关于丢番图方程 kx^2-ly^2=4  

On Diophantine Equation kx^2-ly^2=4

作　　者：吴秋月 罗家贵 WU Qiuyue;LUO Jiagui(School of Mathematics and Information,China West Normal University,Nanchong Sichuan 637009,China)

机构地区：[1]西华师范大学数学与信息学院,四川南充637009

出　　处：《西华师范大学学报（自然科学版）》2020年第1期65-70,共6页Journal of China West Normal University(Natural Sciences)

基　　金：国家自然科学基金项目(10571180);四川省教育厅重大培育项目(16ZA0173)。

摘　　要：利用Stormer定理及其推广和Lehmer序列及其伴随序列的性质,讨论不定方程kx^2-ly^2=4的正整数解(x,y)与其最小解ε的关系。设xn√k+yn√l/2=(ε/2)^ n,我们证明了当x有且只有一个素因子不整除k时,除(k,l,x)=(5,1,x 25)外,x=x 1或x=x q;当x有且只有两个素因子不整除k时,除(k,l,x)=(5,1,x 125)或(k,l,x)=(5,1,x 5q)外,x=x 1或x=x q或x=x q^2。In this paper,the relation between the positive integer solution(x,y)of Diophantine equations kx 2-ly 2=4 and its minimal solutionεis discussed by using Stormer theorem and its generalizations on Pell equation,and fundamental properties of Lehmer sequences and its associated sequences.Let xn√k+yn√l/2=(ε/2)^ n,it is proved that if there is only one prime divisor of x not dividing k,then x=x 1 or x=x q except for(k,l,x)=(5,1,x 25)and if there are only two prime divisors of x not dividing k,then x=x 1 or x=x q or x=x q 2 except for(k,l,x)=(5,1,x 125)or(k,l,x)=(5,1,x 5q).

关 键 词：丢番图方程 PELL方程 Lehmer序列 基本解 最小解 

分 类 号：O156.7[理学—数学]