Fix $a \in \mathbb{Z} \setminus \{0\}$. Find all integers $n$ such that $\frac{n^3+a}{n^2+a}\in \mathbb{Z}? $ May I know what is the correct approach to tackle the a/m problem? Since $n^3+a$ and $n^2+a$ have no common divisors, in order for $\dfrac{n^3+a}{n^2+a}\in \mathbb{Z},$ we must have $ n^3 +a = n^2 + a \implies n = 0 \lor 1 ?$

One of the ways is to find the common divisors of $n^3+a,n^2+a$ 
Let integer $d$ divides  both
$\implies d$ divides $n(n^2+a)-(n^3+a)=an-a$
Again, $d$ will divide $a(n^2+a)-n(an-a)=an+a^2$
Again, $d$ will divide $na+a^2-(an-a)=a^2+a$
One of the necessary condition for  $\displaystyle \frac{n^3+a}{n^2+a}$ be a positive integer is
that    $n^2+a$ must divide $a^2+a$
