A number theory problem by Alejandro Castillo

$\dfrac{1}{m}+\dfrac{1}{n}-\dfrac{1}{mn^2} = \dfrac{3}{4}$

$\dfrac{(m+n)*n-1}{mn^2}=\dfrac{3}{4}$

$4mn+4n^2-4=3mn^2$

$4(mn-1)=n^2(3m-4)$

$n^2=2^2\dfrac{mn-1}{3m-4}$

Here $\dfrac{mn-1}{3m-4}$ must be a perfect square.

Here if $m$ is even, then denominator is always even and numerator is always odd.

Proof, $m=2k$ $\implies$ $\dfrac{mn-1}{3m-4}= \dfrac{2kn-1}{3(2k)-4}$ , As $2kn$ is always even so $2kn-1$ is always odd.

While, $3m-4=3(2k)-4=2(3k-2)$ is always even. So this cannot be possible.

Hence $m$ must be odd.For $m=3$ we get $n=2$ Now for higher values of $m$ , the value of $n^2$ will be in fraction. So it is not possible.

Hence, $m+n = 3+2=\boxed{5}$

$\dfrac{1}{m} + \dfrac{1}{n} - \dfrac{1}{mn^2} = \dfrac{3}{4}$

Clearly, $m,n \neq 0$ .

Rearranging the condition, we get: $(3m - 4)n^2 - (4m)n + 4 = 0$ .

Solving yields: $n = 2*\dfrac{m \pm \sqrt{m^2 - 3m + 4}}{3m - 4}$ .

Therefore, $m^2 - 3m + 4$ has to be a perfect square.

Let us consider an equation:

$m^2 - 3m + 4 = k^2,$ where $k$ is integer.

Calculating the discriminant, we get:

$D = 9 - (4 - k^2) = 5 + k^2$ , which has to be a perfect square as well, and it can only be true if $k = 2$ .

In this case $m = 0$ and $m = 3$ are both solutions, but $m \neq 0$ by condition. So, $m = 3$ .

Finding $n$ yields $n = 2$ and the only pair of integers we get is $(m, n) = (3, 2)$ .

So, the answer is $m + n = 3 + 2 = \boxed{5}$

Consider the equation 1/m+1/n-1/mn^2=3/4. This can be simplified using 1/m+1/n=(m+n)/mn because one can multiply 1/m by n/n and vice versa to keep equality. Now n(m+n)/(mn^2)-1/(mn^2)=3/4. This rearranges to n(m+n)-1=3(mn^2)/4. Nw we have 4(nm+n^2-1)=3(mn^2). This rearranges to 4nm+n^2-3mn^2=4. Now factor out the n. Since we are restricted to integer domain, n must be a positive or negative factor of 4. Trying out values leads to the sole solution (m,n)=(3,2) and the answer is 5.

Note that, $\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2} = \frac{3}{4}$ would imply: $(9mn-12n-16)(3n-4)=28$ (after some re-arranging) .Hence, $(3n-4)$ | $28$ and considering $3n-4 \equiv 2$ $mod(3)$ , it must be the case that $(3n-4)$ is a factor of 28 which is congruent to $2$ modulo 3.

That is, $(3n-4) \in \{-1,2,-7,14,-28\}$

Setting $3n-4$ equal to each of these in turn and then equating $9mn-12n-16$ to the complementary divisor yields a value for $m$ . In only one case the value of m is a non-zero integer and hence the only solution is $m=3 , n=2$ (calculation shows that this is indeed a solution to the original equation).