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Proof of Wilson's Theorem using Fermat's

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  • Jon Perry
    If (p-1)!+1 is co-prime to p, then: [(p-1)!+1]^[p-1]=1modp So, j(p-1)!+1=kp+1 j(p-1)!=kp (p-1)!=xp which is a contradiction, therefore (p-1)!+1 always divides
    Message 1 of 11 , Nov 1, 2001
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      If (p-1)!+1 is co-prime to p, then:

      [(p-1)!+1]^[p-1]=1modp

      So, j(p-1)!+1=kp+1

      j(p-1)!=kp

      (p-1)!=xp

      which is a contradiction, therefore (p-1)!+1 always divides p.

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry
      BrainBench MVP for HTML and JavaScript
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    • Jon Perry
      Fermat s Little Theorem, the well-known a^(p-1)=1modp, holds iff a and p are co-prime. So if a and p are not co-prime then the result is NOT 1modp, but xmodp,
      Message 2 of 11 , Nov 2, 2001
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        Fermat's Little Theorem, the well-known a^(p-1)=1modp, holds iff a and p are
        co-prime.

        So if a and p are not co-prime then the result is NOT 1modp, but xmodp,
        where x<>1.

        [(p-1)!+1]^[p-1] when expanded gives a whole load of powers of (p-1)!, and
        1, so j in:

        j(p-1)!+1=kp+1

        represents a collection of the (p-1)!s. e.g. if p was 3, then we have:

        [(p-1)!+1]^2 = (p-1)!^2 + 4(p-1)! + 1

        so in this case j=(p-1!)+4.

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry
        BrainBench MVP for HTML and JavaScript
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        -----Original Message-----
        From: Marcel Martin [mailto:znz@...]
        Sent: 01 November 2001 21:54
        Cc: Prime Numbers
        Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's



        The proof requires
        (1) The congruence implies the primality
        (2) The primality implies the congruence

        It is not clear what you want prove, (1) or (2)?
        I suppose this is (2), otherwise
        >[(p-1)!+1]^[p-1] = 1 mod p
        would have no reason to be true.

        But where does this come from?
        >So, j(p-1)!+1=kp+1

        Marcel Martin


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      • Jon Perry
        Yes, but p is a prime. Therefore iff (p-1)!+1 is co-prime to p will the RHS be 1modp. ... Ok. I skipped a bit. Jon Perry perry@globalnet.co.uk
        Message 3 of 11 , Nov 3, 2001
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          Yes, but p is a prime. Therefore iff (p-1)!+1 is co-prime to p will the RHS
          be 1modp.

          >j(p-1)! = 0 mod p doesn't imply (p-1)! = 0 mod p

          Ok. I skipped a bit.

          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: 02 November 2001 20:00
          Cc: Prime Numbers
          Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's



          >Fermat's Little Theorem, the well-known a^(p-1)=1modp, holds iff a and p
          are
          >co-prime.

          No, not "if and only if". For instance, 3^(8-1) mod 8 is not equal
          to 1 whereas GCD(3,8)=1.

          >[(p-1)!+1]^[p-1] when expanded gives a whole load of powers of (p-1)!, and
          >1, so j in:
          >j(p-1)!+1=kp+1

          Ok. We can go further.

          j(p-1)! = 0 mod p doesn't imply (p-1)! = 0 mod p

          Using your notation, j(p-1)!=kp =/=> (p-1)!=xp

          For instance, 21 mod 3 = 0 doesn't imply 7 mod 3 = 0.
          If you want use (p-1)!=xp as a contradiction, you have to prove
          first that p doesn't divide j.

          Marcel Martin


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        • Jon Perry
          ... This is going too fast.( and BTW it s lose, not loose. lose means to waste, loose means slack, as in My shirt is loose, but my jeans are tight. ) *** If
          Message 4 of 11 , Nov 4, 2001
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            >Yes but do not loose time to try to prove j <> 0 mod p. Here,
            >we can say p divides j precisely because, p being prime, it
            >cannot divide (p-1)!

            This is going too fast.( and BTW it's lose, not loose. lose means to waste,
            loose means slack, as in 'My shirt is loose, but my jeans are tight.')

            ***
            If (p-1)!+1 is co-prime to p, then: [1]

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

            So, j(p-1)!+1=kp+1 [3]

            j(p-1)!=kp [4]

            (p-1)!=xp [5]
            ***

            [1][2] is true, with p a prime. OK, this doesn't do n composite, but then
            this isn't very difficult.

