Find all $n \in \mathbb{N}^{*}$ such that $n^{2} \mid 2^{n}+1$.

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We observe that $n$ is odd. $n=1$ is a solution. If $n>1$, let $p_{0}$ be the smallest prime divisor of $n$. Let $\omega_{p_{0}}(2)$ be the order of 2 modulo $p_{0}$. We have

$$
\begin{gathered}
\omega_{p_{0}}(2) \mid 2 n \\
\omega_{p_{0}}(2) \nmid n \\
\omega_{p_{0}}(2) \mid p_{0}-1
\end{gathered}
$$

By the minimality of $p_{0}$, $n$ is coprime with $p_{0}-1$, so $2 n \wedge (p_{0}-1)=2$, hence $\omega_{p_{0}}(2) \mid 2$, thus $\omega_{p_{0}}(2)=2$. Therefore, $p_{0} \mid (2^{2}-1)$, so $p_{0}=3$.

According to the Lifting The Exponent (LTE) theorem, $v_{3}(2^{n}+1^{n})=v_{3}(2+1)+v_{3}(n)$, so $v_{3}(n^{2}) \leqslant 1+v_{3}(n)$, hence $v_{3}(n) \leqslant 1$, thus $v_{3}(n)=1$. We denote $n=3 m$. We observe that $m=1$ gives $n=3$, which is indeed a solution.

Otherwise, let $p_{1}$ be the smallest prime divisor of $m$. $m^{2} \left\lvert \frac{8^{m}+1}{9} \right.$. Therefore,

$$
\begin{gathered}
\omega_{p_{1}}(8) \mid 2 m \\
\omega_{p_{1}}(8) \nmid m \\
\omega_{p_{1}}(8) \mid p_{1}-1
\end{gathered}
$$

so, as before, $\omega_{p_{1}}(8)=2$, thus $p_{1} \mid 63$, so $p_{1}=7$.

It suffices to observe that $7 \nmid \frac{8^{m}+1}{9}$.

The only two solutions are therefore $\mathfrak{n}=1$ and $\mathfrak{n}=3$.