將xyz+xy+yz+xz+x+y+z-2變形
xyz+xy+y+xz+z+yz+x-2=xyz+y(x+1)+z(x+1)+yz+x-2
=xyz+(y+z)(x+1)+yz+x-2
=yz(x+1)+(y+z)(x+1)+yz+x-2
=(x+1)(yz+y+z+1-1)+x-2
=(x+1)(y+1)(z+1)-x-1+x-2
=(x+1)(y+1)(z+1)-3
xyz+xy+yz+xz+x+y+z-2=0 所以(x+1)(y+1)(z+1)-3=0
(x+1)(y+1)(z+1)=3
x,y,z均為非負整數(shù),(x+1),(y+1),(z+1)只能為正整數(shù),且不積不能大于3,只有
任意兩項為1,另一項為3.解得x=y=0,z=2(x,y,z的值可以互換)
所以x平方+y平方+z平方的值=4