Find all $n \in \mathbb{N}^{*}$ odd such that $n$ divides $3^{n}+1$.

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First, $n=1$ is a solution.

Moreover, let $q$ be the smallest prime divisor of $n$ and $d$ the order of 3 modulo $q$: $n$ is odd, so $q>2$. On one hand, $d \mid q-1$ and on the other hand $3^{n} \equiv-1(\bmod q)$, so $3^{2 n} \equiv 1(\bmod n)$, thus $d \mid 2 n$.
Therefore, $d$ divides the GCD of $2 n$ and $q-1$, which is 2 for the same reasons as in the previous exercise, so $d$ is 1 or 2, thus $q \mid 3^{2}-1=8$ so $q=2$, which is absurd since $n$ is odd.

Therefore, $n=1$ is the only solution.