椭圆Diophantine方程(x+p)(x^2+p^2)=y^2的本原解

The primitive solutions of elliptic Diophantine equation (x+p)(x^2+p^2)=y^2

设p是素数.在此给出了方程(x+p)(x2+p2)=y2有适合gcd(x,y)=1且y为奇数的正整数解(x,y)的充要条件.

Let p be an odd prime, we give a necessary and sufficient condition for the equation (x+p)(x^2+p^2)=y^2 to have positive integer solutions (x,y) satisfy gcd (x,y)=1 and y is odd.