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I have been trying for several days to prove the equality of the title, but without any positive results. I know I have to first prove that if m divides n (it has to be multiple of m) then $p^{m}-1$ divide $p^{n}-1$.

That is the one that I have been able to demonstrate knowing that there is a quotient for this division that has zero rest and is the following: $\sum_{i=1}^{\alpha}(p^{(\alpha-i)\times m})$.

Someone could help me relate it to the title and get the proof.

Thanks.

Bill Dubuque
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St3g4n0
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    If $a\mid b$ then $(x^a-1)\mid(x^b-1)$ as polynomials. – Angina Seng Jan 01 '19 at 15:53
  • And thus it need no $m\mid n$, just $m\leq n$. – nonuser Jan 01 '19 at 16:04
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    Did you intend to write $, p^{\large m-1}$ divides $,p^{\large n-1}\ $? $ $ If so please correct the question. If not please explain why you think you need to first prove that. – Bill Dubuque Jan 01 '19 at 16:22
  • It´s correct what i wrote in the post. I think that both proves same. – St3g4n0 Jan 01 '19 at 16:25
  • Both are not the same. To find $gcd(x^{p^m-1}-1,x^{p^n-1}-1)$ we can use that $gcd(u^a-1,u^b-1) = u^{gcd(a,b)}-1$ two times, in two different rings. Is it true in every ring $R$ that for $a,b \in \mathbb{Z}_{\ge 1},u \in R$ then as ideals $(u^a-1,u^b-1) = (u^{gcd(a,b)}-1)$ ? – reuns Jan 01 '19 at 16:38
  • @Str0nger If differs from the title so maybe the title is not what you intended. – Bill Dubuque Jan 01 '19 at 16:39

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Write $n=mk$ for $k\in\mathbb{N}$. Now we have $$p^n-1=(p^m)^k-1=(p^m-1)\sum_{j=0}^{k-1}(p^m)^j$$ This is an application of the factorization identity $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+1)$$ which you should definitely memorize.

Ben W
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