<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] We find all integer values of $m$ and $n$ such that $\frac{4n}{(2m+n)^2+3n^2}$ is an integer.  Once this is done, your problem that imposes additional conditions is easily solved. 
If $3n^2 > 4n$, then the bottom has absolute value greater than the absolute value of the top.  Thus all but $n=0$, $n=\pm 1$ are immediately ruled out.
If $n=0$, we are looking at $\frac{0}{4m^2}$, which is an integer for all non-zero $m$. 
If $n=1$, then we are looking at $\frac{4}{4m^2+4m+4}$, or equivalently at $\frac{1}{m^2+m+1}$.  This is an integer only in the cases $m^2+m+1=\pm 1$. The equation $m^2+m+1=1$ has the solutions $m=0$ and $m=-1$.  The equation $m^2+m+1=-1$ has no real solutions, let alone integer solutions.
By a similar argument, or by symmetry, the case $n=-1$ gives the solutions $m=0$ and $m=-1$.
So for integer values of $n$ and $n$, $\frac{4n}{(2m+n)^2+3n^2}$ is an integer precisely in the following cases:
(i) $n=0$, $m$ arbitrary non-zero; (ii) $n=1$, $m=0$ or $m=-1$; and (iii) $n=-1$, $m=0$ or $m=1$.   
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