$(n+1) \mid (n^3-1)$

Note first that $n^{3} - 1 = (n + 1)^{3} - 3n^{2} - 3n - 2 = ((n + 1)^{2} - 3n)(n + 1) - 2$ .

Now clearly $n = -1$ is not a solution since $0$ does not divide $2$ , so dividing through by $n + 1$ yields

$\dfrac{n^{3} - 1}{n + 1} = ((n + 1)^{2} - 3n) - \dfrac{2}{n + 1}$ ,

which will be an integer whenever $(n + 1)|2$ . Since the divisors of $2$ are just $\pm 1$ and $\pm 2$ , there are a total of $\boxed{4}$ possible values of $n$ , namely $-3,-2,0$ and $1$ .

Note that $n^3 + 1 = (n+1)(n^2 - n + 1)$ , so $n+1 \mid n^3 + 1$ . We now have the following

$\begin{aligned} n+1 &\mid n^3 - 1\\ n+1 &\mid n^3 + 1\\ \implies n+1 &\mid (n^3+1)-(n^3-1)=2\\ \end{aligned}$

Therefore, $n+1|2$ . The factors of 2 are $-2, -1, 1, 2$ , which gives the values of $n$ as $-3, -2, 0, 1$ . Therefore, there are 4 different integers which satisfy.