Find all integers $n \geq 1$ such that $n$ divides $2^{n}-1$.
We notice that $n=1$ works. Let $p$ be a prime divisor of $n$. Then $2^{n} \equiv 1 \pmod{p}$ and $2^{p-1} \equiv 1 \pmod{p}$ by Fermat's little theorem. Therefore, considering $p$ minimal, $p \mid 2^{\gcd(n, p-1)}-1=2-1=1$, which is absurd. The answer is therefore 1.