Find all natural numbers $n$ such that $n$ divides $2^{n}-1$.
The number 1 works. Let $p$ be the smallest prime divisor of $n \neq 1$. Since $2^{n} \equiv 1 \pmod{n}$, we also have $2^{n} \equiv 1 \pmod{p}$, and by Fermat's Little Theorem, $2^{p-1} \equiv 1 \pmod{p}$, so the order of 2 modulo $p$ divides the GCD of $n$ and $p-1$, hence it is 1 because $p-1$ is smaller than the smallest prime number dividing $n$, and is coprime with $n$. Thus $2 \equiv 1 \pmod{p}$, so $p=1$, which is absurd. Therefore, 1 is the only solution.

## 2 Arithmetic TD

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