Integer pairs

$\begin{aligned} mn & = 2m + 4n \\ mn - 2m - 4n & = 0 \\ mn - 2m - 4n + 8 & = 8 \\ (m-4)(n-2) & = 8 \end{aligned}$

Since $m$ and $n$ are integers, $(m-4)(n-2)$ are factor pairs of $8$ . And since $8=2^{\red 3}$ , $8$ is $\red 3+1 = 4$ positive ordered factor pairs and $4$ ordered negative factor pairs, that is $(\pm 1, \pm 8)$ , $(\pm 2, \pm 4)$ , $(\pm 4, \pm 2)$ , $(\pm 8, \pm 1)$ , a total of $8$ ordered factor pairs.

When $m=n$ , we have $(n-4)(n-2) = 8 \implies n^2-6n + 8 = 8 \implies n(n-6) = 0 \implies m=n=0$ and $m=n= 6$ from the two factor pairs $(-4, -2)$ and $(2, 4)$ respectively. Therefore, there are $8-2 = \boxed 6$ ordered integer pairs $(m,n)$ , where $m\ne n$ , satisfying the equation.