Clearly an even number $n$ cannot divide $2^{n}-1$, but about odd ones ? If $n$ is an odd prime this cannot happen neither since for an odd prime $p$ we have $2^p\equiv 2\pmod p$ and so $p$ cannot divide $2^p-1$ but what about the general case ($i.e~n$ an odd number larger than 1) ?
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Of course there is the case $n = 1$, but presumably you meant other than that. – Waleed Khan Apr 26 '14 at 17:01
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Yes, I took that into consideration. – Omran Kouba Apr 26 '14 at 17:07
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One thing is clear, $2^p-1|2^n-1$ if $p|n$ and so, $\forall p|n$ where $p$ are primes, $2^p-1|2^n-1$. – Samrat Mukhopadhyay Apr 26 '14 at 17:11
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@SamratMukhopadhyay: true, but that is not useful for this problem because of the way $n$ is specified. – Ross Millikan Apr 26 '14 at 17:13
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Yes, @RossMillikan, I was just stating my observations. – Samrat Mukhopadhyay Apr 26 '14 at 17:15
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The only (odd) value for $n$ for which $n \mid 2^n - 1$ is $n=1$.
Suppose for the sake of contradiction that $n>1$ is such that $n \mid 2^n - 1$ and let $p$ be the smallest prime divisor of $n$. Then we have $p \mid n \mid 2^n - 1$ or $2^{n} \equiv 1 \mod p$. We also have $2^{p-1} \equiv 1 \mod p$ by Fermat's little theorem. It follows that $2^{\gcd(n,p-1)} \equiv 1 \mod p$. Note that $n$ and $p-1$ are coprime since $p$ is the smallest prime divisor of $n$. Thus $2^1 \equiv 1 \mod p$, contradiction.
user133281
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Hint $ $ mod $\rm\color{#c00}{least}$ prime $\,p\mid n\!:\ 2^n \equiv 1\Rightarrow\, 2\,$ has order $\,k\mid n\,\color{#c00}{\Rightarrow}\ k \ge$ $\,p\,\Rightarrow\, 2^{p-1}\not\equiv 1\,\Rightarrow\!\Leftarrow$
Note $ $ Key Idea is: $ $ if $\ a\not\equiv 1,\,\ a^n\equiv 1\,$ then the order of $\,a\,$ is $\ge$ least prime $\,p\mid n.$
Bill Dubuque
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