For what integer $n$ is $1+n$ divisible by $1+n^2$? For what integer $n$ is $1+n$ divisible by $1+n^2$?
I tried it in modulo $3$ and $8$ but to no avail. 
Trial and error gave me answers $0$,$1$,$-2$ and $-3$.

Supposing you mean $n^2+1$ is divisible by $n+1$, we have:
$$n^2+1=(n^2-1)+2= (n-1)(n+1)+2$$
We know that $(n-1)(n+1)$ is divisible by $n+1$, and the problem states that $n^2+1$ is also divisible by $n+1$, thus $n+1$ divides their difference $(n^2+1)-(n^2-1)=2$. So acceptable values for $n+1$ are $1$, $2$, $-1$, $-2$, thus $n=0, 1, -2, -3$ are all integer solutions for this problem.
