摘要
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).
出处
《纯粹数学与应用数学》
CSCD
2011年第5期581-585,共5页
Pure and Applied Mathematics
基金
国家自然科学基金(11071194)