x^3-y^3-z^3=3xyz
x^3-y^3-z^3-3xyz
=[(x-y)^3-3xy(x-y)]-z^3-3xyz
=[(x-y)^3-z^3]-3xy(x-y-z)
=[(x-y-z)^3-3(x-y)z(x-y-z)]-3xy(x-y-z)
=(x-y-z)^3-3(-xy+yz-zx)(x-y-z)
=(x-y-z)[(x-y-z)^2-3(-xy+yz-zx)]
=(x-y-z)(x^2+y^2+z^2+xy-yz+zx)=0
若x^2+y^2+z^2+xy-yz+zx=0
2x^2+2y^2+2z^2+2xy-2yz+2zx=0
(x^2+2xy+y^2)+(y^2-2yz+z^2)+(z^2+2zx+x^2)=0
所以(x+y)^2+(y-z)^2+(z+x)^2=0
所以只有x+y=0,y-z=0,z+x=0
所以y=z=-x
和正整数矛盾
所以x-y-z=0
x=y+z
x^2=2(y+z)
x^2=2x
x>0,x=2
y+z=2
则只有y=z=1
所以xy+yz+zx=2+1+2=5