            [3] is true, although we do not know anything about j.

            [4] is just [3] reduced by +1.

            [5] contains the assumptive step. Here I say (p-1)! obviously doesn't
            contain p as a factor. But I have ignored the fact that in getting from [4]
            to [5] involves assuming that j<>0modp.

            But all hope is not lost. If j was to contain a factor of p, then the proof
            would be saved, but to what purpose? There is an obvious contradiction
            somewhere. So, we may 'safely' assume that j does not contain p as a factor.
            But this is ESTD.

            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 00:09
            Cc: Prime Numbers
            Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's



            >>j(p-1)! = 0 mod p doesn't imply (p-1)! = 0 mod p

            >Ok. I skipped a bit.

            Yes but do not loose time to try to prove j <> 0 mod p. Here,
            we can say p divides j precisely because, p being prime, it
            cannot divide (p-1)!

            Marcel Martin


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          • Phil Carmody
            ... OK, I propose a conjugate theorem to accompany your proof. Phil s Theorem of Binary Bogosity: forall prime p 3, p|2 ... If (2+1) is coprime to p (true
            Message 5 of 11 , Nov 4, 2001
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              On Sun, 04 November 2001, "Jon Perry" wrote:
              > >Yes but do not loose time to try to prove j <> 0 mod p. Here,
              > >we can say p divides j precisely because, p being prime, it
              > >cannot divide (p-1)!

              OK, I propose a conjugate theorem to accompany your proof.

              Phil's Theorem of Binary Bogosity:
              \forall prime p>3, p|2

              > ***
              > If (p-1)!+1 is co-prime to p, then: [1]

              If (2+1) is coprime to p
              (true from p>3)

              > [(p-1)!+1]^[p-1]=1modp [2]

              (2+1)^(p-1) == 1 (mod p)

              > So, j(p-1)!+1=kp+1 [3]

              So, j(2)+1 = kp+1

              > j(p-1)!=kp [4]

              j(2) = kp

              > (p-1)!=xp [5]

              2 = xp

              > ***
              >
              > [1][2] is true, with p a prime. OK, this doesn't do n

              2!+1 = 3 = 1*3
              4!+1 = 25 = 5*5
              6!+1 = 721 = 103*7

              When we can't agree on what 'true' means, then we're going to come up with very different conclusions.

              > composite, but then
              > this isn't very difficult.
              >
              > [3] is true, although we do not know anything about j.
              >
              > [4] is just [3] reduced by +1.
              >
              > [5] contains the assumptive step. Here I say (p-1)! obviously doesn't
              > contain p as a factor. But I have ignored the fact that in getting from [4]
              > to [5] involves assuming that j<>0modp.

              Never ASSUME, as it makes an ASS of U and ME
              (Benny Hill, I believe.)

              > But all hope is not lost. If j was to contain a factor of p, then the proof
              > would be saved, but to what purpose? There is an obvious contradiction
              > somewhere. So, we may 'safely' assume that j does not contain p as a factor.

              I don't like that kind of safety...

              > But this is ESTD.

              Endemic Sexually Transmitted Disease?

              More number-theoretic precautions are required, methinks.

              Phil

              Mathematics should not have to involve martyrdom;
              Support Eric Weisstein, see http://mathworld.wolfram.com
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            • Jon Perry
              ... I didn t mean to - it was an accident. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry BrainBench MVP for HTML and JavaScript
              Message 6 of 11 , Nov 4, 2001
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                >If you wanted to kill all the mathematicians on this list, you
                >succeeded :-)

                I didn't mean to - it was an accident.

                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 18:37
                Cc: Prime Numbers
                Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's



                >>Yes but do not lose time to try to prove j <> 0 mod p. Here,
                >>we can say p divides j precisely because, p being prime, it
                >>cannot divide (p-1)!

                >This is going too fast.

                Yes, you're right. I forgot that gcd((p-1)!+1,p) <> 1. If you could
                prove that p doesn't divide j, your "half" proof could be correct.

                >There is an obvious contradiction somewhere. So, we may 'safely'
                >assume that j does not contain p as a factor.

