# 2023 AMC 10B Problems/Problem 14

## Problem

How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2+mn+n^2 = m^2n^2$?

$\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5$

## Solution 1

Clearly, $m=0,n=0$ is one of the solutions. However, we can be quite sure that there are more, so we apply Simon's Favorite Factoring Trick to get the following:

\begin{align*} m^2+mn+n^2 &= m^2n^2\\ m^2+mn+n^2 +mn &= m^2n^2 +mn\\ (m+n)^2 &= m^2n^2 +mn\\ (m+n)^2 &= mn(mn+1).\\ \end{align*}

Essentially, this says that the product of two consecutive numbers $mn,mn+1$ must be a perfect square. This is practically impossible except $mn=0$ or $mn+1=0$. $mn=0$ gives $(0,0)$. $mn=-1$ gives $(1,-1), (-1,1)$. Answer: $\boxed{\textbf{(C) 3}}.$

~Technodoggo ~minor edits by lucaswujc

## Solution 2

Case 1: $mn = 0$.

In this case, $m = n = 0$.

Case 2: $mn \neq 0$.

Denote $k = {\rm gcd} \left( m, n \right)$. Denote $m = k u$ and $n = k v$. Thus, ${\rm gcd} \left( u, v \right) = 1$.

Thus, the equation given in this problem can be written as $$u^2 + uv + v^2 = k^2 u^2 v^2 .$$

Modulo $u$, we have $v^2 \equiv 0 \pmod{u}$. Because ${\rm gcd} \left( u, v \right) = 1$., we must have $|u| = |v| = 1$. Plugging this into the above equation, we get $2 + uv = k^2$. Thus, we must have $uv = -1$ and $k = 1$.

Thus, there are two solutions in this case: $\left( m , n \right) = \left( 1, -1 \right)$ and $\left( m , n \right) = \left( -1, 1 \right)$.

Putting all cases together, the total number of solutions is $\boxed{\textbf{(C) 3}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~ sravya_m18

## Solution 3 (Discriminant)

We can move all terms to one side and write the equation as a quadratic in terms of $n$ to get $$(1-m^2)n^2+(m)n+(m^2)=0.$$ The discriminant of this quadratic is $$\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).$$ For $n$ to be an integer, we must have $m^2(4m^2-3)$ be a perfect square. Thus, either $(2m)^2-3$ is a perfect square or $m^2 = 0$ and $m = 0$. The first case gives $m=-1,1$ (larger squares are separated by more than 3), which result in the equations $-n+1=0$ and $n-1=0$, for a total of two pairs: $(-1,1)$ and $(1,-1)$. The second case gives the equation $n^2=0$, so it's only pair is $(0,0)$. In total, the total number of solutions is $\boxed{\textbf{(C) 3}}$.

~A_MatheMagician

## Solution 4 (Nice Substitution)

Let $x=m+n, y=mn$ then $$x^2-y=y^2$$

Completing the square in $y$ and multiplying by 4 then gives $$4x^2+1=(2y+1)^2$$

Since the RHS is a square, clearly the only solutions are $x=0,y=0$ and $x=0,y=-1$.

The first gives $(0,0)$.

The second gives $(-1,1)$ and $(1,-1)$ by solving it as a quadratic with roots $m$ and $n$.

Thus there are $\boxed{\textbf{(C) 3}}$ solutions.

~ Grolarbear

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~Interstigation