2021 Fall AMC 10B Problem 12

Let $E=x(x-y)+y(y-z)$ $+z(z-x).$ Expanding gives $$\begin{aligned} 2E={}&(x-y)^2+(y-z)^2\\ &+(z-x)^2. \end{aligned}$$ For the value to be $1,$ the three nonnegative square terms must sum to $2.$ Since $x,y,z$ are integers, this means the squared differences are $1,1,0.$ Thus two of the variables must be equal, and the third must differ from them by $1.$ The condition $x=z$ and $y-1=x$ guarantees exactly that. Thus, the answer is D .