What are the $n \in \mathbb{N}$ such that $n+1 \mid n^{2}+1$?
We know that $\forall n \in \mathbb{N}, n+1 \mid n^{2}-1=n^{2}+1-2$. Therefore, a solution satisfies $n+1 \mid\left(n^{2}+1\right)-\left(n^{2}-1\right)=$ 2. Necessarily $n=0$ or $n=1$, these are the two solutions.