椭圆曲线y2=x3+(p-4)x-2p 的整数点

Integral Points on Elliptic Curve y 2=x 3+(p-4)x-2p

设p=81s^2+10是素数,其中s是使9s^2+2及9s^2+1/2都是素数的正奇数.运用初等数论的方法与技巧及四次丢番图方程的结果,证明了椭圆曲线y^2=x^3+(p-4)x-2p仅有整数点(x,y)=(2,0).

Let p=81s^2+10 be a prime,where s is a positive odd number satisfying that 9s^2+2 and 9s^2+1/2 are primes.By combining some methods and techniques of elementary number theory with some known results of quartic diophantine equations,it is proved that the elliptic curve y^2=x^3+(p-4)x-2p has only the integral point(x,y)=(2,0).