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. 

$4m^2+4mn+4n^2=49m+49n$
$4m^2+4mn+n^2+3n^2-49m-49n=0$
$(2m+n)^2+3n^2-(49/2)(2m+n)-(49/2)n=0$
$16(2m+n)^2-392(2m+n)+48n^2-392n=0$
$(4(2m+n)-49)^2+48n^2-392n=2401$
$3(4(2m+n)-49)^2+144n^2-1176n=7203$ 
$3(4(2m+n)-49)^2+(12n-49)^2=9604$
$3(8m+4n-49)^2+(12n-49)^2=9604$. 
$3x^2+y^2=98^2$ 
Now the last equation has only finitely many integer solutions, with a finite procedure for finding them, then for each solution you can check whether $8m+4n-49=x,12n-49=y$ has integer solutions, and that should do it. 
