Show that $a^{(p-1)!+1} equiv a mod p$












0












$begingroup$


I'm working in the following excercise:




Show that $a^{(p-1)!+1} equiv a mod p$




By Wilson's theorem I have:



$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$



I'm not sure about my proof, is that correct? any help will be really appreciated.






number-theory modular-arithmetic







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$endgroup$








  • 3




    $begingroup$
    Why can you suddenly rid of the minus sign?
    $endgroup$
    – Randall
    Jan 30 at 22:05






  • 1




    $begingroup$
    How can you justify that $-aequiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:06






  • 1




    $begingroup$
    How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:08
















0












$begingroup$


I'm working in the following excercise:




Show that $a^{(p-1)!+1} equiv a mod p$




By Wilson's theorem I have:



$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$



I'm not sure about my proof, is that correct? any help will be really appreciated.






number-theory modular-arithmetic







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Why can you suddenly rid of the minus sign?
    $endgroup$
    – Randall
    Jan 30 at 22:05






  • 1




    $begingroup$
    How can you justify that $-aequiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:06






  • 1




    $begingroup$
    How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:08














0












0








0





$begingroup$


I'm working in the following excercise:




Show that $a^{(p-1)!+1} equiv a mod p$




By Wilson's theorem I have:



$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$



I'm not sure about my proof, is that correct? any help will be really appreciated.






number-theory modular-arithmetic







share|cite|improve this question









$endgroup$




I'm working in the following excercise:




Show that $a^{(p-1)!+1} equiv a mod p$




By Wilson's theorem I have:



$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$



I'm not sure about my proof, is that correct? any help will be really appreciated.






number-theory modular-arithmetic




number-theory modular-arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 30 at 21:57









mrazmraz

450110




450110








  • 3




    $begingroup$
    Why can you suddenly rid of the minus sign?
    $endgroup$
    – Randall
    Jan 30 at 22:05






  • 1




    $begingroup$
    How can you justify that $-aequiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:06






  • 1




    $begingroup$
    How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:08














  • 3




    $begingroup$
    Why can you suddenly rid of the minus sign?
    $endgroup$
    – Randall
    Jan 30 at 22:05






  • 1




    $begingroup$
    How can you justify that $-aequiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:06






  • 1




    $begingroup$
    How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
    $endgroup$
    – fleablood
    Jan 30 at 22:08








3




3




$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05




$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05




1




1




$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06




$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06




1




1




$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08




$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08










1 Answer
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$begingroup$

Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$



[In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]






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    $begingroup$

    Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$



    [In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$



      [In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$



        [In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]






        share|cite|improve this answer









        $endgroup$



        Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$



        [In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 22:07









        MikeMike

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