2023 AMC 10B Problem 14

The left side $m^2 + mn + n^2$ looks almost like $(m + n)^2 = m^2 + 2mn + n^2$. It is just one $mn$ short, so add $mn$ to both sides.

1. Make a square.
$$m^2 + 2mn + n^2 = m^2n^2 + mn, \quad\text{so}\quad (m + n)^2 = (mn)^2 + mn.$$
2. Give $mn$ a name. Let $p = mn$. The left side is a perfect square, so $p^2 + p$ must be a perfect square too.
3. Try small values of $p$.

4. Why bigger values never work.
- If $p$ is $1$ or more: $p^2 + p$ is bigger than $p^2$, but smaller than $p^2 + 2p + 1 = (p + 1)^2$. It sits strictly between two neighboring squares, so it is not a square. (Example: $p = 3$ gives $12$, between $9$ and $16$.)
- If $p$ is $-2$ or less: write $p = -q$ with $q \ge 2$. Then $p^2 + p = q^2 - q$, which is smaller than $q^2$ but bigger than $q^2 - 2q + 1 = (q - 1)^2$. Again it is stuck between two neighboring squares. (Example: $p = -3$ gives $6$, between $4$ and $9$.)
5. So $p = 0$ or $p = -1$. In both cases $p^2 + p = 0$, so $(m + n)^2 = 0$, which means $n = -m$.
- $p = 0$: $mn = -m^2 = 0$, so $m = 0$ and $n = 0$.
- $p = -1$: $mn = -m^2 = -1$, so $m^2 = 1$. That gives $(1, -1)$ and $(-1, 1)$.
6. Check one. $(1, -1)$: left side $1 - 1 + 1 = 1$, right side $1 \cdot 1 = 1$. It works.

There are $3$ ordered pairs: $(0, 0)$, $(1, -1)$, $(-1, 1)$.

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

A useful test for "is this a perfect square?" is to trap the number strictly between two neighboring squares; if it is trapped, it cannot be a square. Completing the square on the left turned the two-letter equation into a question about one number, $p = mn$, and then the trap finished it. When you see $m^2 + mn + n^2$, remember it is close to $(m + n)^2$.

Compare sizes. If both $m$ and $n$ are at least $2$ in size (ignoring sign), let $M$ be the larger of $m^2$ and $n^2$. The right side $m^2n^2$ is at least $4M$, but the left side $m^2 + mn + n^2$ is at most $M + M + M = 3M$. So there are no solutions there. That leaves the cases where $m$ or $n$ is $0$, $1$ or $-1$:

• $m = 0$: the equation is $n^2 = 0$, so $n = 0$.
• $m = 1$: $1 + n + n^2 = n^2$, so $n = -1$.
• $m = -1$: $1 - n + n^2 = n^2$, so $n = 1$.
• $n = 0, 1, -1$ give the same pairs with the roles swapped.

Again the pairs are $(0, 0)$, $(1, -1)$, $(-1, 1)$.