Find all integers $n>0$ such that $n \mid 2^{n}-1$.
First, $n=1$ is a solution.

Suppose now $n>1$ such that $n \mid 2^{n}-1$ and $p$ the smallest prime divisor of $n$.

$$
2^{n} \equiv 1 \quad(\bmod p)
$$

Thus the order $\omega$ of 2 modulo $p$ divides $n$. Moreover,

$$
\omega \leq \phi(p)<p
$$

so $\omega$ is smaller than any prime divisor of $n$, yet $\omega \mid n$ so $\omega=1$. Therefore,

$$
2 \equiv 1 \quad(\bmod p) \Longrightarrow p=1
$$

Impossible, hence $n$ cannot have a smallest prime divisor, thus the only solution to the equation is 1.