Number theory problem, 3rd degree diophantine equation How many positive integers are there that can be written in the form
$$\frac{m^3+n^3}{m^2+n^2+m+n+1}$$
where $m$ and $n$ are positive integers.
I invented this problem and was stuck with it for a long time. It is really interesting to know if there is a solution for this problem. 
 A: A few lines of code trying the first 5000 numbers will give you those answers up to changing order between $m$ and $n$, let $f(m,n) = \frac{m^3 + n^3}{m^2+n^2+m+n+1}$:
$f(1,2) = 1; \qquad f(182,379) = 341; \qquad f(664,4286) = 4200; \qquad f(692, 3847) = 3747$
$f(961,1712) = 1531; \qquad f(996,692) = 896; \qquad f(1112, 270) = 1064$
$f(1712, 961) = 1513; \qquad f(2778,3210) = 3024$
Surely there are more but maybe you could try to find a pattern or see something in common?
As too whether it's interesting, well it's number theory. Fermat's Theorem is hugely famous and Andrew Wiles spent about 5 years to initially find a flawed proof which has then been corrected and yet it has no application whatsoever.
So I guess the interesting bit is purely mental, not practical (or at least not yet).
A: For fixed $n\ge 1$ we can sometimes solve the question as follows. Write
$$
\frac{m^3+n^3}{m^2+m+n^2+n+1}=(m-1) -\frac{(n^2+n)m-(n^3+n^2+n+1)}{m^2+m+(n^2+n+1)}.
$$
The LHS is an integer, if and only if the last fraction on the RHS is an integer. However, for fixed $n$, the denominator grows quadratically, and the nominator only linearly in $m$. For $m$ big enough, the fraction will not be an integer. For small $n$ this works, e.g., say for $n=1$ we have
$$
\frac{m^3+1}{m^2+m+3}=(m-1) -\frac{2m-4}{m^2+m+3},
$$
where the last fraction is an integer if and only if $m=2$. For $n=2$ we obtain
the fraction $\frac{6m-15}{m^2+m+7}$ which is an integer if and only if $m=1$, etc. In general, of course, this still leaves finitely many possible solutions for $m$, if $n$ is fixed, and we have not answered the question.
