Factorization problem Find $m + n$ if $m$ and $n$ are natural numbers such that: $$\frac {m+n} {m^2+mn+n^2} = \frac {4} {49}\;.$$
My reasoning:
Say: $$m+n = 4k$$ $$m^2+mn+n^2 = 49k$$
It follows:$$(m+n)^2 = (4k)^2 = 16k^2 \Rightarrow m^2+mn+n^2 + mn = 16k^2 \Rightarrow mn = 16k^2 - 49k$$
Since: $$mn\gt0 \Rightarrow 16k^2 - 49k\gt0 \Rightarrow k\gt3$$
Then no more progress. 

Observe that $k$ must be a non-zero integer.
We know that $m, n$ are the roots of the quadratic equation 
$$X^2 - 4kX + (16k^2 - 49k)$$
The roots, from the quadratic equation, are
$$ \frac { 4k \pm \sqrt{(4k)^2 - 4(16k^2 - 49k) }} {2}  = 2k \pm \sqrt{ 49k - 12k^2}$$
The expression in the square root must be a perfect square.
Try $k = 1$, $49 k - 12k^2 = 37$ is not a perfect square.
Try $k = 2$, $49k - 12k^2 = 50$ is not a perfect square.
Try $k=3$, $49k-12k^2 = 39$ is not a perfect square.
Try $k=4$, $49k-12k^2 = 4$ is a perfect square. This leads to roots 6, 10, which have sum 16.
For $k\geq 5$, $49k - 12k^2 < 0$ has no solution.
For $k \leq -1$, $49k - 12k^2 < 0$ has no solution.
