关于丢番图方程x2=p2b+2a2t-pb+2at+r+1  

On Diophantine Equation x2=p2b+2a2t-pb+2at+r+1

作　　者：彭燕培 罗家贵 费双林 PENG Yanpei;LUO Jiagui;FEI Shuanglin(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

出　　处：《西华师范大学学报（自然科学版）》2019年第4期367-370,共4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：本文利用同余理论、因式分解、整除性理论等初等方法并结合二元四次不定方程解的性质讨论了与马少麟猜想相关的一类丢番图方程,证明了:如果a>1是奇数,p是素数,那么方程x^2=p^2b+2a^2t-p^b+2^at+r+1,x∈N^+,b,t,r∈N,t≥r有解的充分必要条件是p=2,t=r=1或p=2,t≥r=0,且求出了它的所有解。In combination with the properties of positive integer solutions of binary quartic Diophantine equation,the congruence theory,factorization,divisibility theory and other elementary methods are employed to discuss a class of Diophantine equations related to Ma Shaolin s conjecture in this paper.It is proved that if a is an odd>1,p is a prime,then the equation x^2=p^2b+2a^2t-p^b+2a^t+r+1,x∈N^+.b,t,r∈N,t≥r has positive integer solutions and its all positive integer solutions are obtained if and only if p=2,t=r=1 or p=2,t≥r=0.

关 键 词：丢番图方程 McFarland s猜想 PELL方程 基本解 

分 类 号：O156.7[理学—数学]