2021 AMC Fall 10B Problem 12

Expand the left side: $x^2 + y^2 + z^2 - xy - yz - zx$. Doubling it gives
$$2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = (x-y)^2 + (y-z)^2 + (z-x)^2.$$
So the equation says $(x-y)^2 + (y-z)^2 + (z-x)^2 = 2$. Three squares of integers summing to $2$ must be $1, 1, 0$ in some order: two of the numbers are equal and the third is exactly $1$ away from them.

Check the choices against this description.

• (A) $x > y = z$: the differences are $x - y, 0, x - y$, giving $2(x-y)^2$, which is $2$ only if $x - y = 1$. Not guaranteed.
• (B) $x, y, z$ consecutive ($z = x + 2$): squares $1, 1, 4$, total $6$. Fails.
• (C) $z, x, y$ consecutive: total $6$ again. Fails.
• (D) $x = z$ and $y = x + 1$: squares $1, 1, 0$, total $2$. Always works.
• (E) $x + y + z = 1$: for example $x = 1, y = z = 0$ works but $x = 3, y = -1, z = -1$ gives $(x-y)^2 + \dots = 16 + 0 + 16 = 32$. Fails.

The answer is $\boxed{\textbf{(D)}\ x=z \text{ and } y-1=x}$.

The expression $x^2 + y^2 + z^2 - xy - yz - zx$ appears constantly (it is the second factor in $x^3 + y^3 + z^3 - 3xyz$), and its half-sum-of-squares form makes its sign and size transparent. Turning "$= 1$" into a statement about differences reduces a symbolic question to a finite check.

Plug numbers into each condition, but use two different triples per condition. For (A), $x = 2, y = z = 1$ gives $1$ but $x = 3, y = z = 1$ gives $4$, so (A) is not sufficient. Only (D) survives such double-checking, since $x = z = t$, $y = t + 1$ gives $t(-1) + (t+1)(1) + t(0) = 1$ for every $t$.

• Testing one convenient triple for a condition (such as x = 2, y = z = 1 for choice A), seeing it work, and calling the condition sufficient.
• Misreading (B) or (C) as "two equal, third adjacent"; both describe three consecutive integers, which give $6$, not $1$.