2023 AMC 10B Problem 14

Add $mn$ to both sides to complete a square on the left:
$$(m + n)^2 = m^2 n^2 + mn.$$
Let $p = mn$. Then $(m+n)^2 = p^2 + p$ must be a perfect square.

If $p \ge 1$, then $p^2 < p^2 + p < p^2 + 2p + 1 = (p+1)^2$, so $p^2 + p$ is strictly between consecutive squares: impossible.

If $p \le -2$, then $(p+1)^2 = p^2 + 2p + 1 < p^2 + p < p^2$ (the left inequality is $p + 1 < 0$, the right is $p < 0$), so $p^2 + p$ lies strictly between the consecutive squares $(|p| - 1)^2$ and $|p|^2$: impossible again.

So $p = 0$ or $p = -1$, and in both cases $(m+n)^2 = 0$, so $n = -m$.

• $p = 0$: $mn = 0$ with $n = -m$ forces $m = n = 0$.
• $p = -1$: $mn = -m^2 = -1$ gives $m = \pm 1$, so $(1, -1)$ and $(-1, 1)$.

Three ordered pairs satisfy the equation: $(0,0)$, $(1,-1)$, $(-1,1)$.

The answer is $\boxed{\textbf{(C)}\ 3}$.

Integer equations with a square on one side are attacked by squeezing: if an expression sits strictly between two consecutive perfect squares, it cannot be a square. The symmetric combination $m^2 + mn + n^2$ is one $mn$ short of $(m+n)^2$, which converts the equation into "$p^2 + p$ is a square," a one-variable bounding problem.

Small cases: $m = 0$ forces $n = 0$. For $|m|, |n| \ge 2$, $m^2 n^2 \ge 4\max(m^2, n^2) > m^2 + mn + n^2$, so no solutions. That leaves $|m| = 1$ or $|n| = 1$, e.g. $m = 1$: $1 + n + n^2 = n^2$ gives $n = -1$. Checking the few remaining cases yields the same three pairs.

• Forgetting the solution (0, 0) or counting (1, -1) and (-1, 1) as one, giving 1 or 2.
• Handling only positive $p = mn$ in the squeeze and missing the negative case, which is where the nonzero solutions live.