                In short, the fact we know this is provable otherwise would allow us
                to accept lame proofs?
                If you wanted to kill all the mathematicians on this list, you
                succeeded :-)

                Marcel Martin


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              • Jon Perry
                ... Funny. Easier Stated Than Demonstrated. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry BrainBench MVP for HTML and JavaScript
                Message 7 of 11 , Nov 4, 2001
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                  > But this is ESTD.

                  >Endemic Sexually Transmitted Disease?

                  Funny. Easier Stated Than Demonstrated.

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


                  -----Original Message-----
                  From: Phil Carmody [mailto:fatphil@...]
                  Sent: 04 November 2001 12:44
                  To: primenumbers@yahoogroups.com
                  Subject: RE: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's


                  On Sun, 04 November 2001, "Jon Perry" wrote:
                  > >Yes but do not loose time to try to prove j <> 0 mod p. Here,
                  > >we can say p divides j precisely because, p being prime, it
                  > >cannot divide (p-1)!

                  OK, I propose a conjugate theorem to accompany your proof.

                  Phil's Theorem of Binary Bogosity:
                  \forall prime p>3, p|2

                  > ***
                  > If (p-1)!+1 is co-prime to p, then: [1]

                  If (2+1) is coprime to p
                  (true from p>3)

                  > [(p-1)!+1]^[p-1]=1modp [2]

                  (2+1)^(p-1) == 1 (mod p)

                  > So, j(p-1)!+1=kp+1 [3]

                  So, j(2)+1 = kp+1

                  > j(p-1)!=kp [4]

                  j(2) = kp

                  > (p-1)!=xp [5]

                  2 = xp

                  > ***
                  >
                  > [1][2] is true, with p a prime. OK, this doesn't do n

                  2!+1 = 3 = 1*3
                  4!+1 = 25 = 5*5
                  6!+1 = 721 = 103*7

                  When we can't agree on what 'true' means, then we're going to come up with
                  very different conclusions.

                  > composite, but then
                  > this isn't very difficult.
                  >
                  > [3] is true, although we do not know anything about j.
                  >
                  > [4] is just [3] reduced by +1.
                  >
                  > [5] contains the assumptive step. Here I say (p-1)! obviously doesn't
                  > contain p as a factor. But I have ignored the fact that in getting from
                  [4]
                  > to [5] involves assuming that j<>0modp.

                  Never ASSUME, as it makes an ASS of U and ME
                  (Benny Hill, I believe.)

                  > But all hope is not lost. If j was to contain a factor of p, then the
                  proof
                  > would be saved, but to what purpose? There is an obvious contradiction
                  > somewhere. So, we may 'safely' assume that j does not contain p as a
                  factor.

                  I don't like that kind of safety...

                  > But this is ESTD.

                  Endemic Sexually Transmitted Disease?

                  More number-theoretic precautions are required, methinks.

                  Phil

                  Mathematics should not have to involve martyrdom;
                  Support Eric Weisstein, see http://mathworld.wolfram.com
                  Find the best deals on the web at AltaVista Shopping!
                  http://www.shopping.altavista.com


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                  The Prime Pages : http://www.primepages.org



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                • Jon Perry
                  ... (A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2 [(p-1)!+1]^(p-1) - 1 = x^2-1 = (x-1)(x+1) Therefore either x-1 or x+1 contains p as a factor. As p is odd,
                  Message 8 of 11 , Nov 4, 2001
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                    Mark II:

                    ---

                    (A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2

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

                    Therefore either x-1 or x+1 contains p as a factor. As p is odd, let 2z=p-1:

                    {[(p-1)!+1]^z - 1}{[(p-1)!+1]^z + 1} = (x-1)(x+1)

                    Therefore x=[(p-1)!+1]^z.

                    Note that:

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

                    if (x-1) contains the factor of p.

                    Therefore (A-B):

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

                    2.[(p-1)!+1]^z = xp --> # (contradiction)

                    as [(p-1)!+1] doesn't contain a factor of p.

                    And :

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

                    if (x+1) contains the factor of p.

                    Therefore (A-C):

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

                    -2.[(p-1)!+1]^z = xp --> # (contradiction)

                    as [(p-1)!+1] doesn't contain a factor of p.

                    Q.e.d

                    ---

                    Jon Perry
                    perry@...
                    http://www.users.globalnet.co.uk/~perry
                    BrainBench MVP for HTML and JavaScript
                    http://www.brainbench.com
                  • 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 9 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 10 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 11 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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