To solve $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ How do we find all positive integers $(m,n)$ such that $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ ? 
 A: Since this is equivalent to the integer equation $$4n^2+4mn-4=3mn^2$$ then $n$ must divide $4$. Similar to @MarkBennet's answer, there are only a small number of cases to check. 
If $n=1$, then $4m=3m$, so this doesn't give any solutions.
If $n=2$, then $12+8m=12m$, and $m=3$.
If $n=4$, then $60+16m=48m$, which gives no solutions.

You can check for negative solutions too this way, but it turns out there are none:
If $n=-1$, then $-4m=3m$, so this doesn't give any solutions.
If $n=-2$, then $12-8m=12m$, which gives no solutions.
If $n=-4$, then $60-16m=48m$, which gives no solutions.
A: Since you need $\frac 1m+\frac 1n\gt \frac 34$ you need either $\frac 1m\ge \frac 38\gt \frac 13$ or alternatively $\frac 1n \gt \frac 13$
The number of cases is small and they can be checked quickly.

To be more concrete about the cases involved, we must have either $m\leq 2$ or $n\leq 2$. 
$m=1$ gives $\frac 1n-\frac 1{n^2}=-\frac 14$, or $4(n-1)=-n^2$which is impossible for positive integer $n$. Note that $n^2+4n-4=0$ has roots $n=-2\pm 2\sqrt 2$.
$m=2$ gives $\frac 1n-\frac 1{2n^2}=\frac 14$, which becomes $4n-2=n^2$ and this is impossible in integers since an even square is divisible by $4$. Also $n^2-4n+2=0$ has roots $n=2\pm \sqrt 2$
$n=1$ gives $\frac 1m+1-\frac 1m=\frac 34$ which is false.
$n=2$ gives $\frac 1m-\frac 1{4m}=\frac 14$ or $m=3$
