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Prove if $p∣m$ then $2^p−1∣2^m-1.$

I need this to understand this:

In general, if $p=\gcd(m,n)$ then $p=mx+ny$ for some integers $x,y$.

Now, if $d = \gcd(2^m-1,2^n-1)$ then $2^m \equiv 1 \pmod d$ and $2^n \equiv 1\pmod d$ so $$2^p = 2^{mx+ny} = (2^m)^x(2^n)^y \equiv 1 \pmod d$$

So $d\mid 2^p-1$.

On the other hand, if $p\mid m$ then $2^p-1\mid 2^m-1$ so $2^p-1$ is a common factor.

And yes, you can replace $2$ with any $a$.

In this context. Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$

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Jack Pan
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  • What specifically are you not understanding? – Riley Dec 03 '17 at 05:09
  • I do not understand why if $p|m$ then $2^p-1|2^m-1$, where it comes from at all. – Jack Pan Dec 03 '17 at 05:11
  • So $m=pn$ and you are wondering why $a^{n}\equiv1\pmod{a-1}$ where $a=2^p$? – Angina Seng Dec 03 '17 at 05:14
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    $a^k-1=(a-1)(a^{k-1}+a^{k-2}+\cdots+1),$. So, if $m=kp$ then $2^m-1=\left(2^{p}\right)^k-1=(2^p-1)(\dots)$. – dxiv Dec 03 '17 at 05:15
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    If it looks reasonable to you that $111$ (three ones) is a divisor of $111111111111111$ (fifteen ones), it should not come as a surprise that $2^3-1$ is a divisor of $2^{15}-1$: just think to base-2 representations. – Jack D'Aurizio Dec 03 '17 at 05:46

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A simple way to visualize it is to set $2^p=a$ and $m=kp $. In this way we are left with the trivial result

$$a−1∣a^k-1$$

Anatoly
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