- Open Access
A note on the equation
© Lu and Li; licensee Springer. 2014
Received: 11 March 2014
Accepted: 7 April 2014
Published: 7 May 2014
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 .
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.
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. □
has only the solutions .
Lemma 2.4 ([, Theorem 8.4])
has only the solution .
Lemma 2.5 ([, Theorem 8.4])
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 .
But, by Lemma 2.2, (2.5) is impossible for . Thus, the lemma is proved. □
Proof This lemma is the special case of [, Theorems 1 and 2] for and .
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 ()
Lemma 2.9 ()
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 ()
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 .
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.
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 .
In addition, by Lemmas 3.1 and 3.2, x, y and z are distinct.
which contradicts (3.5). Similarly, we can remove the case that .
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 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 .
But, since and , we get from (4.11) that and , a contradiction.
But recall that , by Lemma 2.6, (4.12) is impossible. Thus, (1.1) has no solution with . The theorem is proved.
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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