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RE: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

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  • Jon Perry
    (A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2 (B) {[(p-1)!+1]^z - 1}^2 = 0modp If you expand B, you get [(p-1)!+1]^(p-1) - 2[(p-1)!+1]^z + 1 as 2z=p-1.
    Message 1 of 11 , Nov 5, 2001
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      (A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2

      (B) {[(p-1)!+1]^z - 1}^2 = 0modp

      If you expand B, you get [(p-1)!+1]^(p-1) - 2[(p-1)!+1]^z + 1

      as 2z=p-1.

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com


      -----Original Message-----
      From: Marcel Martin [mailto:znz@...]
      Sent: 04 November 2001 23:25
      Cc: primenumbers@yahoogroups.com
      Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's



      >Therefore (A-B):
      >2.[(p-1)!+1]^z - 1 = -1modp

      How do you obtain this using (A) and (B)?

      Marcel Martin

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    • Jon Perry
      I think I have fixed all remaining bugs: http://www.users.globalnet.co.uk/~perry/maths/wilsonfermat/wilsonfermat.htm (or:
      Message 2 of 11 , Nov 10, 2001
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        I think I have fixed all remaining bugs:

        http://www.users.globalnet.co.uk/~perry/maths/wilsonfermat/wilsonfermat.htm

        (or:

        http://www.users.globalnet.co.uk/~perry/maths/othermaths.htm

        [Wilson's from Fermat's]

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry
        BrainBench MVP for HTML and JavaScript
        http://www.brainbench.com


        -----Original Message-----
        From: Marcel Martin [mailto:znz@...]
        Sent: 05 November 2001 21:05
        Cc: primenumbers@yahoogroups.com
        Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's



        >(A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2
        >(B) {[(p-1)!+1]^z - 1}^2 = 0modp
        >If you expand B, you get [(p-1)!+1]^(p-1) - 2[(p-1)!+1]^z + 1
        >as 2z=p-1.

        That's the first thing I did. But that doesn't answer my question:

        How do you obtain this using (A) and (B)?
        >2.[(p-1)!+1]^z - 1 = -1modp

        Marcel Martin

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      • Jon Perry
        I ve changed the proof. (n-1)!+1=0modn iff n is prime , is known. If n is prime, then (p-1)!+1=0modp, if not, (n-1)!+1 does not divide n. Jon Perry
        Message 3 of 11 , Nov 10, 2001
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          I've changed the proof.

          "(n-1)!+1=0modn iff n is prime",

          is known. If n is prime, then (p-1)!+1=0modp, if not, (n-1)!+1 does not
          divide n.

          Jon Perry
          perry@...
          http://www.users.globalnet.co.uk/~perry
          BrainBench MVP for HTML and JavaScript
          http://www.brainbench.com


          -----Original Message-----
          From: Marcel Martin [mailto:znz@...]
          Sent: 10 November 2001 11:51
          Cc: primenumbers@yahoogroups.com
          Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's



          >(B) [(p-1)!+1]^(2z) - 2[(p-1)!+1]^z + 1 = jp
          >Now consider (A)-(B). As 2z = p-1, we get:
          >2[(p-1)!+1]^z + 1 = xp + 1

          Once more, how do you get it? (I assume the variable x is not the one
          you previously defined to be equal to [(p-1)!+1]^z).

          [(p-1)!+1]^(2z) - 2[(p-1)!+1]^z + 1 = jp
          [(p-1)!+1]^(2z) - 2[(p-1)!+1]^z + 1 = 0 mod p
          1 - 2[(p-1)!+1]^z + 1 = 0 mod p (using (A))
          - 2[(p-1)!+1]^z + 1 = -1 mod p
          2[(p-1)!+1]^z - 1 = 1 mod p
          2[(p-1)!+1]^z - 1 = xp + 1

          I am very curious to see how you got

          2[(p-1)!+1]^z + 1 = xp + 1

          Moreover, in order to state "(n-1)!+1=0modn iff n is prime", you
          must also prove that the congruence implies the primality. Up to now,
          you only try to prove 'the primality implies the congruence'.

          Marcel Martin


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