Question

解方程组：y2=x3-3x2+2x;x2=y3-3y2+2y

Answer 1

y^2=x^3-3x^2+2x

x^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^2-2x-2y+2=0 ,化为x^2+x(y-2)+y^2-2y+2=0

delta=(y-2)^2-4(y^2-2y+2)=y^2-4y+4-4y^2+8y-8=-3y^2+4y-4=-(3y^2-4y+4)