A different way to show that $a^k-1 mid a^n-1$ if $kmid n$?
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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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$begingroup$
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
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What other way did he have in mind? Both directions seem pretty straightforward to me.
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– AlexR
Apr 22 '15 at 19:47
add a comment |
$begingroup$
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
$endgroup$
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
edited Mar 23 at 11:18
user477343
3,60831243
3,60831243
asked Apr 22 '15 at 19:43
SarahSarah
824821
824821
2
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What other way did he have in mind? Both directions seem pretty straightforward to me.
$endgroup$
– AlexR
Apr 22 '15 at 19:47
add a comment |
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
add a comment |
1 Answer
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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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Feb 18 at 17:31
answered Apr 22 '15 at 19:49
Bill DubuqueBill Dubuque
213k29195654
213k29195654
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What other way did he have in mind? Both directions seem pretty straightforward to me.
$endgroup$
– AlexR
Apr 22 '15 at 19:47