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Re: [PrimeNumbers] Prime equivalence theorem

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  • 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 1 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 2 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 3 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 4 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 5 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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