Question about Wilson's theoremHelp manipulating Wilson's Theorem.Wilson's Theorem and related identitiesWilson's theoremPart of a proof for Wilson's TheoremWilson's theorem intuition(Elementary Number Theory) Using Wilson's Theorem ..??Using Wilson's Theorem to prove Fermat's Little TheoremProve $717$ is not prime using Wilson's TheoremProving Wilson's theoremQuestion about Wilson's theorem, when $n = 4$.

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Question about Wilson's theorem


Help manipulating Wilson's Theorem.Wilson's Theorem and related identitiesWilson's theoremPart of a proof for Wilson's TheoremWilson's theorem intuition(Elementary Number Theory) Using Wilson's Theorem ..??Using Wilson's Theorem to prove Fermat's Little TheoremProve $717$ is not prime using Wilson's TheoremProving Wilson's theoremQuestion about Wilson's theorem, when $n = 4$.













1












$begingroup$


For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?



Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.



However is $p in mathbbZ$? Or is $p in mathbbN$?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    It's $(p-1)!$ : how do you define factorials for negative integers ?
    $endgroup$
    – Max
    Mar 11 at 10:08















1












$begingroup$


For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?



Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.



However is $p in mathbbZ$? Or is $p in mathbbN$?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    It's $(p-1)!$ : how do you define factorials for negative integers ?
    $endgroup$
    – Max
    Mar 11 at 10:08













1












1








1





$begingroup$


For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?



Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.



However is $p in mathbbZ$? Or is $p in mathbbN$?










share|cite|improve this question











$endgroup$




For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?



Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.



However is $p in mathbbZ$? Or is $p in mathbbN$?







abstract-algebra elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 11 at 9:55









Andrews

1,2691421




1,2691421










asked Mar 11 at 9:45









J.BryJ.Bry

486




486







  • 2




    $begingroup$
    It's $(p-1)!$ : how do you define factorials for negative integers ?
    $endgroup$
    – Max
    Mar 11 at 10:08












  • 2




    $begingroup$
    It's $(p-1)!$ : how do you define factorials for negative integers ?
    $endgroup$
    – Max
    Mar 11 at 10:08







2




2




$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08




$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08










1 Answer
1






active

oldest

votes


















2












$begingroup$

I assume you mean $(p-1)!$



It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:




$(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).




More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.






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    1 Answer
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    1 Answer
    1






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    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    I assume you mean $(p-1)!$



    It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:




    $(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).




    More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.






    share|cite|improve this answer








    New contributor




    nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$

















      2












      $begingroup$

      I assume you mean $(p-1)!$



      It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:




      $(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).




      More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.






      share|cite|improve this answer








      New contributor




      nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$















        2












        2








        2





        $begingroup$

        I assume you mean $(p-1)!$



        It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:




        $(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).




        More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.






        share|cite|improve this answer








        New contributor




        nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$



        I assume you mean $(p-1)!$



        It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:




        $(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).




        More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.







        share|cite|improve this answer








        New contributor




        nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer






        New contributor




        nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered Mar 11 at 10:06









        nammienammie

        1899




        1899




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