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I came across a problem where i had to tell the number of divisors of $2^i-1$ which are of the form $2^j-1$. I saw many contestants using the fact that if $i$ is divisible by $j$ then $2^i-1$ is divisible by $2^j-1$. How is that true ? I could not find a proof for this. Please help

$i>j\ge1$

Example for $i=6$ has $3$ factors $1,2,3 \lt 6$

and $2^6-1 =63$ has $3$ factor of form $2^j-1$

$1,3,7$

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Hint: How can you factorize $x^{ab}-1$ (see $x^{ab}$ as $(x^{a})^b$)

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  • $\begingroup$ But how can we factorise x^(abc*d...)-1 so that it is divisible by x^a-1,x^b-1,x^c-1.P.S-Sorry for asking stupid questions.Am in a learning phase now $\endgroup$ – ayushrocker92 Sep 2 '14 at 10:28
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    $\begingroup$ You have to make use of the formula $x^n-1=(x-1)(x^{n-1}+x^{n-2}\ldots+x+1)$. $\endgroup$ – Marc Bogaerts Sep 2 '14 at 11:44
  • $\begingroup$ oh got it thanks !! $\endgroup$ – ayushrocker92 Sep 2 '14 at 13:47

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