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Prime equivalence theorem

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  • Sebastian Martin Ruiz
    Hello all: Theorem: Let p and q consecutive prime numbers =5 Let m=(p+q)/2 m+2 is prime if and only if m-2 is prime. Sincerely Sebastian Martin Ruiz [Non-text
    Message 1 of 6 , Jun 16, 2012
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      Hello all:

      Theorem:


      Let p and q consecutive prime numbers >=5


      Let m=(p+q)/2

      m+2 is prime if and only if m-2 is prime.


      Sincerely

      Sebastian Martin Ruiz


      [Non-text portions of this message have been removed]
    • Sebastian Martin Ruiz
      m +/-1, m+/-2, m+/-3, m+/-6 are 4 values for the theorem. It is possible to find others?  ________________________________ De: Maximilian Hasler
      Message 2 of 6 , Jun 16, 2012
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        m +/-1, m+/-2, m+/-3, m+/-6 are 4 values for the theorem. It is possible to find others? 


        ________________________________
        De: Maximilian Hasler <maximilian.hasler@...>
        Para: Sebastian Martin Ruiz <s_m_ruiz@...>
        Enviado: Sábado 16 de junio de 2012 23:21
        Asunto: Re: [PrimeNumbers] Prime equivalence theorem


        Don't you have the same also for m +/- 1  and m +/- 6 (p>7)?

        Maximilian




        On Sat, Jun 16, 2012 at 11:37 AM, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:



        >Hello all:
        >
        >Theorem:
        >
        >Let p and q consecutive prime numbers >=5
        >
        >Let m=(p+q)/2
        >
        >m+2 is prime if and only if m-2 is prime.
        >
        >Sincerely
        >
        >Sebastian Martin Ruiz
        >
        >[Non-text portions of this message have been removed]
        >
        >
        >

        [Non-text portions of this message have been removed]
      • Mark
        As Norman has intimated, here is what is occurring: Choose consecutive odd primes p and q. Let m be the number exactly between them, at a distance x from p
        Message 3 of 6 , Jun 16, 2012
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          As Norman has intimated, here is what is occurring:

          Choose consecutive odd primes p and q. Let m be the number exactly between them, at a distance x from p and q. Thus m-x = p and m+x = q. So of course m +/- x are primes, because they are the p and q we originally chose! So we must mind our p's and q's !

          (2x represents the gap between p and q. Depending on the p and q chosen, x can be any number > 1. For instance if p is 89 and q is 97, we have m = 93, and 93 +/- 4 is prime. )

          Mark



          --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
          >
          > m +/-1, m+/-2, m+/-3, m+/-6 are 4 values for the theorem. It is possible to find others? 
          >
          >
          > ________________________________
          > De: Maximilian Hasler <maximilian.hasler@...>
          > Para: Sebastian Martin Ruiz <s_m_ruiz@...>
          > Enviado: Sábado 16 de junio de 2012 23:21
          > Asunto: Re: [PrimeNumbers] Prime equivalence theorem
          >
          >
          > Don't you have the same also for m +/- 1  and m +/- 6 (p>7)?
          >
          > Maximilian
          >
          >
          >
          >
          > On Sat, Jun 16, 2012 at 11:37 AM, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
          >
          >
          > > 
          > >Hello all:
          > >
          > >Theorem:
          > >
          > >Let p and q consecutive prime numbers >=5
          > >
          > >Let m=(p+q)/2
          > >
          > >m+2 is prime if and only if m-2 is prime.
          > >
          > >Sincerely
          > >
          > >Sebastian Martin Ruiz
          > >
          > >[Non-text portions of this message have been removed]
          > >
          > >
          > >
          >
          > [Non-text portions of this message have been removed]
          >
        • Maximilian Hasler
          I m not sure if I understand completely what you meant to say, because you use the same x for (q-p)/2 (which varies with each pair (p,q)) and for the m +/- x
          Message 4 of 6 , Jun 16, 2012
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            I'm not sure if I understand completely what you meant to say, because you
            use "the same x" for (q-p)/2 (which varies with each pair (p,q))
            and for the m +/- x (which is constant for the whole procedure /
            "theorem", e.g. x=2 in the theorem of this thread).

            But indeed these "theorems" can easily be proved by analysing the possible
            values in a similar way :

            q-p > 2 d => m +/- d are both composite, since strictly comprised between
            the two consecutive primes.

            q-p = 2 d => m +/- d are the two primes p & q.

