Prove that $(p - 1)! equiv (p - 1) pmod{p(p - 1)}$












3












$begingroup$


Prove that: $$(p - 1)! equiv p - 1 pmod{p(p - 1)}$$



In text it's not mentioned that $p$ is prime, but I checked and this doesn't hold for non-prime, so I guess $p$ is prime ..
I know that $(p - 1)! equiv -1 equiv p - 1 pmod p$,
and $(p - 1)! equiv 0 equiv (p - 1) pmod{p - 1}$,
the problem is that I don't know how to combine these.
Is it true that then we have: $(p - 1)! equiv (p - 1)^2 = p^2 - p - p + 1 equiv - p + 1 = - (p - 1) pmod{p(p - 1)}$, which is not the desired result .. ?
Or I need to multiply both sides of congruences ?
What are the laws for congruences that allow me to do this ? (I hope you get the idea of what I'm trying to ask).






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    What you need is the so called Chinese Remainder Theorem: en.wikipedia.org/wiki/Chinese_remainder_theorem
    $endgroup$
    – Ingix
    Dec 20 '18 at 16:06


















3












$begingroup$


Prove that: $$(p - 1)! equiv p - 1 pmod{p(p - 1)}$$



In text it's not mentioned that $p$ is prime, but I checked and this doesn't hold for non-prime, so I guess $p$ is prime ..
I know that $(p - 1)! equiv -1 equiv p - 1 pmod p$,
and $(p - 1)! equiv 0 equiv (p - 1) pmod{p - 1}$,
the problem is that I don't know how to combine these.
Is it true that then we have: $(p - 1)! equiv (p - 1)^2 = p^2 - p - p + 1 equiv - p + 1 = - (p - 1) pmod{p(p - 1)}$, which is not the desired result .. ?
Or I need to multiply both sides of congruences ?
What are the laws for congruences that allow me to do this ? (I hope you get the idea of what I'm trying to ask).






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    What you need is the so called Chinese Remainder Theorem: en.wikipedia.org/wiki/Chinese_remainder_theorem
    $endgroup$
    – Ingix
    Dec 20 '18 at 16:06
















3












3








3


1



$begingroup$


Prove that: $$(p - 1)! equiv p - 1 pmod{p(p - 1)}$$



In text it's not mentioned that $p$ is prime, but I checked and this doesn't hold for non-prime, so I guess $p$ is prime ..
I know that $(p - 1)! equiv -1 equiv p - 1 pmod p$,
and $(p - 1)! equiv 0 equiv (p - 1) pmod{p - 1}$,
the problem is that I don't know how to combine these.
Is it true that then we have: $(p - 1)! equiv (p - 1)^2 = p^2 - p - p + 1 equiv - p + 1 = - (p - 1) pmod{p(p - 1)}$, which is not the desired result .. ?
Or I need to multiply both sides of congruences ?
What are the laws for congruences that allow me to do this ? (I hope you get the idea of what I'm trying to ask).






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$




Prove that: $$(p - 1)! equiv p - 1 pmod{p(p - 1)}$$



In text it's not mentioned that $p$ is prime, but I checked and this doesn't hold for non-prime, so I guess $p$ is prime ..
I know that $(p - 1)! equiv -1 equiv p - 1 pmod p$,
and $(p - 1)! equiv 0 equiv (p - 1) pmod{p - 1}$,
the problem is that I don't know how to combine these.
Is it true that then we have: $(p - 1)! equiv (p - 1)^2 = p^2 - p - p + 1 equiv - p + 1 = - (p - 1) pmod{p(p - 1)}$, which is not the desired result .. ?
Or I need to multiply both sides of congruences ?
What are the laws for congruences that allow me to do this ? (I hope you get the idea of what I'm trying to ask).






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 16:07









user10354138

7,4322925




7,4322925










asked Dec 20 '18 at 16:01







user626177



















  • $begingroup$
    What you need is the so called Chinese Remainder Theorem: en.wikipedia.org/wiki/Chinese_remainder_theorem
    $endgroup$
    – Ingix
    Dec 20 '18 at 16:06




















  • $begingroup$
    What you need is the so called Chinese Remainder Theorem: en.wikipedia.org/wiki/Chinese_remainder_theorem
    $endgroup$
    – Ingix
    Dec 20 '18 at 16:06


















$begingroup$
What you need is the so called Chinese Remainder Theorem: en.wikipedia.org/wiki/Chinese_remainder_theorem
$endgroup$
– Ingix
Dec 20 '18 at 16:06






$begingroup$
What you need is the so called Chinese Remainder Theorem: en.wikipedia.org/wiki/Chinese_remainder_theorem
$endgroup$
– Ingix
Dec 20 '18 at 16:06












2 Answers
2






active

oldest

votes


















4












$begingroup$

Apply Chinese remainder theorem to the coprime moduli $p$ and $p-1$.



However, if you can't use CRT, you can simply observe $(p-1)!-(p-1)$ is divisible by $p-1$ (obvious factor) and by $p$ (Wilson's), so it is divisible by their least common multiple $operatorname{lcm}(p,p-1)=p(p-1)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Oooh right, so in general, if $x equiv y (mod a)$ and $x equiv y (mod b)$ then $x equiv y (mod [a, b])$, right ?
    $endgroup$
    – user626177
    Dec 20 '18 at 16:23








  • 1




    $begingroup$
    That is correct.
    $endgroup$
    – user10354138
    Dec 20 '18 at 16:25



















0












$begingroup$

1) Theorem. If $a|c$, $b|c$ and $(a,b)=1$ then $ab|c$



We know that $ (p,p-1)=1$ and also $p-1|(p-1)!-(p-1)$



By wilson theorem $p|(p-1)!+1$. Then $p|[(p-1)!+1]-p$



Therefore $p(p-1)| (p-1)!-(p-1)$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Apply Chinese remainder theorem to the coprime moduli $p$ and $p-1$.



