Prove that $\forall n > 1, \quad2^n - 1 \pmod n \neq 0$
I've thought of the induction but I can't figure out how to prove the step. Fermat's theorem (and its variations) aren't particularly useful as well. Any hint would be greatly appreciated!
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Not sure how to solve this, but perhaps this will be useful: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2005&task=show_msg&msg=1441.0004 – Math1000 Apr 07 '15 at 21:11
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I edited my answer to make it clearer and more general. If anything is still not clear please feel welcome to ask questions. Note that Bezout's Lemma is not needed, only the fact that if $,a^n\equiv 1,$ then the order of $,a,$ divides $,n.\ \ $ – Bill Dubuque Apr 07 '15 at 23:26
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Assume you have got a $n>1$, so that $2^n \equiv 1 \pmod{n}$. Take the smallest prime divisor $p$ of $n$.
You have $p\;|\;n\;|\;2^n-1$, hence $2^n\equiv 1 \pmod{p}$.
But you have also, from Fermat's little theorem, that $2^{p-1}\equiv 1\pmod{p}$.
Since $p-1$ and $n$ are coprime, you have thus by Bézout's identity, some $a,b$ such that $a(p-1)+bn=1$, and then
$$1\equiv(2^{p-1})^a\cdot (2^n)^b \equiv 2^{a(p-1)+bn}\equiv 2^1 \pmod{p}$$
Thus $2^1\equiv 1\pmod{p}$, or equivalently $p\;|\;1$, but this is not possible.
Jean-Claude Arbaut
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A counterexample $\,n\,$ i.e. $\ n\mid 2^n-1\,$ would contradicts the more general theorem below.
Theorem $\,\ \ \ m,n>1,\,\ m\mid 2^{\large n}-1\,\Rightarrow\, \ell(m) > \ell(n),\ \ $ $\ell(x) =\,$ least prime factor of $\,x $
Proof $\quad\ \, {\rm mod}\,\ p = \ell(m)\!:\,\ 2^{\large n}\equiv 1\equiv\, 2^{\large\color{}{ p-1}}\,$ $\Rightarrow$ $\,\ \ell(m) = p > {\rm ord}\,2\color{#80f}{ \ge \ell(n)}$
Remark $\,\ $ The $ $ key dea $ $ is: $\ \ 2^n\equiv 1,\,\ \color{#0a0}{2\not\equiv 1}\,$ implies the order of $\,2\,$ is $\,\color{#80f}{\ge\ {\rm least\ prime}\,\ q\mid n},\ $ because the order must divide $\,n,\,$ and is $\color{}{\neq \color{#c00}1}$ (else $\,\color{#0a0}{2^{\color{#c00}{\large 1}}\equiv 1})$.
Bill Dubuque
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