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. 

It is useful to write the stipulation of the problem in a few different ways:
$$\frac{49}{4}=\frac{(m+n)^2-mn}{m+n}=m+n-\frac{mn}{m+n},$$
$$\frac{49}{4}=\frac{(m+n)n+m^2}{m+n}=n+\frac{m^2}{m+n},$$
$$\frac{49}{4}=\frac{(m+n)m+n^2}{m+n}=m+\frac{n^2}{m+n}.$$
From the first, we glean that $m+n>12$ because $49/4=12+1/4$ and $m,n>0$. From the second and third, we see that $n\le 12$ and $m\le 12$.
Another thing to notice is that $mn/(m+n)$ must reduce to a fraction with a denominator of $4$. This means that
$$4mn=c(m+n)$$
for some $c$ relatively prime to $4$. This implies that $4$ divides $m+n$, so there are only three cases to consider: $m+n=16$, $m+n=20$, $m+n=24$. From the first, we have
$$12+\frac{1}{4}=16-\frac{m(16-m)}{16}$$
which easily gives the solution $\{m,n\}=\{6,10\}$. The other two cases result in quadratic equations with no real solutions.
