The Exponential Diophantine Equation4m2+1x+5m2-1y=(3m)z

Juanli Su,Xiaoxue Li

doi:10.1155/2014/670175

Abstract

Letmbe 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 ifm>90and3|m, then the equation4m2+1x + 5m2-1y=(3m)zhas only the positive integer solution(x,y,z)=(1,1,2).

Highlights

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 m > 90 and 3|m, the equation (4m2 + 1)x + (5m2 − 1)y = (3m)z has only the positive integer solution (x, y, z) = (1, 1, 2)
Let Z, N be the sets of all integers and positive integers, respectively
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

Summary

Introduction

Given a triple (a, b, c) of coprime positive integers with min{a, b, c} > 1, there are many papers that investigated the equation ax + by = cz, x, y, z ∈ N (1). Let a, b, c be positive integers satisfying (2). Combining Terai [8] and our Theorem, we know that only the values 20 < m ≤ 90, 3 | m are left to investigate (2). In this case, the equation (4m2 + 1)x + (5m2 − 1)y = (3m)z has only finitely many solutions in (x, y, z). We would be able to solve completely the equation

Preliminaries

Proof of Theorem