<s>[INST] 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 $k=m,n$
 [/INST] Since $(2m+n)^2 + 3n^2 = 4m^2 + 4mn + n^2 + 3n^2 = 4m^2 + 4mn + 4n^2$, then
$$\frac{4(m+n)}{(2m+n)^2+3n^2} = \frac{4(m+n)}{4m^2+4mn+4n^2} = \frac{m+n}{m^2+mn+n^2}.$$
Similarly,
$$\frac{4n}{(2m+n)^2+3n^2} = \frac{4n}{4m^2+4mn+4n^2} = \frac{n}{m^2+mn+n^2}.$$
For both to be integers, you need $m^2+mn+n^2$ to divide both $m+n$ and $n$; hence, it must divide both $m$ and $n$. 
If $mn\gt 0$, then this is clearly impossible, since then $m^2+mn+n^2$ is strictly larger than each of $m$ and $n$ (as they are all integers). So $m$ and $n$ cannot both be positive or both be negative.
If $m=0$ or $n=0$, then you have that $m^2$ divides $m$, or $n^2$ divides $n$, hence you must have $(m,n) = (0,\pm 1)$ or $(\pm 1,0)$.
So we are now reduced to the case where one of $m,n$ is negative and one is positive. Let us say, without loss of generality that $m\lt 0\lt n$. If $|m|\lt n$, then $mn\lt n^2$, so $m^2+mn+n^2 \gt m^2 \geq |m|$; so $m^2+mn+n^2$ cannot divide $m$. If $|m|\gt |n|$, then $-m\gt n$, so $-m^2\lt mn$; hence $m^2+mn\gt 0$, so $m^2+mn+n^2\gt n^2\geq n$; hence $m^2+mn+n^2$ cannot divide $n$.
Finally, if $|m|=n$, then $mn=-n^2$, so $m^2+mn+n^2 = m^2 = n^2$; this divides $n$ if and only if $n=1$. So another solution is $(m,n) = (\pm 1,\mp 1)$, in which case your first fraction is $0$, and your second fraction is $1$. 
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