Let $\sigma(n)$ be the sum of the prime divisors of $n$ (counted without multiplicity). Find all $n$ such that:

$$
\sigma\left(2^{n}+1\right)=\sigma(n)
$$
(Solved by Elias Caeiro)

Let us note that by Zsigmondy, if $n \neq 3, 2^{n}+1$ admits a primitive prime factor which is therefore congruent to 1 modulo $2n$. Indeed, if we denote $q$ this factor, $q \neq 2$ and the order of 2 modulo $q$ divides $2n$, but does not divide $n$. In particular, the order can be written as $2a$ with $a$ dividing $n$. We know that $p$ divides $\frac{2^{2a}-1}{2^{a}-1}=2^{a}+1$ so we must have $a=n$ because $q$ is primitive.

In particular, $\sigma\left(2^{n}+1\right)>2n$. Now, if we denote $p_{1}, \ldots, p_{k}$ the prime factors of $n$, we have $n \geqslant p_{1} \ldots p_{k}$ and $\sigma(n)=p_{1}+\cdots+p_{k} \leqslant 2 p_{1} \times \cdots \times p_{k}$ (we can easily show this by induction on $k$ since all the $p_{i}$ are at least 2).

It remains to consider the case $n=3$: we have $2^{n}+1=9$ so $\sigma\left(2^{n}+1\right)=\sigma(n)=3$. Thus, only $n=3$ is a solution.