Find all $n \in \mathbb{Z}_{>0}$ such that $n^2+a \mid n^3+a$ Could anyone advise me how to find all $n \in \mathbb{Z}_{>0}$ such that $n^2+a $ divides $ n^3+a,$ where $a \in \mathbb{Z} \setminus \{0\}$ is fixed ?
I have checked it is necessary that $n^2+a $ divides $ a^2+a,$ i.e $-a \leq n \leq a.$ Am I on the right track? Thank you. 

$n^2+a$ divides $n^3+na$ and so it divides $n^3+a$ if and only if $n^2+a$ divides the difference, which is $na-a$. Of course you want positive integer solutions, and for $n>a$ we have $n^2+a\geq na+a> na-a$.
Hence you have to try for values between $1$ and $a$.
