方程問題(15分)

7. 設,x,y是整數，且，xy+x+y+19=0,x^2y+xy^2+20=0
求x^2+y^2=
由x^2y+xy^2+20=0
xy(y+x)=-20
又xy+x+y+19=0
xy+x+y=-19
xy+x+y-1=-20
所以
xy+x+y-1=xy(y+x)
xy+x+y-1-xy(y+x)=0
xy(x+y-1)+(x+y-1)=0
(xy+1)(x+y-1)=0
即
xy=-1 或 x+y=1
若xy=-1
則因xy+x+y+19=0
-1+x+y+19=0
x+y=-18
x^2+y^2
=(x+y)^2-2xy
=(-18)^2-2(-1)
=326
若x+y=1
則因xy+x+y+19=0
xy+1+19=0
xy=-20
x^2+y^2
=(x+y)^2-2xy
=(1)^2-2(-20)
=41