A different way to show that $a^k-1 mid a^n-1$ if $kmid n$?












3












$begingroup$


A student that I'm tutoring showed me the following problem:




Let $agt 1,$ and $k,ngt 0.$ Prove that $a^k-1mid a^n-1$ if and only if $kmid n$.



Solution: Since $k$ divides $n$, we have $n=ks$ for some integer $s$. $a^n-1=a^{ks}-1=(a^k)^s-1=(a^k-1)((a^k)^{s-1}+(a^k)^{s-2}+dots+(a^k)^2+a^k+1)$. Thus, $a^k-1$ divides $a^n-1$.



Now suppose that $a^k-1$ divides $a^n-1$. Let $n=ks+r$ where $r (0le rle k-1)$ is a remainder when we divide $n$ by $k$. $a^n-1=a^{ks}a^r-1=((a^k)^s-1)a^r+(a^r-1)$. By the argument above, $a^k-1$ divides $a^{ks}-1.$ Thus, $a^k-1$ must divide $a^r-1$. Since $rlt k,$ we have $r=0.$




He wanted to see another way to prove the converse and I got stumped! Can anyone think of an easy proof of why $a^k-1 mid a^n-1$ if $kmid n$?






number-theory elementary-number-theory







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  • 2




    $begingroup$
    What other way did he have in mind? Both directions seem pretty straightforward to me.
    $endgroup$
    – AlexR
    Apr 22 '15 at 19:47


















3












$begingroup$


A student that I'm tutoring showed me the following problem:




Let $agt 1,$ and $k,ngt 0.$ Prove that $a^k-1mid a^n-1$ if and only if $kmid n$.



Solution: Since $k$ divides $n$, we have $n=ks$ for some integer $s$. $a^n-1=a^{ks}-1=(a^k)^s-1=(a^k-1)((a^k)^{s-1}+(a^k)^{s-2}+dots+(a^k)^2+a^k+1)$. Thus, $a^k-1$ divides $a^n-1$.



Now suppose that $a^k-1$ divides $a^n-1$. Let $n=ks+r$ where $r (0le rle k-1)$ is a remainder when we divide $n$ by $k$. $a^n-1=a^{ks}a^r-1=((a^k)^s-1)a^r+(a^r-1)$. By the argument above, $a^k-1$ divides $a^{ks}-1.$ Thus, $a^k-1$ must divide $a^r-1$. Since $rlt k,$ we have $r=0.$




He wanted to see another way to prove the converse and I got stumped! Can anyone think of an easy proof of why $a^k-1 mid a^n-1$ if $kmid n$?






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What other way did he have in mind? Both directions seem pretty straightforward to me.
    $endgroup$
    – AlexR
    Apr 22 '15 at 19:47
















3












3








3


2



$begingroup$


A student that I'm tutoring showed me the following problem:




Let $agt 1,$ and $k,ngt 0.$ Prove that $a^k-1mid a^n-1$ if and only if $kmid n$.



Solution: Since $k$ divides $n$, we have $n=ks$ for some integer $s$. $a^n-1=a^{ks}-1=(a^k)^s-1=(a^k-1)((a^k)^{s-1}+(a^k)^{s-2}+dots+(a^k)^2+a^k+1)$. Thus, $a^k-1$ divides $a^n-1$.



Now suppose that $a^k-1$ divides $a^n-1$. Let $n=ks+r$ where $r (0le rle k-1)$ is a remainder when we divide $n$ by $k$. $a^n-1=a^{ks}a^r-1=((a^k)^s-1)a^r+(a^r-1)$. By the argument above, $a^k-1$ divides $a^{ks}-1.$ Thus, $a^k-1$ must divide $a^r-1$. Since $rlt k,$ we have $r=0.$




He wanted to see another way to prove the converse and I got stumped! Can anyone think of an easy proof of why $a^k-1 mid a^n-1$ if $kmid n$?






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$




A student that I'm tutoring showed me the following problem:




Let $agt 1,$ and $k,ngt 0.$ Prove that $a^k-1mid a^n-1$ if and only if $kmid n$.



Solution: Since $k$ divides $n$, we have $n=ks$ for some integer $s$. $a^n-1=a^{ks}-1=(a^k)^s-1=(a^k-1)((a^k)^{s-1}+(a^k)^{s-2}+dots+(a^k)^2+a^k+1)$. Thus, $a^k-1$ divides $a^n-1$.



Now suppose that $a^k-1$ divides $a^n-1$. Let $n=ks+r$ where $r (0le rle k-1)$ is a remainder when we divide $n$ by $k$. $a^n-1=a^{ks}a^r-1=((a^k)^s-1)a^r+(a^r-1)$. By the argument above, $a^k-1$ divides $a^{ks}-1.$ Thus, $a^k-1$ must divide $a^r-1$. Since $rlt k,$ we have $r=0.$




He wanted to see another way to prove the converse and I got stumped! Can anyone think of an easy proof of why $a^k-1 mid a^n-1$ if $kmid n$?






number-theory elementary-number-theory




number-theory elementary-number-theory






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edited Mar 23 at 11:18









user477343

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asked Apr 22 '15 at 19:43









SarahSarah

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  • 2




    $begingroup$
    What other way did he have in mind? Both directions seem pretty straightforward to me.
    $endgroup$
    – AlexR
    Apr 22 '15 at 19:47
















  • 2




    $begingroup$
    What other way did he have in mind? Both directions seem pretty straightforward to me.
    $endgroup$
    – AlexR
    Apr 22 '15 at 19:47










2




2




$begingroup$
What other way did he have in mind? Both directions seem pretty straightforward to me.
$endgroup$
– AlexR
Apr 22 '15 at 19:47






$begingroup$
What other way did he have in mind? Both directions seem pretty straightforward to me.
$endgroup$
– AlexR
Apr 22 '15 at 19:47












1 Answer
1






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4












$begingroup$

A modular way: $ {rm mod} a^{large k}!-!1!:, color{#c00}{a^{large k}equiv 1},Rightarrow, a^{large r+kq}equiv a^{large r}(color{#c00}{a^{large k}})^{large q}equiv a^{large r}color{#c00}1^{large q}equiv a^{large r},$



This uses standard Congruence Rules.



