Find all $n\in \mathbb{Z}$ such that $n+1$ divides $n^3+2$.
Notice that $n+1$ divides $n^3+1$ so using linear properties of divisibility we have : $(n+1) \mid (n^3+2-n^3-1)$. So $(n+1)\mid 1$. So $n=0,-2$ are the only possibility.
Are there an other methods ? Thanks in advance !
 A: The systematic method is the Euclidean algorithm, which in this case has only one step:
$$x^3 + 2 = (x^2 - x + 1)(x + 1) + 1$$
A: $n+1|n^3+2\implies  n+1|(n^3+1)+1 \implies n+1|1\implies n+1=\pm1\implies n=-2,0$
A: One way to see things clearly is to let $n+1=m$, so you are asking for $m$'s such that $m$ divides $(m-1)^3+2=(m^3-3m^2+3m-1)+2=m^3-3m^2+3m+1$.  This implies $m$ divides $1$, so $m$ can only be $\pm1$, and thus $n=m-1$ can only be $0$ or $-2$.
A: This answer is written mostly because I believe that you really want to know are there any other methods (no matter how they look like) , and this is a method that is highly impractical and tedious but it is a method, although I really do not know will it lead to an answerable answer.
We have $(n+1) | (n^3+2)$, that is we have $n^3+2=k(n+1)$ for some integer $k$. From this we obtain $n^3-kn+(2-k)=0$.
Insert values of coefficients of this polynomial into formula for solutions of cubic equation.
This will give a relatively complicated expression that looks like it can be by cubing and squaring turned into a polynomial. 
Then that polynomial can be investigated to see for which $n$´s it can have an integer solution.
