<s>[INST] Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$ I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine.

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

Initially it looked like a typical quadratic problem, however I hit a dead end each time I solve it. 
My methodology is as follows, 
$$\frac{n^2+mn-1}{mn^2}=\frac{3}{4}$$
$$\implies 4n^2+4mn-4 = 3mn^2$$
$$\implies (4-3m)n^2+4m \cdot n-4=0$$
I used the quadratic formula, and got, 
$$n = \frac{-4m \pm \sqrt{(4m)^2-4\cdot(4-3m)\cdot(-4)}}{2(4-3m)}$$
I do the usual algebraic manipulations and drop at, 
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
I am unsure how I go ahead from this. 
Some help would be much appreciated. 
Cheers!
 [/INST] The other answers may be more elegant, but here is a way to continue from the point you reached. Assume $m$ and $n$ are integers. From
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
you know that $-2m \pm 2\sqrt{m^2-3m+4}$ must be a multiple of 
the integer $4-3m$; in particular, it must be an integer. 
Therefore $2\sqrt{m^2-3m+4}$ is an integer
(since it is a difference of the integer $-2m$ and another integer), 
which implies that $4(m^2-3m+4)$ is the square of an integer.
Suppose $4(m^2-3m+4) = k^2$ for some integer $k.$
Complete the square:
\begin{align}
k^2 &= 4(m^2-3m+4)\\ &= 4m^2 - 12m + 16 \\
&= (2m - 3)^2 + 7
\end{align}
and therefore the difference between the two squares $(2m - 3)^2$
and $k^2$ is $7.$
This is possible only if the two squares are $9$ and $16.$
Therefore $k^2 = 16$ and $$(2m - 3)^2 = 9.$$
This quadratic equation in $m$ has two roots, $m = 0$ and $m = 3,$
but $m = 0$ cannot be true for any solution to the original problem
(since we require $1/m$ to be defined); therefore
$$ m = 3.$$
Plug this into your equation for $n$ and confirm that the only possible integer value of $n$ is $n = 2.$

Alternative method:
Suppose that you solved for $m$ as in several other answers,
so that you found the equation
$$ m = \frac{4(n^2 - 1)}{3n^2 - 4n}. $$
If you know a few facts about rational functions (polynomials divided by polynomials), you can sketch $m$ as a function of $n$ free-hand
(without the assistance of a graphing calculator or software)
and solve the problem graphically.
First, plot the functions $m = 4(n^2 - 1)$ and $m = 3n^2 - 4n$
as shown below.

Because $3n^2 - 4n = 0$ at $n=0$ and $n = \frac43,$ we know that the rational function has vertical asymptotes at the lines $n=0$ and $n = \frac43.$
By looking at the highest-order coefficients of $4(n^2 - 1)$ and $3n^2 - 4n,$
we know the rational function has a horizontal asymptote at $m = \frac43.$
We know the rational function has the same zeros as $4(n^2 - 1),$
namely $n = \pm1.$
And by carefully accounting for the direction in which $3n^2 - 4n$ crosses the $n$-axis and the sign of $4(n^2 - 1)$ at each crossing,
we can find out how the rational function approaches each of its asymptotes.
The result is something like the following graph:

Assuming we sketched this free-hand, we do not yet know how close the curve is to various points with integer coordinates.
But since the middle branch has asymptotes $n=0$ and $n = \frac43,$ its only possible integer solution is at $n=1$ (which indeed occurs at $(1,0)$).
Since the left branch passes through $(-1,0),$ its only possible other
integer solution is at $m=1$ (at which we can solve for $n,$ discovering that $n$ is not an integer in that case).
We can then chip away at the right branch, either by finding $m$ when
$n = 2$ (the smallest possible integer value of $n$ on that branch)
or by solving for $n$ when $m = 2$ 
(the smallest possible integer value of $m$ on that branch).
Either way, we quickly find out that $(2,3)$ is a solution,
that the only other possible integer solution would be at $m = 2,$
and that in fact the curve crosses $m = 2$ at a non-integer value of $n$;
therefore there are no other solutions.
We eliminate $(1,0)$ and $(-1,0)$ by referring to the original problem statement, leaving only $n = 2, m = 3.$
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