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A note on the equation
Journal of Inequalities and Applications volume 2014, Article number: 170 (2014)
Abstract
In this paper, we shall use some simple inequalities and a deep result on the existence of primitive divisors of Lucas numbers to prove that the exponential Diophantine equation has no positive integer solution with .
MSC:11D61.
1 Introduction
Let ℤ, ℕ be the sets of all integers and positive integers, respectively. Recently, Zhang and Yuan [1] were interested in the equation
Using the Gel’fond-Baker method, they proved that all solutions of (1.1) satisfy . This upper bound is far beyond the computable scope at present. In this paper, we shall use some simple inequalities and a deep result on the existence of primitive divisors of Lucas numbers to prove the following result.
Theorem Equation (1.1) has no solution with .
In addition, it is obvious that is a solution of (1.1). Because one have not found the other solutions, we propose a conjecture as follows:
Conjecture Equation (1.1) has only the solution .
Our theorem supports the above mentioned conjecture.
2 Preliminaries
Lemma 2.1 Let , where X is a real number. Then is an increasing function for .
Proof Since , we have for . Thus, the lemma is proved. □
Lemma 2.2 Let , where X is a real number. Then we have for .
Proof Since for , g(X) is an increasing function satisfying for . The lemma is proved. □
The equation
has only the solutions .
Lemma 2.4 ([[4], Theorem 8.4])
The equation
has only the solution .
Lemma 2.5 ([[4], Theorem 8.4])
The equation
has only the solution .
Let D be a positive integer, and let denote the class number of positive binary quadratic primitive forms of discriminant .
Lemma 2.6 .
Proof Notice that , , , . The lemma holds for . By Theorems 11.4.3, 12.10.1, and 12.14.3 of [5], if , then
Therefore, if , then from (2.4) we get
But, by Lemma 2.2, (2.5) is impossible for . Thus, the lemma is proved. □
Lemma 2.7 Let k be a positive integer with . Every solution of the equation
can be expressed as
where , , are positive integers satisfying
Proof This lemma is the special case of [[6], Theorems 1 and 2] for and .
Let α, β be algebraic integers. If and αβ 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 p is called a primitive divisor of () if and . A Lucas pair such that has no primitive divisor will be called an n-defective Lucas pair. Further, a positive integer n is called totally non-defective if no Lucas pair is n-defective. □
Lemma 2.8 ([7])
Let n satisfy and . Then, up to equivalence, all parameters of n-defective Lucas pairs are given as follows:
-
(i)
, .
-
(ii)
, .
-
(iii)
, .
-
(iv)
, .
-
(v)
, .
-
(vi)
, .
Lemma 2.9 ([8])
If , then n is totally non-defective.
3 Further lemmas on the solutions of (1.1)
Throughout this section, we assume that is a solution of (1.1) with .
Lemma 3.1 ([1])
x, y and z are coprime.
Lemma 3.2 .
Proof Since , we have . If , since , then and , a contradiction. Similarly, if , then and , a contradiction. Therefore, we have .
If , then
Further, by Lemma 3.1, y and z are odd integers with . Hence, we see from (3.1) that (2.3) has the solution . But, by Lemma 2.5, it is impossible.
Similarly, if or , then we have
or
But, by Lemmas 2.3 and 2.4, (3.2) and (3.3) are impossible. Thus, we get . The lemma is proved. □
Lemma 3.3 .
Proof By (1.1), we have and . Hence,
and
In addition, by Lemmas 3.1 and 3.2, x, y and z are distinct.
If , by Lemma 3.2, then . Hence, by Lemma 2.1, we get
which contradicts (3.5). Similarly, we can remove the case that .
If , then and
which contradicts (3.4). Thus, we get . The lemma is proved. □
4 Proof of theorem
We now assume that is a solution of (1.1) with . Since , by Lemmas 3.1, 3.2 and 3.3, we have , , and .
We see from (1.1) that the equation
has the solution
Applying Lemma 2.7 to (4.1) and (4.2), we have
where , , are positive integers satisfying
and
Let
We see from (4.5) and (4.7) that and are coprime nonzero integers, is not a root of unity. Hence, is a Lucas pair with parameters . Further, Let () denote the corresponding Lucas numbers. By (4.4) and (4.7), we have
We find from (4.7) and (4.8) that the Lucas number has no primitive divisor. Therefore, by Lemma 2.9, we have . Further, since and by (4.3), it is easy to remove all cases in Lemma 2.8 and conclude that .
If , then from (4.4) we get
Let . Since , we have and . Further, since , we get , and by (4.9). Therefore, we have and, by (4.9), and
It implies that
But, since and , we get from (4.11) that and , a contradiction.
If , then from (4.3) and (4.6) that , and
But recall that , by Lemma 2.6, (4.12) is impossible. Thus, (1.1) has no solution with . The theorem is proved.
References
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Acknowledgements
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 P. S. F. (2013JZ001) and N.S.F. (11371291) of P.R. China.
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YL obtained the theorems and completed the proof. XL corrected and improved the final version. Both authors read and approved the final manuscript.
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Lu, Y., Li, X. A note on the equation . J Inequal Appl 2014, 170 (2014). https://doi.org/10.1186/1029-242X-2014-170
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DOI: https://doi.org/10.1186/1029-242X-2014-170