How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2+mn+n^2 = m^2n^2$?
$\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5$
Clearly, $m=0,n=0$ is 1 solution. However there are definitely more, so we apply [Simon's Favorite Factoring Expression](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick) to get this:
\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*}
This basically say that the product of two consecutive numbers $mn,mn+1$ must be a perfect square which 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