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 $k=m,n$
On
I wrote small Maple program that search for possible solutions :
for n from -1000 to 1000 do
for m from -1000 to 1000 do
if not(m = 0 and n = 0) then
a:=(4*(m+n))/((2*m+n)^2+3*n^2):
b:=(4*n )/((2*m+n)^2+3*n^2):
if type(a,integer) and type(b,integer) then
print(m,n);
end if;
end if;
end do;
end do;
It seems that there are only four pairs :
$(m,n)=(0,-1) , (m,n)=(-1,0) , (m,n)=(1,0) , (m,n)=(0,1)$
EDIT :
As Arturo rightly observed pairs :
$(m,n)=(-1,1) , (m,n)=(1,-1)~$ are also solutions .
On
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 $1$.
On
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 $(m,n)=(1,-1),(-1,1)$.
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 $m=1$.