt3 + 2 = ax4 + 6abx2 y 2 t2 + abt4 , then (3.12) implies y = 4ax3 t and hence ax4 |abt4 −2 and abt4 |ax4 −2, which is impossible since x > 1. Thus from (3.13) one has either 4abxyt3 + 2  ax4 + 6abx2 y 2 t2 + abt4 + by 3 or 4abxyt3 + 2 + by 3  ax4 + 6abx2 y 2 t2 + abt4 . In the first case, 4abxyt3 > by 3 , by combining this with (3.10) yields t > 9ax5 /4. The second inequality together with (3.10) implies by 3 < 6abx2 y 2 t2 , thus y < 6ax2 t2 . Combining this with (3.10) implies t > x/2. Then, from (3.5) one has v > 44 abu3 t4 > 16abu3 x4 > 16bu4 . If x = 2 and a = 1, then 15y 4 − z 4 =

14 = 1, 2, 7, 14. b

Similar to the case k = 6, x = 2 and a = 1, we can exclude b = 1, 14 by the results of Bennett and the remaining cases by the Legendre symbol equalities 

−7 5



 =

−2 5

 = −1.

If x  3 and a = 1 or x  2 and a  2, then combining (3.9) and (3.11), we get  u12−5λ <

16 a

λ  16λ ,

contradicting (2.2) provided x  35 (i.e. u  1 500 624). From the equation  4   ax − 1 by 4 − 1 = abz 4 − 1, we get 

 1 4 ax4 − 2 x − y − z4 = . a ab 4

Then one has

(3.14)

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  4 1/4  4 ((ax4 − 2)/a)1/4  x4 − 1 − z  < ((ax − 2)/ab)  .  a y 4y 4 4y 4 Hence,  from (3.10), we have that z/y is a convergent in the continued fraction expansion to 4 x4 − 1/a, that is z/y = pj /qj where, additionally, since z/y <

 4 x4 − 1/a,

j is even. We write  4 x4 − 1/a = [a0 , a1 , a2 , . . .],

pi /qi = [a0 , . . . , ai ],

where the a i denote the partial quotients in the infinite simple continued fraction expansion to 4 x4 − 1/a and the pi /qi the corresponding convergent. From the classical theory of continued fractions,   4  1  x4 − 1 − p j  > .  a qj  (aj+1 + 2)qj2 Since t > x/2, we have   1/4 9a2 x8 < 4y 2 < (aj+1 + 2) ax4 − 2 /a and hence aj+1 > 9a2 x7 − 2. The case a = 1 and 3  x  34 has treated by Bennett [3]. Now we assume a  2 and 2  x  34. We have aj+1  9 × 22 × 27 − 2 = 4606

(3.15)

with j even. On the other hand, from (3.14) we obtain λ

16 > u

12−4λ

 λ a > u12−4λ x−4λ , u

that is (2x)4λ − u12−4λ > 0. Let f (x, a) = (2x)4λ − u12−4λ , then when x  2 is fixed, this is a decreasing function in a  1. Thus from f (x, a) > 0, it is easy to bound a, that is the following table. x

a

x

a

x

a

x

a

x

a

x

a

2 8 14

205 16 5

3 9 15,16

98 12 4

4 10 17, 18, 19

58 10 3

5 11 20, . . . , 23

38 8 2

6 12 24, . . . , 34

27 7 1

7 13

20 6

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259

From the table, we get x  23 when a  2. From the inequality (3.8) we have z 4−λ <

16u2 16u2  . ab a

Combining this with the table we get z < 8.1 × 107 when a  2. We calculate that, in each case in the table with a  2, one has p14 > 9.5 × 107 , while for even j < 16, aj+1  423, a contradiction to (3.15). To complete the proof of Theorem 1.1, we need to treat Eqs. (3.2) and (3.3), in positive integers x, y, z and k with x > 1, y > 1 and k = 9 or k  5 a prime. Solutions to these equations imply the existence of positive integers u and v for which u − 1 = axk ,

v − 1 = by k ,

uv + 1 = abz k ,

(3.16)

u − 1 = axk ,

v + 1 = by k ,

uv − 1 = abz k ,

(3.17)

respectively. Similar discussion to Eq. (3.1), we get no solutions for Eqs. (3.2) and (3.3), in positive integers x, y, z and k with x > 1, y > 1 and k = 9 or k  5 a prime. In fact, in case (3.16) we can assume v  u, and then  0<

z xy

k −

1.1(u − 1) u < . u−1 ab(xy)k

In case (3.17), if v  u, then 0<

u − u−1



z xy

k <

1.1(u − 1) ab(xy)k

and v+1 − 0< v



xy z

k <

v abz k

if u > v. This completes the proof. 2 Acknowledgment The author is grateful to the referee for his/her carefully reading the manuscript and making valuable suggestions. References [1] M.A. Bennett, Explicit lower bounds for rational approximation to algebraic numbers, Proc. Lond. Math. Soc. 75 (1997) 63–78.

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[2] M.A. Bennett, Rational approximation to algebraic numbers of small height: the Diophantine equation |axn − by n | = 1, J. Reine Angew. Math. 535 (2001) 1–49. [3] M.A. Bennett, The Diophantine equation (xk − 1)(y k − 1) = (z k − 1)t , Indag. Math. 18 (2007) 507–525. [4] Y. Bugeaud, On the Diophantine equation (xk − 1)(y k − 1) = (z k − 1), Indag. Math. 15 (2004) 21–28.