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Research Article | Open Access

Volume 2014 |Article ID 670175 | https://doi.org/10.1155/2014/670175

Juanli Su, Xiaoxue Li, "The Exponential Diophantine Equation ", Abstract and Applied Analysis, vol. 2014, Article ID 670175, 5 pages, 2014. https://doi.org/10.1155/2014/670175

# The Exponential Diophantine Equation

Accepted18 Mar 2014
Published09 Apr 2014

#### 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 ). Recently, Terai  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  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 ). 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 ). 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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