Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation
\[x(x-y)+y(y-z)+z(z-x) = 1?\]
$\textbf{(A)} \: x>y$ and $y=z$
$\textbf{(B)} \: x=y-1$ and $y=z-1$
$\textbf{(C)} \: x=z+1$ and $y=x+1$
$\textbf{(D)} \: x=z$ and $y-1=x$
$\textbf{(E)} \: x+y+z=1$
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 $\boxed{D}$.
~[isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)