Find all integers $n \geqslant 1$ such that $n$ divides $3^{n}+1$ and $11^{n}+1$.
First, $n=1$ and $n=2$ work. Conversely, suppose we have a solution $n \geqslant 3$. We write $n$ in the form $n=2^{\alpha} m$, where $m$ is odd.

- If $\alpha=0$, let $p$ be the smallest prime factor of $n$, and $\omega$ the multiplicative order of 3 modulo $p$. Since $\omega$ divides $p-1$ and $2 n$, it also divides $(2 n) \wedge(p-1)$, so it divides $2(n \wedge(p-1))=2$, which means $p$ divides $3^{2}-1=2^{3}$. This case is therefore impossible.
- If $\alpha \geqslant 1$, then $3^{n}+1 \equiv 9^{n / 2}+1 \equiv 1^{n / 2}+1 \equiv 2(\bmod 8)$, so $\alpha=1$ and $m \geqslant 3$. Let $p$ be the smallest prime factor of $m$, and $\omega$ the multiplicative order of 3 modulo $p$. Again, $\omega$ divides $p-1$ and $2 n$, so it divides $(2 n) \wedge(p-1)=4 \wedge(p-1)$, which means $p$ divides $3^{4}-1=2^{5} \times 5$. We conclude that $p=5$, and thus 5 divides $11^{n}+1$, which is also impossible.

In conclusion, the solutions are $n=1$ and $n=2$.