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.

Hint $\rm\ \ n^2\!-\color{#0A0}7\mid n^3\!-\color{#C00}3\ \Rightarrow\ mod\,\ n^2\!-7\!:\,\ \color{#0A0}7^3\! = n^6 = \color{#C00}3^2\:\Rightarrow\: n^2\!-7\mid 7^3\!-3^2 =\, 2\cdot 167\ $ 
$\rm\,2,\,167\,$ are prime so, by unique factorization, $\rm\:n^2\!-7\mid\, 2\cdot 167\,$ $\Rightarrow$  $\rm\, n^2\!-7 = \pm\{1, 2,167, 334\},\,$ so $\ldots$ 
