show that the remainder upon dividing $a^n-1$ by $a^m-1$ is $a^r-1$












2












$begingroup$


If $n=qm+r, 0 leq r <m$, then show that the remainder upon dividing $a^n-1$ by $a^m-1$ is $a^r-1$.



Answer:



If $q$ is a positive integer, then $a^{m}$ divides $large a^{qm}-1$. For,



$a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$.



But I can not finish the problem.



Help me.






elementary-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    See this popular duplicate.
    $endgroup$
    – Dietrich Burde
    Jan 18 at 19:43
















2












$begingroup$


If $n=qm+r, 0 leq r <m$, then show that the remainder upon dividing $a^n-1$ by $a^m-1$ is $a^r-1$.



Answer:



If $q$ is a positive integer, then $a^{m}$ divides $large a^{qm}-1$. For,



$a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$.



But I can not finish the problem.



Help me.






elementary-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    See this popular duplicate.
    $endgroup$
    – Dietrich Burde
    Jan 18 at 19:43














2












2








2


1



$begingroup$


If $n=qm+r, 0 leq r <m$, then show that the remainder upon dividing $a^n-1$ by $a^m-1$ is $a^r-1$.



Answer:



If $q$ is a positive integer, then $a^{m}$ divides $large a^{qm}-1$. For,



$a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$.



But I can not finish the problem.



Help me.






elementary-number-theory







share|cite|improve this question









$endgroup$




If $n=qm+r, 0 leq r <m$, then show that the remainder upon dividing $a^n-1$ by $a^m-1$ is $a^r-1$.



Answer:



If $q$ is a positive integer, then $a^{m}$ divides $large a^{qm}-1$. For,



$a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$.



But I can not finish the problem.



Help me.






elementary-number-theory




elementary-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 19:18









M. A. SARKARM. A. SARKAR

2,2121619




2,2121619












  • $begingroup$
    See this popular duplicate.
    $endgroup$
    – Dietrich Burde
    Jan 18 at 19:43


















  • $begingroup$
    See this popular duplicate.
    $endgroup$
    – Dietrich Burde
    Jan 18 at 19:43
















$begingroup$
See this popular duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 19:43




$begingroup$
See this popular duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 19:43










2 Answers
2






active

oldest

votes


















4












$begingroup$

Hint: If $n=qm+r$, then $a^n-1 = a^r(a^{qm} - 1) + (a^r - 1)$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true?
    $endgroup$
    – M. A. SARKAR
    Jan 18 at 19:24






  • 2




    $begingroup$
    @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+cdots+x^{n-1})$$ with $x=a^m$ and $n=q$.
    $endgroup$
    – Clive Newstead
    Jan 18 at 19:26





















1












$begingroup$

$bmod, color{#c00}{a^{large m}!-!1}!:, a^{large r+mq}!= a^{large r} (color{#c00}{a^{large m}})^{large q}!equiv a^{large r} color{#c00}1^{q}!equiv a^{large r}$



i.e. if $ a^{large m}equiv 1$ then $, a^{large n}!equiv a^{large nbmod m},,$ i.e. $ $ expts on $,a,$ can be reduced $!bmod m$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Hint: If $n=qm+r$, then $a^n-1 = a^r(a^{qm} - 1) + (a^r - 1)$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true?
      $endgroup$
      – M. A. SARKAR
      Jan 18 at 19:24






    • 2




      $begingroup$
      @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+cdots+x^{n-1})$$ with $x=a^m$ and $n=q$.
      $endgroup$
      – Clive Newstead
      Jan 18 at 19:26


















    4












    $begingroup$

    Hint: If $n=qm+r$, then $a^n-1 = a^r(a^{qm} - 1) + (a^r - 1)$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true?
      $endgroup$
      – M. A. SARKAR
      Jan 18 at 19:24






    • 2




      $begingroup$
      @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+cdots+x^{n-1})$$ with $x=a^m$ and $n=q$.
      $endgroup$
      – Clive Newstead
      Jan 18 at 19:26
















    4












    4








    4





    $begingroup$

    Hint: If $n=qm+r$, then $a^n-1 = a^r(a^{qm} - 1) + (a^r - 1)$.






    share|cite|improve this answer









    $endgroup$



    Hint: If $n=qm+r$, then $a^n-1 = a^r(a^{qm} - 1) + (a^r - 1)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 18 at 19:21









    Clive NewsteadClive Newstead

    51.6k474135




    51.6k474135








    • 1




      $begingroup$
      Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true?
      $endgroup$
      – M. A. SARKAR
      Jan 18 at 19:24






    • 2




      $begingroup$
      @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+cdots+x^{n-1})$$ with $x=a^m$ and $n=q$.
      $endgroup$
      – Clive Newstead
      Jan 18 at 19:26
















    • 1




      $begingroup$
      Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true?
      $endgroup$
      – M. A. SARKAR
      Jan 18 at 19:24






    • 2




      $begingroup$
      @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+cdots+x^{n-1})$$ with $x=a^m$ and $n=q$.
      $endgroup$
      – Clive Newstead
      Jan 18 at 19:26










    1




    1




    $begingroup$
    Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true?
    $endgroup$
    – M. A. SARKAR
    Jan 18 at 19:24




    $begingroup$
    Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true?
    $endgroup$
    – M. A. SARKAR
    Jan 18 at 19:24




    2




    2




    $begingroup$
    @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+cdots+x^{n-1})$$ with $x=a^m$ and $n=q$.
    $endgroup$
    – Clive Newstead
    Jan 18 at 19:26






    $begingroup$
    @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+cdots+x^{n-1})$$ with $x=a^m$ and $n=q$.
    $endgroup$
    – Clive Newstead
    Jan 18 at 19:26













    1












    $begingroup$

    $bmod, color{#c00}{a^{large m}!-!1}!:, a^{large r+mq}!= a^{large r} (color{#c00}{a^{large m}})^{large q}!equiv a^{large r} color{#c00}1^{q}!equiv a^{large r}$



    i.e. if $ a^{large m}equiv 1$ then $, a^{large n}!equiv a^{large nbmod m},,$ i.e. $ $ expts on $,a,$ can be reduced $!bmod m$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $bmod, color{#c00}{a^{large m}!-!1}!:, a^{large r+mq}!= a^{large r} (color{#c00}{a^{large m}})^{large q}!equiv a^{large r} color{#c00}1^{q}!equiv a^{large r}$



      i.e. if $ a^{large m}equiv 1$ then $, a^{large n}!equiv a^{large nbmod m},,$ i.e. $ $ expts on $,a,$ can be reduced $!bmod m$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $bmod, color{#c00}{a^{large m}!-!1}!:, a^{large r+mq}!= a^{large r} (color{#c00}{a^{large m}})^{large q}!equiv a^{large r} color{#c00}1^{q}!equiv a^{large r}$



        i.e. if $ a^{large m}equiv 1$ then $, a^{large n}!equiv a^{large nbmod m},,$ i.e. $ $ expts on $,a,$ can be reduced $!bmod m$






        share|cite|improve this answer











        $endgroup$



        $bmod, color{#c00}{a^{large m}!-!1}!:, a^{large r+mq}!= a^{large r} (color{#c00}{a^{large m}})^{large q}!equiv a^{large r} color{#c00}1^{q}!equiv a^{large r}$



        i.e. if $ a^{large m}equiv 1$ then $, a^{large n}!equiv a^{large nbmod m},,$ i.e. $ $ expts on $,a,$ can be reduced $!bmod m$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 18 at 22:07

























        answered Jan 18 at 21:59









        Bill DubuqueBill Dubuque

        210k29192645




        210k29192645






























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