2021 AMC Fall 10B Problem 12

"Sufficient to guarantee" means the condition must work for every triple that fits it, not just one. So first find out exactly which triples make the equation true.

1. Multiply out. The left side is
$$x^2 - xy + y^2 - yz + z^2 - zx.$$
2. Double it and regroup. Doubling gives $2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx$. Hand out the squares in pairs:
$$(x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2) = (x-y)^2 + (y-z)^2 + (z-x)^2.$$
3. Restate the equation. The original equation says the left side is $1$, so the doubled version is $2$:
$$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2.$$
4. Which integers do that? Three perfect squares adding to $2$ must be $1, 1, 0$. So one difference is $0$ (two of the numbers are equal) and the third number is exactly $1$ away from them. Example: $5, 5, 6$ gives $0 + 1 + 1 = 2$.
5. Check each choice.
- (A) $x > y$ and $y = z$: differences $x - y$, $0$, $x - y$. That works only if $x - y = 1$. With $x = 3$, $y = z = 1$ the sum is $8$. Not guaranteed.
- (B) $x = y - 1$, $y = z - 1$: three numbers in a row, like $1, 2, 3$. Squares $1 + 1 + 4 = 6$. Fails.
- (C) $x = z + 1$, $y = x + 1$: again three in a row, like $z = 1, x = 2, y = 3$. Sum $6$. Fails.
- (D) $x = z$ and $y = x + 1$: two equal, third one more. Squares $1 + 1 + 0 = 2$. Works every time.
- (E) $x + y + z = 1$: $x = 3, y = -1, z = -1$ fits, but the sum of squares is $16 + 0 + 16 = 32$. Fails.

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

The expression $x^2 + y^2 + z^2 - xy - yz - zx$ shows up often, and doubling it always turns it into a sum of three squared differences. Once it is written that way, "equals $1$" becomes a small puzzle about three squares, which you can solve completely. That tells you exactly what the numbers look like, so checking the choices is quick.

Test each choice with two different triples, not one. For (A), $x = 2, y = z = 1$ gives $2 + 0 - 1 = 1$, but $x = 3, y = z = 1$ gives $6 + 0 - 2 = 4$, so (A) is out. The same kind of second test knocks out (B), (C) and (E). For (D), put $x = z = t$ and $y = t + 1$ for any integer $t$:
$$t \cdot (-1) + (t+1) \cdot 1 + t \cdot 0 = -t + t + 1 = 1.$$
It works for every $t$.

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