Mordell方程y^3=x^2+2p^4的正整数解

The Positive Integer Solutions of Mordell's Equation y^3=x^2+2p^4

设p是奇素数.对于非负整数r,设U_(2r+1)=(α^(2r+1)+β^(2r+1))/2^(1/2),V_(2r+1)=(α^(2r+1)-β^(2r+1))/6^(1/2),其中α=(1+3^(1/2))/2^(1/2),β=(1-3^(1/2))/2^(1/2).运用初等数论方法证明了:方程y^3=x^2+2p^4有适合gcd(x,y)=1的正整数解(x,y)的充要条件是p=U_(2m+1),其中m是正整数.当上述条件成立时,方程仅有正整数解(x,y)=(V(2m+1)(V_(2m+1)~2-6),V_(2m+1)~2+2)适合gcd(x,y)=1.由此可知:当p<10000时,方程仅有正整数解(p,x,y)=(5,9,11),(19,1265,123),(71,68675,1683)和(3691,9677201305,4541163)适合gcd(x,y)=1.

Let p be an odd prime. For any nonnegative integer r, let U2r＋1 = （α^2r＋1 ＋ β^2r＋1）/√2 and V2r＋1 = （α^2r＋1 - β^2r＋1）/√6, where α = （1 ＋ x√3）√2,β ： （1 -√3）/√2.In this paper, using some elementary number theory methods, we prove that the equation y^3 = x^2＋ 2p^4 has positive integer solutions ix, y） with gcd（x,y） = 1 if and only if p = U2m＋1, where m is a positive integer. Moreover, if p = U2m＋1 then the equation has only the positive integer solution （x,y） = （V2m＋1（V2m＋1^2- 6）, V2m＋1^2, ＋ 2） with gcd（x,y） = 1. Thus it can be seen that if p 〈 10000, then the equation has only the positive integer solutions （p,x,y） = （5,9,11）,（19, 1265, 123）, （71, 68675, 1683） and （3691, 9677201305, 4541163） with gcd（x, y） = 1.