# Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$? [duplicate]

Clearly $2^n - 1$ is an odd integer, therefore, let $n$ be an odd integer and it divides $2^n - 1$. We can write $2^n - 1 = (2-1)(2^{n-1} + \cdots \cdots + 1) = (2^{n-1} + \cdots \cdots + 1)$

From here on I don't know how to proceed?

Suppose contrary that an integer $$n>1$$ exists such that $$n\mid 2^n-1$$. Clearly, $$n$$ is odd. Take $$p$$ to be the smallest prime dividing $$n$$. Then, $$p\mid 2^n-1$$ and $$p\mid 2^{p-1}-1$$. Hence, $$p\mid 2^d-1$$, where $$d:=\gcd\big(n,p-1\big)$$. However, as $$p$$ is the smallest prime divisor of $$n$$, we have $$d=1$$. Hence, $$p\mid 2^d-1=1$$, a contradiction. Hence, $$n$$ does not exist.
Related Question for Interested Party: There exist infinitely many $$n\in\mathbb{N}$$ such that $$n\mid 2^n+1$$. Examples are $$n=3^k$$, $$n=3^{2+k}\cdot 19$$, and $$n=3^{4+k}\cdot 19\cdot 163$$, where $$k\in\mathbb{N}_0$$. My question is as follows.
For any given $$r\in\mathbb{N}$$, does there exist $$n\in\mathbb{N}$$ with exactly $$r$$ prime divisors such that $$n\mid 2^n+1$$? (If such an $$n$$ exists for a particular $$r$$, it can be easily shown that $$3^kn$$ is also a solution for every $$k\in\mathbb{N}_0$$.)
• Sorry for making a comment on your problem 1 year ago, but why does $p | 2^d - 1$ – stenvik team Aug 23 '16 at 10:33
• @stenvikteam It is well known that $\gcd(a^k-1,a^l-1)=a^{\gcd(k,l)}-1$ for any positive integers $k$, $l$, and $a>1$. – user593746 Oct 24 '18 at 14:40