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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 $p=1$.

MyNameIs
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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 $\,p\mid n.$

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Bill Dubuque
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  • Are you a bot? That answer came so damn quick, i couldn't even type it that fast. – MyNameIs Jul 22 '16 at 14:53
  • @MyNameIs It's one of my old sci.math answers, where it was a FAQ. – Bill Dubuque Jul 22 '16 at 14:55
  • 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
  • At least he didn't answer it before it was asked. Actually, he did. – marty cohen Jul 22 '16 at 15:22
  • @MyNameIs By definition, order = smallest power $,k,$ such $,2^k\equiv 1,,$ so all smaller powers are $\not\equiv 1,,$ so this is true for the smaller power $,p-1 < k,\ $ i.e. $,2^{p-1}\not\equiv 1,,$ contra little Fermat. – Bill Dubuque Jul 22 '16 at 15:26