Let $p$ be prime, $k ∈ mathbb{N}$ and suppose that $gcd(k,p-1)=d$. Show that $x^k ≡ 1pmod{p}$ has $d$...












1












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I'm lost here. Any help would be great.




Let $p$ be prime, $k ∈ mathbb{N}$ and suppose that $gcd(k,p-1)=d$. Show that $x^k ≡ 1pmod{p}$ has $d$ distinct solutions modulo $p$.







discrete-mathematics modular-arithmetic greatest-common-divisor







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    $begingroup$
    Do you know any group theory?
    $endgroup$
    – Arthur
    Jan 30 at 11:53










  • $begingroup$
    Immediate consequence of this result by taking $g$ to be a primitive root.
    $endgroup$
    – Bill Dubuque
    Jan 30 at 16:44
















1












$begingroup$


I'm lost here. Any help would be great.




Let $p$ be prime, $k ∈ mathbb{N}$ and suppose that $gcd(k,p-1)=d$. Show that $x^k ≡ 1pmod{p}$ has $d$ distinct solutions modulo $p$.







discrete-mathematics modular-arithmetic greatest-common-divisor







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know any group theory?
    $endgroup$
    – Arthur
    Jan 30 at 11:53










  • $begingroup$
    Immediate consequence of this result by taking $g$ to be a primitive root.
    $endgroup$
    – Bill Dubuque
    Jan 30 at 16:44














1












1








1





$begingroup$


I'm lost here. Any help would be great.




Let $p$ be prime, $k ∈ mathbb{N}$ and suppose that $gcd(k,p-1)=d$. Show that $x^k ≡ 1pmod{p}$ has $d$ distinct solutions modulo $p$.







discrete-mathematics modular-arithmetic greatest-common-divisor







share|cite|improve this question











$endgroup$




I'm lost here. Any help would be great.




Let $p$ be prime, $k ∈ mathbb{N}$ and suppose that $gcd(k,p-1)=d$. Show that $x^k ≡ 1pmod{p}$ has $d$ distinct solutions modulo $p$.







discrete-mathematics modular-arithmetic greatest-common-divisor




discrete-mathematics modular-arithmetic greatest-common-divisor






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 12:06









egreg

185k1486206




185k1486206










asked Jan 30 at 11:44









SomethingSomething

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817








  • 1




    $begingroup$
    Do you know any group theory?
    $endgroup$
    – Arthur
    Jan 30 at 11:53










  • $begingroup$
    Immediate consequence of this result by taking $g$ to be a primitive root.
    $endgroup$
    – Bill Dubuque
    Jan 30 at 16:44














  • 1




    $begingroup$
    Do you know any group theory?
    $endgroup$
    – Arthur
    Jan 30 at 11:53










  • $begingroup$
    Immediate consequence of this result by taking $g$ to be a primitive root.
    $endgroup$
    – Bill Dubuque
    Jan 30 at 16:44








1




1




$begingroup$
Do you know any group theory?
$endgroup$
– Arthur
Jan 30 at 11:53




$begingroup$
Do you know any group theory?
$endgroup$
– Arthur
Jan 30 at 11:53












$begingroup$
Immediate consequence of this result by taking $g$ to be a primitive root.
$endgroup$
– Bill Dubuque
Jan 30 at 16:44




$begingroup$
Immediate consequence of this result by taking $g$ to be a primitive root.
$endgroup$
– Bill Dubuque
Jan 30 at 16:44










1 Answer
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$begingroup$

If you know that there is a primitive root mod $p$, then this is easy: the solutions are $g^{jfrac{p-1}{d}}$ for $j=0,dots,d-1$, where $g$ is a primitive root.



Start by proving that $x^k ≡ 1bmod{p} iff x^d ≡ 1bmod{p}$.






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    $begingroup$

    If you know that there is a primitive root mod $p$, then this is easy: the solutions are $g^{jfrac{p-1}{d}}$ for $j=0,dots,d-1$, where $g$ is a primitive root.



    Start by proving that $x^k ≡ 1bmod{p} iff x^d ≡ 1bmod{p}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If you know that there is a primitive root mod $p$, then this is easy: the solutions are $g^{jfrac{p-1}{d}}$ for $j=0,dots,d-1$, where $g$ is a primitive root.



      Start by proving that $x^k ≡ 1bmod{p} iff x^d ≡ 1bmod{p}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If you know that there is a primitive root mod $p$, then this is easy: the solutions are $g^{jfrac{p-1}{d}}$ for $j=0,dots,d-1$, where $g$ is a primitive root.



        Start by proving that $x^k ≡ 1bmod{p} iff x^d ≡ 1bmod{p}$.






        share|cite|improve this answer









        $endgroup$



        If you know that there is a primitive root mod $p$, then this is easy: the solutions are $g^{jfrac{p-1}{d}}$ for $j=0,dots,d-1$, where $g$ is a primitive root.



        Start by proving that $x^k ≡ 1bmod{p} iff x^d ≡ 1bmod{p}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 12:44









        lhflhf

        167k11172404




        167k11172404






























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