In the same way $, bequiv 1,Rightarrow, f(b)equiv f(1),$ for any polynomial $,f(x),$ with integer coefficients (above is $,f(x) = x^q, b = a^k).$ So the result, is a special case of the Factor Theorem $,b!-!1mid f(b)!-!f(1),,$ which is a special case of the linked Polynomial Congruence Rule. The advantage of the modular method is that one doesn't need to compute any quotients (only remainders, or congruent numbers).



Remark $ $ One can prove further that $, (a^k!-!1,a^n!-!1) = a^{(n,k)}!-!1,$ where $,(x,y):=gcd(x,y).,$ This is a special case of a strong divisibility sequence.






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    $begingroup$

    A modular way: $ {rm mod} a^{large k}!-!1!:, color{#c00}{a^{large k}equiv 1},Rightarrow, a^{large r+kq}equiv a^{large r}(color{#c00}{a^{large k}})^{large q}equiv a^{large r}color{#c00}1^{large q}equiv a^{large r},$



    This uses standard Congruence Rules.



    In the same way $, bequiv 1,Rightarrow, f(b)equiv f(1),$ for any polynomial $,f(x),$ with integer coefficients (above is $,f(x) = x^q, b = a^k).$ So the result, is a special case of the Factor Theorem $,b!-!1mid f(b)!-!f(1),,$ which is a special case of the linked Polynomial Congruence Rule. The advantage of the modular method is that one doesn't need to compute any quotients (only remainders, or congruent numbers).



    Remark $ $ One can prove further that $, (a^k!-!1,a^n!-!1) = a^{(n,k)}!-!1,$ where $,(x,y):=gcd(x,y).,$ This is a special case of a strong divisibility sequence.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      A modular way: $ {rm mod} a^{large k}!-!1!:, color{#c00}{a^{large k}equiv 1},Rightarrow, a^{large r+kq}equiv a^{large r}(color{#c00}{a^{large k}})^{large q}equiv a^{large r}color{#c00}1^{large q}equiv a^{large r},$



      This uses standard Congruence Rules.



      In the same way $, bequiv 1,Rightarrow, f(b)equiv f(1),$ for any polynomial $,f(x),$ with integer coefficients (above is $,f(x) = x^q, b = a^k).$ So the result, is a special case of the Factor Theorem $,b!-!1mid f(b)!-!f(1),,$ which is a special case of the linked Polynomial Congruence Rule. The advantage of the modular method is that one doesn't need to compute any quotients (only remainders, or congruent numbers).



      Remark $ $ One can prove further that $, (a^k!-!1,a^n!-!1) = a^{(n,k)}!-!1,$ where $,(x,y):=gcd(x,y).,$ This is a special case of a strong divisibility sequence.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        A modular way: $ {rm mod} a^{large k}!-!1!:, color{#c00}{a^{large k}equiv 1},Rightarrow, a^{large r+kq}equiv a^{large r}(color{#c00}{a^{large k}})^{large q}equiv a^{large r}color{#c00}1^{large q}equiv a^{large r},$



        This uses standard Congruence Rules.



        In the same way $, bequiv 1,Rightarrow, f(b)equiv f(1),$ for any polynomial $,f(x),$ with integer coefficients (above is $,f(x) = x^q, b = a^k).$ So the result, is a special case of the Factor Theorem $,b!-!1mid f(b)!-!f(1),,$ which is a special case of the linked Polynomial Congruence Rule. The advantage of the modular method is that one doesn't need to compute any quotients (only remainders, or congruent numbers).



        Remark $ $ One can prove further that $, (a^k!-!1,a^n!-!1) = a^{(n,k)}!-!1,$ where $,(x,y):=gcd(x,y).,$ This is a special case of a strong divisibility sequence.






        share|cite|improve this answer











        $endgroup$



        A modular way: $ {rm mod} a^{large k}!-!1!:, color{#c00}{a^{large k}equiv 1},Rightarrow, a^{large r+kq}equiv a^{large r}(color{#c00}{a^{large k}})^{large q}equiv a^{large r}color{#c00}1^{large q}equiv a^{large r},$



        This uses standard Congruence Rules.



        In the same way $, bequiv 1,Rightarrow, f(b)equiv f(1),$ for any polynomial $,f(x),$ with integer coefficients (above is $,f(x) = x^q, b = a^k).$ So the result, is a special case of the Factor Theorem $,b!-!1mid f(b)!-!f(1),,$ which is a special case of the linked Polynomial Congruence Rule. The advantage of the modular method is that one doesn't need to compute any quotients (only remainders, or congruent numbers).



        Remark $ $ One can prove further that $, (a^k!-!1,a^n!-!1) = a^{(n,k)}!-!1,$ where $,(x,y):=gcd(x,y).,$ This is a special case of a strong divisibility sequence.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 18 at 17:31

























        answered Apr 22 '15 at 19:49









        Bill DubuqueBill Dubuque

        213k29195654




        213k29195654






























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