smc

It is obvious $x$, $y$, and $z$ are symmetrical. We are going to solve the problem by Completing the Square. $x^2+y^2+z^2-xy-yz-zx=1$ $2x^2+2y^2+2z^2-2xy-2yz-2zx=2$ $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=2$ Because $x,y,z$ are integers, $(x-y{)}^2$, $(y-z{)}^2$, and $(z-x{)}^2$ can only equal $0,1,1$. So one variable must equal another, and the third variable is $1$ different from those $2$ equal variables. So the answer is $\boxed{x=z\text{ and }y-1=x}$.