On a Diophantine equation related to the difference of two Pell numbers

Abstract In this paper, the Diophantine equation Pn − Pm = 3 is considered and all solutions for this equation are obtained. In the proof of the main theorem, lower bounds for the absolute value of linear combinations of logarithms and a version of the Baker-Davenport reduction method are used.


Introduction
In recent years, many researchers investigated the solutions of Diophantine equations of the form u n ± u m = p a where (u n ) is a fixed linear recurrence sequence and p is a prime. For example, Bravo and Luca [3,4] solved the equation u n + u m = 2 a for the cases when (u n ) is the Fibonacci sequence and when (u n ) is the Lucas sequence. Also, Bitim and Keskin [1] found all the solutions of the equation u n − u m = 3 a for the case when (u n ) is the Fibonacci sequence. Also, many other researches on this topic, such as [6], have been carried out.
In this paper, we search all the solutions of the Diophantine equation where P n is the Pell sequence and n, m and a are nonnegative integers such that n ≥ m. The main argument used for the solution of such problems is Baker's theory (lower bound for the absolute value of linear combinations of logarithms of algebraic numbers) and a version of the Baker-Davenport reduction method.

Preliminaries
A linear recurrence sequence of order k is a sequence whose general term is (a n ) = L (a n−1 , a n−2 , . . . , a n−k ) where k is a fixed positive integer and L is a linear function. A linear recurrence sequence of order 2 is known as a binary recurrence sequence. Pell sequence, one of the most familiar binary recurrence sequence, is defined by P 0 = 0, P 1 = 1 and P n = 2P n−1 + P n−2 . Some of the terms of the Pell sequence are given by 0, 1, 2, 5, 12, 29, 70, . . . . Its characteristic polynomial is of the form x 2 − 2x − 1 = 0 whose roots are α = 1 + √ 2 and β = 1 − √ 2. Binet's formula enables us to rewrite the Pell sequence by using the roots α and β as Also, it is known that α n−2 ≤ P n ≤ α n−1 .
We give the definition of the logarithmic height of an algebraic number and some of its properties.
where the a i 's are relatively prime integers with a 0 > 0 and the ξ i 's are conjugates of ξ. Then is called the logarithmic height of ξ.
Proposition 2.1. Let ξ, ξ 1 , ξ 2 , . . . , ξ t be elements of an algebraic closure of Q and m ∈ Z. Then We will use the following theorem (see [8] or Theorem 9.4 in [5]) and lemma (see [2] which is a variation of the result due to [7]) for proving our results.   Proof. In the case that n = m, it is obvious that there exists no solution for the Diophantine equation (1). So we consider the case that n > m in the rest of the paper. By a simple computation, we observe that all triples (n, m, a) with 0 ≤ m < n ≤ 200 satisfying the equation (1)  Assume that n > 200. From (1) and (3), we get 3 a = P n − P m ≤ P n ≤ α n−1 < 3 n and so a < n. When we replace P n in the equation (1) with its closed form, we obtain
Now let's improve the upper bound on n a little bit more. Set The inequality (4) can be also written as |1 − e z1 | < 2 α n−m . By using (1) and (2), we get α n 2 Therefore, we have It is easy to see that 2 α n−m < 0.829 for all n − m ≥ 1. Therefore we have e |z1| < 5.85. Then we get by dividing both sides of the inequality above by log α. From Lemma 2.1, we have the irrational number γ = log 3 log α with µ = log 2 √ 2 log α , A = 12 log α , B = α, w = n − m.
On the other hand, we recall that a < n < 1.63 · 10 29 . From Lemma 2.1, we can set M := 1.63 · 10 29 and if we take the denominator of the 58th convergent of γ, then we get q = 15.50 · 10 29 > 6M . By using Mathematica Script Language, we obtain ε = µq − M γq = 0.184766 > 0. Applying Lemma 2.1 to the above parameters, we conclude that there is no solution to the inequality (9) for the values n − m with n − m ≥ log (Aq/ε) log B = 81.788.
Therefore, for the inequality (9) to be solvable, our upper limit for n − m must be at most 81. By substituting the upper bound value for n − m in the inequality (7), we get n < 1.211 · 10 16 . Let us improve this upper bound value on n a little more. Put z 2 := a log 3 − n log α + log 2 √ 2 1 − α m−n −1 .
Thus, n must be less than or equal to 50 for a solution which contradicts our assumption. This completes the proof.

Conclusion
We obtain all solutions of the Diophantine equation P n − P m = 3 a . Linear forms in logarithms and Baker's theory are the main tools used in our proofs. The method used in this paper may be applied to other Diophantine equations.