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. 

Finishing the original reasoning started in the question: 
$$(\frac {m+n}{2})^2 -mn \ge0 \Rightarrow (\frac {4k}{2})^2 - (16k^2 -49k) \ge0 $$
$$ 49k - 12k^2 \ge 0 \Rightarrow k\le4$$
If: $$3\lt k \le 4 \Rightarrow k=4$$
Finally, substituting $k=4$ in $m+n = 4k$, we get $m+n=16$. 
