y^2=x^3-3x^2+2x x^2=y^3-3x^2+2y 解方程组
问题描述：
y^2=x^3-3x^2+2x x^2=y^3-3x^2+2y 解方程组
rt
答
y^2=x^3-3x^2+2xx^2=y^3-3y^2+2y两式相减得：y^2-x^2=(x^3-y^3)-3(x^2-y^2)+2(x-y)(x-y)(x^2+xy+y^2-2x-2y+2)=0所以x-y=0,再代入1）式得：x^3-2x^2+2x=0,即x(x^2-2x+2)=0,得：x=0,故y=0,即(0,0)为一组解或x^2+xy+y^2...