x^3=y^3+2y^2+1求所有整数数对（x,y)

Question

x^3=y^3+2y^2+1 求所有整数数对（x,y)

Answer 1

2y^2+1>0
所以x^3>y^3,x>y
假设x=y+t,t为正整数
(y+t)^3=y^3+2y^2+1
y^3+3y^2*t+3y*t^2+t^3=y^3+2y^2+1
(3t-2)y^2+3t^2*y+(t^3-1)=0
判别式=(3t^2)^2-4(3t-2)(t^3-1)=-3t^4+8t^3+12t-8为完全平方数
t=1时,判别式=9,方程变为y^2+3y=0,y=0或-3,相应的x=1或-2
t=2时,判别式=32,不是完全平方数
t=3时,判别式=1,方程变为7y^2+27y+26=0,y不为整数,舍去
t>=4时,判别式=-3t^4+8t^3+12t-8=t^3(8-3t)+12t-8