# $n$ divides $2^n-1$ $\implies n=1$

If $n\mid (2^n-1)$, then $n=1$.

Somehow I am unsure if I got this right, my 'proof' seems to 'easy'. Can you please give me feedback?

So I take a prime divisor $p\mid n$. Then $p\mid (2^n-1)$, hence $2^n\equiv 1\mod p.$

So $2$ has multiplicative order $n$ in $\Bbb F_p^\times$ and therefore, by Lagrange's theorem, $n\mid (p-1)$. But since we also have $p\mid n$, this is only possible for $p=1$.

• No, $x^n=1$ in a group does not imply that the order of $x$ is $n$; it only implies that the order of $x$ divides $n$. – David C. Ullrich Jul 22 '16 at 14:53
• Right that's more reasonable – MyNameIs Jul 22 '16 at 14:57

Hint  mod $\rm\color{#c00}{least}$ prime $\,p\mid n\!:\ 2^n \equiv 1\Rightarrow\, 2\,$ has order $\,k\mid n\,\color{#c00}{\Rightarrow}\ k \ge$ $\,p\,\Rightarrow\, 2^{p-1}\!\not\equiv 1\,\Rightarrow\!\Leftarrow$
The key Idea is:  if $\ a\not\equiv 1,\,\ a^n\equiv 1\,$ then the order of $\,a\,$ is $\,\ge\,$ least prime $\,p\mid n.$
• So from $2^n\equiv 1\mod p$ we can only conclude that $2$ has order a divisor of $n$ in $\Bbb F_p^\times$? – MyNameIs Jul 22 '16 at 14:55
• @MyNameIs That's all we need here. Every divisor $>1$ of $\,n\,$ is $\ge$ the least prime factor $\,p\,$ of $\,n\,$ (by uniqueness of prime factorizations). – Bill Dubuque Jul 22 '16 at 14:58
• Thank you. In the last step you said $k\geq p$ implies $2^{p-1}\not\equiv1$. How did you conclude that? – MyNameIs Jul 22 '16 at 15:09