Find all positive integers $n$ such that $n+1$ divides $n^{2}+1$.
Let $n$ be a potential solution. Since $n \equiv-1(\bmod n+1)$, we know that $0 \equiv n^{2}+1 \equiv(-1)^{2}+$ $1 \equiv 2(\bmod n+1)$. This means that $n+1$ divides 2. Since $n+1$ is positive, and the only positive divisors of 2 are 1 and 2, this means that $n=0$ or $n=1$.

Conversely, if $n=0$ or $n=1$, then $n=n^{2}$, so $n+1$ does indeed divide $n^{2}+1=n+1$.