Find all integers such that $\frac{n^3-3}{n^2-7}$ is an integer Find all integers such that $\frac{n^3-3}{n^2-7}$ is an integer.
I have no idea how to approach these types of proofs.
But I tried a few things, did not get me anywhere.
$n^3 -3 = an^2-7a$ then $n^3-an^2 = 3-7a$, and hence $n^2(n-a) = 3-7a$
And then I have no where to go...
Any help is appreciated thanks.

Note that
$$
\frac{n^3-3}{n^2-7} = n + \frac{7n-3}{n^2-7}.
$$
So it suffices to determine when $n^2-7$ divides $7n-3$. It is not hard to show that $|n^2-7| > |7n-3|$ when $|n|>8$; so for those $n$, it is impossible for $n^2-7$ to divide $7n-3$. That leaves only the cases $n=-8,-7,\dots,8$ to check, revealing the two solutions $n=-3$ and $n=3$.
