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Re: Prime equivalence theorem

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  • 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 1 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 2 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 3 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 4 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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