关于Diophantine方程x^3+1=py^2  

On the Diophantine equation x～3+1 = py～2

作　　者：高洁[1] 袁进[1] 

机构地区：[1]西北大学数学系

出　　处：《纯粹数学与应用数学》2010年第4期687-690,共4页Pure and Applied Mathematics

摘　　要：在素数p=3(8t+4)(8t+5)+1和p=3(8t+3)(8t+4)+1的情形下,运用初等数论的方法给出了丢番图方程x3+1=py2无正整数解的充分条件,并得到无数个6k+1型的素数p使得方程x3+1=py2无正整数解.Using elementary theory of numbers methods,a sufficient condition is obtained that the Diophantine equation x3＋1 = py2 has no positive integer solution,where p = 3（8t＋4）（8t＋5）＋1 and p = 3（8t＋3）（8t＋4）＋1.And get countless p,where p is an odd prime of the form 6k＋1,meet the equation x3＋1 = py2 without positive integer solution.

关 键 词：丢番图方程 正整数解 奇素数 同余 

分 类 号：O156.1[理学—数学]