2021 Fall AMC 12B Problem 7

The expression satisfies $$\small \begin{aligned} &2\bigl[x(x-y) + y(y-z) + z(z-x)\bigr] \\ &= (x-y)^2 + (y-z)^2 + (z-x)^2. \end{aligned}$$ So the equation holds exactly when this sum of squares equals $2.$ Since the three differences sum to $0,$ this requires two of them to be $\pm 1$ and one to be $0.$ Option D gives $z - x = 0,$ $x - y = -1,$ and $y - z = 1,$ so the squares are $1 + 1 + 0 = 2.$ This works for all such integers. Thus, the correct answer is D.