How can I make the following 2 fractions integers? Let $m,n$ be integers. I want to find the possible values of $m,n$ such that $4(m+n)\over (2m+n)^2+3n^2$ and $4n\over (2m+n)^2+3n^2$ are both integers too. Would someone please help? Of course letting $(2m+n)^2+3n^2=4$ gives some good values, but is this all the $m,n$ I can get? 
Added: I can see that the problem can be reduced to  asking for $4k\over (2m+n)^2+3n^2$ to be an integer for both $k=m,n$

We can't have $m=n=0$. Let's suppose for a moment that $n=0$ then (using Graphth's answer) $n/(m^2+mn+n^2)=0$ and $m/(m^2+mn+n^2)=1/m$ and $1/m$ is integer only if $m=-1$ or $m=1$. Since the problem is symmetric in $m$ and $n$ we found the solutions $(m,n)=(0,-1),(0,1),(1,0),(-1,0)$.
We know that if $a|b$ and $a,b\neq0$ then $|a|\le|b|$. Since $m^2+mn+n^2$ is always positive, we can say $m^2+mn+n^2\le |n|$ and since $|n|\le n^2$ for $n\in\mathbb{Z}$ we have $m^2+mn+n^2\le n^2$ then $m^2+mn\le 0$. Analogously $mn+n^2\le 0$. Summing up we get $m^2+2mn+n^2\le0$ or $(m+n)^2\le 0$. This inequality is only true if $(m+n)=0$. Now suppose $m=-n$.
We get that $m^2=m^2+mn+n^2|m$ then $m=1$ or $m=-1$. Using the symmetry we get two more solutions $(m,n)=(1,-1),(-1,1)$.
