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. 

$n^2+a \mid n^3+a-n(n^2+a)$, therefore the divisibility is equivalent to
$$
n^2+a \mid a(n-1).
$$
Define the integer $k$ such that $a=n^2-k$. Then, equivalently
$$
k\mid (n^2-k)(n-1) \Leftrightarrow k\mid n^3-n^2.
$$
It follows that all solutions $(n,a)$ are in the form $(n,n^2-k)$, where $k$ stands for a divisor of $n^3-n^2$.
