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I know a elementary way to prove that for $n>1$, $n$ never divides $2^n-1$:

Suppose that $n$ does divide $2^n - 1$. Then $n$ must be odd. Let $p$ be the smallest prime dividing $n$. Then $2^{p-1}\equiv1 \pmod p$. Let $m$ be the smallest divisor of $p - 1$ such that $2^m \equiv1 \pmod p$. Since $m$ is smaller than $p$ it must be coprime to $n$, so $n = qm + r$ with $0 < r < m$. Hence $2^r \equiv 1 \pmod p$. Contradiction.

One can find this solution here (it's from an old Putnam exam). But I heard that there is a nice way to prove the claim utilizing group theory and Lagrange theorem. Could somebody provide such a proof?

  • Lagrange's Theorem is a weird thing to use here, because it concludes that $n||G|$, rather than has it as an assumption. It seems more natural to use the "converse" of Lagrange's Theorem, except that converse isn't true in general. – Stella Biderman Apr 30 '17 at 13:53
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    $2^{p-1} \equiv 1 \pmod p$ (more generally $a^{p-1} \equiv 1 \pmod p$ for any $a$ not divisible by $p$) can be proved using Lagrange's theorem. – Derek Holt Apr 30 '17 at 14:04

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As in your given solution, assume that $n$ is odd. EDIT : For any prime divisor $p$ of $n$, $2^n \equiv 1$ mod $n$ implies $2^n \equiv 1$ mod $p$. This in turn implies that the order $m$ of $2$ mod $p$ in the group $(\mathbf Z/p\mathbf Z)^*$ must divide both $p-1$ (Lagrange !) and $n$ (by definition of the order). If you choose $p$ to be the smallest prime divisor of $n$, $p-1$ and $n$ are coprime, and you are done. Note that this is just a reformulation of your argument, using Lagrange as you wished.

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