    However, if you can't use CRT, you can simply observe $(p-1)!-(p-1)$ is divisible by $p-1$ (obvious factor) and by $p$ (Wilson's), so it is divisible by their least common multiple $operatorname{lcm}(p,p-1)=p(p-1)$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Oooh right, so in general, if $x equiv y (mod a)$ and $x equiv y (mod b)$ then $x equiv y (mod [a, b])$, right ?
      $endgroup$
      – user626177
      Dec 20 '18 at 16:23








    • 1




      $begingroup$
      That is correct.
      $endgroup$
      – user10354138
      Dec 20 '18 at 16:25
















    4












    $begingroup$

    Apply Chinese remainder theorem to the coprime moduli $p$ and $p-1$.



    However, if you can't use CRT, you can simply observe $(p-1)!-(p-1)$ is divisible by $p-1$ (obvious factor) and by $p$ (Wilson's), so it is divisible by their least common multiple $operatorname{lcm}(p,p-1)=p(p-1)$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Oooh right, so in general, if $x equiv y (mod a)$ and $x equiv y (mod b)$ then $x equiv y (mod [a, b])$, right ?
      $endgroup$
      – user626177
      Dec 20 '18 at 16:23








    • 1




      $begingroup$
      That is correct.
      $endgroup$
      – user10354138
      Dec 20 '18 at 16:25














    4












    4








    4





    $begingroup$

    Apply Chinese remainder theorem to the coprime moduli $p$ and $p-1$.



    However, if you can't use CRT, you can simply observe $(p-1)!-(p-1)$ is divisible by $p-1$ (obvious factor) and by $p$ (Wilson's), so it is divisible by their least common multiple $operatorname{lcm}(p,p-1)=p(p-1)$.






    share|cite|improve this answer











    $endgroup$



    Apply Chinese remainder theorem to the coprime moduli $p$ and $p-1$.



    However, if you can't use CRT, you can simply observe $(p-1)!-(p-1)$ is divisible by $p-1$ (obvious factor) and by $p$ (Wilson's), so it is divisible by their least common multiple $operatorname{lcm}(p,p-1)=p(p-1)$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 20 '18 at 16:13

























    answered Dec 20 '18 at 16:06









    user10354138user10354138

    7,4322925




    7,4322925








    • 1




      $begingroup$
      Oooh right, so in general, if $x equiv y (mod a)$ and $x equiv y (mod b)$ then $x equiv y (mod [a, b])$, right ?
      $endgroup$
      – user626177
      Dec 20 '18 at 16:23








    • 1




      $begingroup$
      That is correct.
      $endgroup$
      – user10354138
      Dec 20 '18 at 16:25














    • 1




      $begingroup$
      Oooh right, so in general, if $x equiv y (mod a)$ and $x equiv y (mod b)$ then $x equiv y (mod [a, b])$, right ?
      $endgroup$
      – user626177
      Dec 20 '18 at 16:23








    • 1




      $begingroup$
      That is correct.
      $endgroup$
      – user10354138
      Dec 20 '18 at 16:25








    1




    1




    $begingroup$
    Oooh right, so in general, if $x equiv y (mod a)$ and $x equiv y (mod b)$ then $x equiv y (mod [a, b])$, right ?
    $endgroup$
    – user626177
    Dec 20 '18 at 16:23






    $begingroup$
    Oooh right, so in general, if $x equiv y (mod a)$ and $x equiv y (mod b)$ then $x equiv y (mod [a, b])$, right ?
    $endgroup$
    – user626177
    Dec 20 '18 at 16:23






    1




    1




    $begingroup$
    That is correct.
    $endgroup$
    – user10354138
    Dec 20 '18 at 16:25




    $begingroup$
    That is correct.
    $endgroup$
    – user10354138
    Dec 20 '18 at 16:25











    0












    $begingroup$

    1) Theorem. If $a|c$, $b|c$ and $(a,b)=1$ then $ab|c$



    We know that $ (p,p-1)=1$ and also $p-1|(p-1)!-(p-1)$



    By wilson theorem $p|(p-1)!+1$. Then $p|[(p-1)!+1]-p$



    Therefore $p(p-1)| (p-1)!-(p-1)$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      1) Theorem. If $a|c$, $b|c$ and $(a,b)=1$ then $ab|c$



      We know that $ (p,p-1)=1$ and also $p-1|(p-1)!-(p-1)$



      By wilson theorem $p|(p-1)!+1$. Then $p|[(p-1)!+1]-p$



      Therefore $p(p-1)| (p-1)!-(p-1)$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        1) Theorem. If $a|c$, $b|c$ and $(a,b)=1$ then $ab|c$



        We know that $ (p,p-1)=1$ and also $p-1|(p-1)!-(p-1)$



        By wilson theorem $p|(p-1)!+1$. Then $p|[(p-1)!+1]-p$



        Therefore $p(p-1)| (p-1)!-(p-1)$






        share|cite|improve this answer









        $endgroup$



        1) Theorem. If $a|c$, $b|c$ and $(a,b)=1$ then $ab|c$



        We know that $ (p,p-1)=1$ and also $p-1|(p-1)!-(p-1)$



        By wilson theorem $p|(p-1)!+1$. Then $p|[(p-1)!+1]-p$



        Therefore $p(p-1)| (p-1)!-(p-1)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 2:36









        Julio Trujillo GonzalezJulio Trujillo Gonzalez

        856




        856






























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