Fall 2021 AMC 10 B

Answer (D): The given equation is equivalent to $2x(x-y)+2y(y-z)+2z(z-x)=2$, which can be rewritten as $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=2$. This equation has an integer solution if and only if two of the squares are 1 and one is 0. This in turn means that two of the variables must be equal and the third must differ from this common value by 1. Choice (D) gives one instance of this, and the other choices do not imply this condition. Specifically, choice (A) fails when $x=2$, $y=0$, and $z=0$ because the left-hand side of the original equation equals 4; choice (B) fails when $x=1$, $y=2$, and $z=3$ because the left-hand side equals 3; choice (C) fails when $x=1$, $y=2$, and $z=0$ because the left-hand side equals 3; and choice (E) fails when $x=2$, $y=0$, and $z=-1$ because the left-hand side equals 7.