proof of this problem : if q|((b^m)-1) and (m|t) then q|((b^t)-1)

**mostafashaeri** · November 26th 2011, 07:15 AM

hello.
please help to proof this problem :

if m is least number that q|((b^m)-1) and (m|t) then q|((b^t)-1).

i don't know where to start.
tanx

**princeps** · November 26th 2011, 09:25 PM

Use following Lemma :

$a^n-1=(a-1)\cdot \displaystyle \sum_{i=0}^{n-1} a^{i}$

$b^t-1=b^{m\cdot k}-1=(b^{m})^{k}-1=(b^m-1)\cdot \displaystyle \sum_{i=0}^{k-1} b^{m\cdot i}$

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proof of this problem : if q|((b^m)-1) and (m|t) then q|((b^t)-1)

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