Prove that if $m$ divide $n$ then $x^{p^{m-1}}-1$ divide $x^{p^{n-1}}-1$ [duplicate]












0












$begingroup$



This question already has an answer here:




  • How to prove $,x^a-1 mid x^b-1 iff amid b$

    5 answers




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.






elementary-number-theory proof-explanation







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marked as duplicate by Bill Dubuque, Community♦ Jan 1 at 17:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    If $amid b$ then $(x^a-1)mid(x^b-1)$ as polynomials.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 15:53










  • $begingroup$
    And thus it need no $mmid n$, just $mleq n$.
    $endgroup$
    – greedoid
    Jan 1 at 16:04






  • 1




    $begingroup$
    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.
    $endgroup$
    – Bill Dubuque
    Jan 1 at 16:22












  • $begingroup$
    It´s correct what i wrote in the post. I think that both proves same.
    $endgroup$
    – Str0nger
    Jan 1 at 16:25










  • $begingroup$
    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 $(u^a-1,u^b-1) = (u^{gcd(a,b)}-1)$ ?
    $endgroup$
    – reuns
    Jan 1 at 16:38


















0












$begingroup$



This question already has an answer here:




  • How to prove $,x^a-1 mid x^b-1 iff amid b$

    5 answers




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.






elementary-number-theory proof-explanation







share|cite|improve this question











$endgroup$



marked as duplicate by Bill Dubuque, Community♦ Jan 1 at 17:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    If $amid b$ then $(x^a-1)mid(x^b-1)$ as polynomials.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 15:53










  • $begingroup$
    And thus it need no $mmid n$, just $mleq n$.
    $endgroup$
    – greedoid
    Jan 1 at 16:04






  • 1




    $begingroup$
    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.
    $endgroup$
    – Bill Dubuque
    Jan 1 at 16:22












  • $begingroup$
    It´s correct what i wrote in the post. I think that both proves same.
    $endgroup$
    – Str0nger
    Jan 1 at 16:25










  • $begingroup$
    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 $(u^a-1,u^b-1) = (u^{gcd(a,b)}-1)$ ?
    $endgroup$
    – reuns
    Jan 1 at 16:38
















0












0








0





$begingroup$



This question already has an answer here:




  • How to prove $,x^a-1 mid x^b-1 iff amid b$

    5 answers




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.






elementary-number-theory proof-explanation







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • How to prove $,x^a-1 mid x^b-1 iff amid b$

    5 answers




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.





This question already has an answer here:




  • How to prove $,x^a-1 mid x^b-1 iff amid b$

    5 answers







elementary-number-theory proof-explanation




elementary-number-theory proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 16:46









Bill Dubuque

209k29190632




209k29190632










asked Jan 1 at 15:52









Str0ngerStr0nger

12




12




marked as duplicate by Bill Dubuque, Community♦ Jan 1 at 17:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Bill Dubuque, Community♦ Jan 1 at 17:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    If $amid b$ then $(x^a-1)mid(x^b-1)$ as polynomials.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 15:53










  • $begingroup$
    And thus it need no $mmid n$, just $mleq n$.
    $endgroup$
    – greedoid
    Jan 1 at 16:04






  • 1




    $begingroup$
    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.
    $endgroup$
    – Bill Dubuque
    Jan 1 at 16:22












  • $begingroup$
    It´s correct what i wrote in the post. I think that both proves same.
    $endgroup$
    – Str0nger
    Jan 1 at 16:25










  • $begingroup$
    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 $(u^a-1,u^b-1) = (u^{gcd(a,b)}-1)$ ?
    $endgroup$
    – reuns
    Jan 1 at 16:38
















  • 1




    $begingroup$
    If $amid b$ then $(x^a-1)mid(x^b-1)$ as polynomials.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 15:53










  • $begingroup$
    And thus it need no $mmid n$, just $mleq n$.
    $endgroup$
    – greedoid
    Jan 1 at 16:04






  • 1




    $begingroup$
    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.
    $endgroup$
    – Bill Dubuque
    Jan 1 at 16:22












  • $begingroup$
    It´s correct what i wrote in the post. I think that both proves same.
    $endgroup$
    – Str0nger
    Jan 1 at 16:25










  • $begingroup$
    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 $(u^a-1,u^b-1) = (u^{gcd(a,b)}-1)$ ?
    $endgroup$
    – reuns
    Jan 1 at 16:38










1




1




$begingroup$
If $amid b$ then $(x^a-1)mid(x^b-1)$ as polynomials.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:53




$begingroup$
If $amid b$ then $(x^a-1)mid(x^b-1)$ as polynomials.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:53












$begingroup$
And thus it need no $mmid n$, just $mleq n$.
$endgroup$
– greedoid
Jan 1 at 16:04




$begingroup$
And thus it need no $mmid n$, just $mleq n$.
$endgroup$
– greedoid
Jan 1 at 16:04




1




1




$begingroup$
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.
$endgroup$
– Bill Dubuque
Jan 1 at 16:22






$begingroup$
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.
$endgroup$
– Bill Dubuque
Jan 1 at 16:22














$begingroup$
It´s correct what i wrote in the post. I think that both proves same.
$endgroup$
– Str0nger
Jan 1 at 16:25




$begingroup$
It´s correct what i wrote in the post. I think that both proves same.
$endgroup$
– Str0nger
Jan 1 at 16:25












$begingroup$
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 $(u^a-1,u^b-1) = (u^{gcd(a,b)}-1)$ ?
$endgroup$
– reuns
Jan 1 at 16:38






$begingroup$
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 $(u^a-1,u^b-1) = (u^{gcd(a,b)}-1)$ ?
$endgroup$
– reuns
Jan 1 at 16:38












1 Answer
1






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$begingroup$

Write $n=mk$ for $kinmathbb{N}$. 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.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Write $n=mk$ for $kinmathbb{N}$. 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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Write $n=mk$ for $kinmathbb{N}$. 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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Write $n=mk$ for $kinmathbb{N}$. 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.






        share|cite|improve this answer









        $endgroup$



        Write $n=mk$ for $kinmathbb{N}$. 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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 16:03









        Ben WBen W

        2,105615




        2,105615















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