            Thereafter, there may be remaining cases (for d>1) :

            If d=2,
            q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)

            If d=3,
            q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)
            q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)

            If d=4 or d=5, exceptions are possible.

            If d=6,
            q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)
            q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)
            q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)
            q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)
            q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)

            This shows that the Theorem holds for d=1, 2, 3 and 6.

            I think that for all other d there are infinitely many counter-examples.

            Maximilian


            [Non-text portions of this message have been removed]
          • Norman Luhn
            Another argument is: m-k must be p and m+k must be q,  otherwise  m-k is the consecutive prime to p .. and that is q (q-p)=2k
            Message 5 of 6 , Jun 17, 2012
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              Another argument is:

              m-k must be p and m+k must be q,  otherwise  m-k is the consecutive prime to p .. and that is q

              (q-p)=2k


              ________________________________
              Von: Maximilian Hasler <maximilian.hasler@...>
              An: Mark <mark.underwood@...>
              CC: primenumbers@yahoogroups.com
              Gesendet: 8:19 Sonntag, 17.Juni 2012
              Betreff: Re: [PrimeNumbers] Re: Prime equivalence theorem


               
              I'm not sure if I understand completely what you meant to say, because you
              use "the same x" for (q-p)/2 (which varies with each pair (p,q))
              and for the m +/- x (which is constant for the whole procedure /
              "theorem", e.g. x=2 in the theorem of this thread).

              But indeed these "theorems" can easily be proved by analysing the possible
              values in a similar way :

              q-p > 2 d => m +/- d are both composite, since strictly comprised between
              the two consecutive primes.

              q-p = 2 d => m +/- d are the two primes p & q.

              Thereafter, there may be remaining cases (for d>1) :

              If d=2,
              q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)

              If d=3,
              q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)
              q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)

              If d=4 or d=5, exceptions are possible.

              If d=6,
              q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)
              q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)
              q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)
              q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)
              q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)

              This shows that the Theorem holds for d=1, 2, 3 and 6.

              I think that for all other d there are infinitely many counter-examples.

              Maximilian

              [Non-text portions of this message have been removed]




              [Non-text portions of this message have been removed]
            • Mark
              Maximilian is correct in that he actually shows why m-k must be p and m+k must be q. It really has to do with admissible 3 - prime constellations of a given
              Message 6 of 6 , Jun 17, 2012
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                Maximilian is correct in that he actually shows why m-k must be p and m+k must be q. It really has to do with admissible 3 - prime constellations of a given length.

                Mark


                --- In primenumbers@yahoogroups.com, Norman Luhn <n.luhn@...> wrote:
                >
                > Another argument is:
                >
                > m-k must be p and m+k must be q,  otherwise  m-k is the consecutive prime to p .. and that is q
                >
                > (q-p)=2k
                >
                >
                > ________________________________
                > Von: Maximilian Hasler <maximilian.hasler@...>
                > An: Mark <mark.underwood@...>
                > CC: primenumbers@yahoogroups.com
                > Gesendet: 8:19 Sonntag, 17.Juni 2012
                > Betreff: Re: [PrimeNumbers] Re: Prime equivalence theorem
                >
                >
                >  
                > I'm not sure if I understand completely what you meant to say, because you
                > use "the same x" for (q-p)/2 (which varies with each pair (p,q))
                > and for the m +/- x (which is constant for the whole procedure /
                > "theorem", e.g. x=2 in the theorem of this thread).
                >
                > But indeed these "theorems" can easily be proved by analysing the possible
                > values in a similar way :
                >
                > q-p > 2 d => m +/- d are both composite, since strictly comprised between
                > the two consecutive primes.
                >
                > q-p = 2 d => m +/- d are the two primes p & q.
                >
                > Thereafter, there may be remaining cases (for d>1) :
                >
                > If d=2,
                > q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)
                >
                > If d=3,
                > q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)
                > q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)
                >
                > If d=4 or d=5, exceptions are possible.
                >
                > If d=6,
                > q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)
                > q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)
                > q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)
                > q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)
                > q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)
                >
                > This shows that the Theorem holds for d=1, 2, 3 and 6.
                >
                > I think that for all other d there are infinitely many counter-examples.
                >
                > Maximilian
                >
                > [Non-text portions of this message have been removed]
                >
                >
                >
                >
                > [Non-text portions of this message have been removed]
                >
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