n divides 2ⁿ − 1 only for n = 1

Answer. Only $n=1$.

Idea. The order of $2$ modulo the smallest prime factor of $n$ must be a divisor of $n$ smaller than that prime.

Proof. Let $n>1$ with $n\mid2^n-1$ and let $p$ be the smallest prime divisor of $n$. Since $2^n-1$ is odd, $p$ is odd. Let $d$ be the order of $2$ modulo $p$, i.e. the least $d\ge1$ with $2^d\equiv1\pmod p$.

• From $2^n\equiv1\pmod p$ we get $d\mid n$.
• From Fermat's little theorem $2^{p-1}\equiv1$ we get $d\mid p-1$.

So $d$ divides both $n$ and $p-1$. Now $d\ne1$ because $2\not\equiv1\pmod p$. Hence $d>1$ has a prime divisor $q$ with $q\le d\le p-1<p$, and $q\mid d\mid n$. This contradicts the minimality of $p$.

Check. $n=1$ works trivially; for instance $n=3$ fails since $2^3-1=7$.

Remark. The same argument shows that $n\mid2^n-1$ cannot hold for $n>1$, but $n\mid2^n+1$ has infinitely many solutions (e.g. $n=3,9,27,\dots$).

Lesson. "Take the smallest prime divisor and use orders" is the standard way to show that a divisibility $n\mid a^n-1$ is impossible.