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How to prove:

If $a\mid b$, then $2^b \equiv 1 \pmod{2^a-1}$.

I learned wilson's theorem, fermat's little theorem. I have no idea how to provide these 2 into the prove.

Jyrki Lahtonen
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user378456
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    You have $(m^x-1)| (m^{xy}-1)$ by factorising – N.S.JOHN Feb 17 '17 at 04:55
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    This is easier than it looks and you learned how to do it years before you ever took a number theory course. What is $\frac {x^n -1}{x-1} $? Just replace $x $ with $2^a $ and $n$ with $\frac bn $. – fleablood Feb 17 '17 at 04:58
  • Learn to search the site. Your question is a special case of this and covered in many other questions linked to that one. – Jyrki Lahtonen Feb 17 '17 at 05:51
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    And the basic ingredient is that for a number $q > 1$, the remainder of the division of $q^{b} - 1$ by $q^{a} - 1$ is $q^{r} - 1$, where $r$ is the remainder of the division of $b$ by $a$. – Andreas Caranti Feb 17 '17 at 10:40

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Note that since $b = ka$, $2^b = (2^a)^k$

$\begin{align}\text{Then }2^b-1 &= (2^a)^k-1 \\&= (2^a-1)\big((2^a)^{k-1}+(2^a)^{k-2}+\cdots+2^a+1\big)\end{align}$
and $2^a{-}1\mid 2^b{-}1$ as required.

Joffan
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