Abstract and Applied Analysis

Volume 2014 (2014), Article ID 670175, 5 pages

http://dx.doi.org/10.1155/2014/670175

Research Article

The Exponential Diophantine Equation

Juanli Su¹ and Xiaoxue Li²

¹School of Arts and Sciences, Yangling Vocational and Technical College, Yangling, Shaanxi 712100, China

²School of Mathematics, Northwest University, Xi'an, Shaanxi 710127, China

Received 28 January 2014; Accepted 18 March 2014; Published 9 April 2014

Academic Editor: Youyu Wang

Copyright © 2014 Juanli Su and Xiaoxue Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

Let m be a positive integer. In this paper, using some properties of exponential diophantine equations and some results on the existence of primitive divisors of Lucas numbers, we prove that if and , then the equation has only the positive integer solution .

1. Introduction

Let , be the sets of all integers and positive integers, respectively. Given a triple of coprime positive integers with , there are many papers that investigated the equation

(see [1–8]). Recently, Terai [8] proved that if , , satisfy

then (1) has only the solution , provided that or . The proof of this result is based on elementary methods and Baker’s method. In this paper, using some properties of exponential diophantine equations and some results on the existence of primitive divisors of Lucas numbers, we prove a general result as follows.

Theorem 1. Let be positive integers satisfying (2). If and , then (1) has only the solution .

Some Note. Combining Terai [8] and our Theorem, we know that only the values , are left to investigate (2). In this case, the equation has only finitely many solutions in . Moreover, they are not only effectively but also practically solvable. With some computer assistance, we would be able to solve completely the equation.

2. Preliminaries

In this section, we assume that are positive integers satisfying (2). Then, (1) can be written as

Further, let be a solution of (3) with .

Lemma 2. If , then , and

Proof. Since , by (3), we have and . Since , we get

Since , we have and . It implies that

Hence, by (3), (5), and (6), we get

If , then from (7) we get , which contradicts (5). So we have

Since is a positive integer, by (7), we have

On the other hand, by (3), we have and . It implies that

whence we obtain

Since , we have . Thus, by (9) and (11), we get (4). The lemma is proved.

Let , , be positive integers such that and .

Lemma 3 ([9, Lemmas 1 and 6]). If is a fixed solution of the equation

then there exists a unique positive integer such that

The positive integer is called the characteristic number of the solution and is denoted by the symbol .

Lemma 4 ([9, Theorems 1 and 3]). For a fixed characteristic number , let be the set of all solutions of (13) with . Then one has the following.

(i) has a unique solution satisfying , and , where through all solutions of . Such is called the least solution of . (ii)Every element of can be expressed as

Lemma 5. If and , then the equality

cannot hold.

Proof. If (15) holds, then

By (15) and (17), we have

We may assume that ; then, by (18), we have

Since and , we see from (19) that . Further, since , by (16), we get

It implies that .

Let be an odd prime divisor of . Since , we have . Further let

Since , by (21), we have

Further, by (21) and (22), we get

Therefore, by (16), (21), and (23), we obtain

Putting through all odd prime divisors of , we get from (24) that and

By (19) and (25), we have and

But (26) is impossible for any positive integer . Thus, (15) is false. The lemma is proved.

Let be algebraic integers. If and are nonzero coprime integers and is not a root of unity, then is called a Lucas pair. Further, let and . Then we have

where . We call the parameters of the Lucas pair . Two Lucas pairs and are equivalent if . Given a Lucas pair , one defines the corresponding sequence of Lucas numbers by

For equivalent Lucas pairs and , we have for any . A prime is called a primitive divisor of if and . A Lucas pair such that has no primitive divisors will be called an -defective Lucas pair. Further, a positive integer is called totally nondefective if no Lucas pair is -defective.

Lemma 6 (see [10]). Let satisfy and . Then, up to equivalence, all parameters of n-defective Lucas pair are given as follows:

(i), , , , , , , ,(ii), ,(iii), ,(iv), ,(v), ,(vi), .

Lemma 7 (see [11]). If , then is totally nondefective.

Let be positive integers such that and .

Lemma 8 (see [9, Theorems 1 and 3]). Every solution of the equation

can be expressed as

where are positive integers satisfying

where is the class number of positive binary quadratic primitive forms of discriminant .

For any positive integer , let denote the set of distinct prime divisors of .

Lemma 9. If is a solution of (29) with , then , except the possibility of the following cases:

(i), where is defined as in (30),(ii), , .

Proof. Let be a solution of (29) with . By Lemma 8, , and satisfy (30) and (31), where , and satisfy (32). Let

By (32) and (33), we have that , , , and satisfies . It implies that is a Lucas pair with parameters . Let denote the corresponding Lucas numbers. By (28), (31), and (33), we get

Since , by the definition of primitive divisors, we see from (34) that either or and the Lucas number has no primitive divisor.

If , then from (30) and (32) we get . If , by Lemmas 6 and 7, using an easy computation, the solution satisfies the case (i) or (ii). Thus, the lemma is proved.

Lemma 10 ([12, Theorems and ]). For any positive integer , one has

3. Proof of Theorem

We now assume that is a solution of (3) with . By Lemma 2, we have

We first consider the case that . Then, by (3) and (36), the equation

has a solution

Let . Since , we have . Hence, by Lemma 3, we get

In addition, (37) has another solution

Let . We have

By (39) and (41), we get . It implies that the solutions (38) and (40) belong to a same class of solutions of (37). Further, since , (40) is the least solution of . Therefore, applying Lemma 4 to (38), we get , , and

However, since and , we have and . By Lemma 5, (42) is false.

We finally consider the case that . Then the equation

has a solution

Since , applying Lemma 8 to (44), we have

where , are positive integers satisfying

If , then from (46) we get

where are integers satisfying

By (49), we have

Further, since , we see from (50) that , and

Furthermore, by (51), we get ; whence we obtain . But, since , it is impossible. So we have .

If , then from (46) we get

Since , by (47), we see from (53) that and

Further, since and , by (54), we have

But, since , we get from (55) that , , and , a contradiction. So we have .

Notice that the solution (44) satisfies . Therefore, since and , by Lemma 9, we have and

Further, applying Lemma 10 to (56), we get

On the other hand, by Lemma 2, satisfies (4). The combination of (4) and (57) yields

But (58) is false for . Thus, the solution with does not exist. The theorem is proved.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to thank the referee for his very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the N.S.F. (11071194) of China, the Scientific Research Program Funded by Shaanxi Provincial Education Department (Program no. 12JK0871), and the Scientific Research Program Funded by Yangling Vocational and Technical College (Program no. A2013027).

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