Diophantine方程(a^m-1)(b^n-1)=x^2的一点注记

A note on the exponential Diophantine equation(a^m-1)(b^n-1)=x^2

设a,b是适合min(a,b)>1,2|a,2 b以及v(b 1)是正奇数,其中v(b 1)表示整除b 1的2的最高次数.本文运用初等方法以及同余性质,研究了方程(am 1)(bn 1)=x2的可解性.对某些特殊素数p,证明了该方程无解.证明了如果存在适合p≡±3(mod 8)的奇素数p,可使a≡1(mod p)以及b≡0(mod p),则方程(am 1)(bn 1)=x2无正整数解(x,m,n).

Let a, b be positive integers satisfying min（a, b） 〉 1, 2 ｜ a, 2 ＋ b and v（b - 1） is odd, where v（b - 1） denote the highest power of 2 dividing b - 1. The main purpose of this paper is using the elementary method and the properties of congruence to study the solvability of the equation （am - 1）（bn - 1） = x2. Proved that the equation has no positive integer solution for some special prime p. If there exist an odd prime p such that p ≡±3（mod 8）, a = -1（mod p） and b ≡0（mod p）, then the equation （am - 1）（bn - 1） =x2 has no positive integer solution （x, m, n）.