Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$

Hint: express as: $$4m+4n=3mn+\frac{4}{n}.$$

HINT: solving for $m$ is better, we do not need a quadratic equation: $$m=\frac{4(n^2-1)}{3n^2-4n}$$ We get $$m=3,n=2$$

As an alternative to Dr. Sonnhard Graubner's answer on how to find the complete set of solutions from

$m=\frac{4(n^{2}−1)}{3n^{2}−4n}$

consider the following:

We can factor that expression further to

$m=\frac{4(n+1)(n-1)}{n(3n−4)}$

That tells us a lot about $n$. First off, if $n$ is 1, then $m$ must be 0, but that is not a valid solution because the original expression involves a division by $m$. So $n$ cannot be 1. For the same reason, $n$ cannot be -1.

Since $n$ is an integer, both the numerator and denominator of that expression are integers. In order for their quotient $m$ to also be an integer, the numerator must be divisible by the denominator. Since $n\neq\pm1$, both $(n+1)$ and $(n-1)$ must be relatively prime to $n$, so neither $n$ nor any of its prime factors can divide either of these terms. Therefore, if $m$ is an integer, 4 must be divisible by $n$. The only candidates are $\pm1$, $\pm2$, and $\pm4$, and we already excluded $\pm1$.

Substituting $n=2$ gives us $m=3$, so $(m, n) = (3, 2)$ is a valid solution.

Substituting $n=4$ gives us $m=\frac{15}{8}$, which is not an integer and therefore not a valid solution. Likewise, $n=-2$ yields $m=\frac{3}{5}$ and $n=-4$ yields $m=\frac{15}{64}$.