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Integers then Equals

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  • Sebastian Martin Ruiz
    Hello all:     Let P(n) the n-th prime number. Let Sqrt(P)=p^(1/2)   If Sqrt(P(n-1)+7) and Sqrt(P(n+1)-1) are both integers then:   1)
    Message 1 of 7 , Mar 14, 2009
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      Hello all:

       

       

      Let P(n)
      the n-th prime number. Let Sqrt(P)=p^(1/2)

       

      If
      Sqrt(P(n-1)+7) and Sqrt(P(n+1)-1) are both integers then:

       

      1)
      Sqrt(P(n-1)+7)=Sqrt(P(n+1)-1) 

       

      2)
      P(n)=P(n-1)+2  (Twin Primes) 

       

      3) P(n+1)=P(n-1)+8
      Sincerely
      Sebastián Martín Ruiz

       




      [Non-text portions of this message have been removed]
    • Kermit Rose
      1. Integers then Equals Posted by: Sebastian Martin Ruiz s_m_ruiz@yahoo.es s_m_ruiz Date: Sat Mar 14, 2009 12:11 pm ((PDT)) Hello all: Hello Sebastián Let
      Message 2 of 7 , Mar 15, 2009
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        1. Integers then Equals
        Posted by: "Sebastian Martin Ruiz" s_m_ruiz@... s_m_ruiz
        Date: Sat Mar 14, 2009 12:11 pm ((PDT))

        Hello all:

        Hello Sebastián

        Let P(n)
        the n-th prime number. Let Sqrt(P)=p^(1/2)

        Kermit:
        ****
        p1 = 2
        p2 = 3
        p3 = 5
        2 + 7 = 9
        5 - 1 = 4
        sqrt(p1+7) = 3
        sqrt(p3-1) = 2
        ****

        If
        Sqrt(P(n-1)+7) and Sqrt(P(n+1)-1) are both integers then:

        Kermit:
        *****
        If sqrt(p(n-1) + 7) is an integer, then p(n-1) = 0 or 2 mod 3.
        If sqrt(p(n+1) -1) is an integer, then p(n+1) = 1 or 2 mod 3.
        p(n-1) = 0 mod 3 implies p = 3.

        sqrt(3 + 7) is not an integer,

        Thus

        If sqrt(p(n-1) + 7) is an integer, then p(n-1) = 2 mod 3
        if sqrt(p(n+1) -1) is an integer, then p(n+1) = 1 or 2 mod 3.

        If sqrt(p(n-1) + 7) is an integer > 3,
        and
        sqrt(p(n+1) - 1) is an integer, then

        p(n+1) = p(n-1) + 8

        and p(n) might be p(n-1) + 2 or p(n-1) + 6.



        sqrt(p1+7) = 3
        sqrt(p3-1) = 2
        *****

        1)
        Sqrt(P(n-1)+7)=Sqrt(P(n+1)-1)

        Kermit:
        ****
        Provided n is sufficiently large.
        3**2 - 7 = 2
        4**2 - 7 = 9
        5**2 - 7 = 19
        2,3,5,7,11,13,17,19
        p8 = 19
        p9 = 23
        p10 = 29
        sqrt(28) > sqrt(25)


        ******

        2)
        P(n)=P(n-1)+2 (Twin Primes)

        Kermit:
        ***
        I expect there to be exceptions to this also.

        Because

        If p(n+1) = p(n-1) + 8,

        p(n) might be p(n-1) + 2,
        or
        p(n) might be p(n-1) + 6

        ******




        3) P(n+1)=P(n-1)+8
        Sincerely
        Sebastián Martín Ruiz
      • Maximilian Hasler
        ... no: p(n+1)=m²+1 = p(n-1)+6 = m²-1 = (m+1)(m-1) can t be prime. M.
        Message 3 of 7 , Mar 15, 2009
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          --- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
          > If
          > Sqrt(P(n-1)+7) and Sqrt(P(n+1)-1) are both integers then:
          > 2)
          > P(n)=P(n-1)+2 (Twin Primes)
          > I expect there to be exceptions to this also.
          > Because
          > If p(n+1) = p(n-1) + 8,
          > p(n) might be p(n-1) + 2,
          > or
          > p(n) might be p(n-1) + 6

          no:
          p(n+1)=m²+1 => p(n-1)+6 = m²-1 = (m+1)(m-1) can't be prime.

          M.
        • Werner D. Sand
          In words: If N is a square (N=m²) and N+1 and N-7 are primes, then N-5 is a prime, too. 3) is superfluous, because it follows from 1) by simple
          Message 4 of 7 , Mar 18, 2009
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            In words: If N is a square (N=m²) and N+1 and N-7 are primes, then N-5 is a prime, too.

            3) is superfluous, because it follows from 1) by simple transformation.

            Proof: Let N be a square and p1=N-7 and p3=N+1 primes. Let p2 be the single prime between p1 and p3: p1 < p2 < p3. Then the gap p3-p1=8. Then there are 3 possibilities for p2:
            1.) p2=p1+2 (gaps 2,6)
            2.) p2=p1+4 (gaps 4,4)
            3.) p2=p1+6 (gaps 6,2)

            2.) cannot occur, for there are no equal neighbouring gaps besides 6n,6n.
            3.) From p1=N-7 and p2=p1+6 follows p2=N-1=m²-1=(m+1)*(m-1), not prime.

            Thus 1.) is the only possibility.

            Btw: The final digits of p1,p2,p3 are 9,1,7. Who proves?

            Werner D. Sand
          • Maximilian Hasler
            ... Stated like this, it is indeed more or less trivial. As I see it, the nontrivial part of the assertion is: There is NO m such that * m²-7 is prime *
            Message 5 of 7 , Mar 18, 2009
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              --- In primenumbers@yahoogroups.com, "Werner D. Sand" wrote:
              >
              > In words: If N is a square (N=m²) and N+1 and N-7 are primes,
              > then N-5 is a prime, too.

              Stated like this, it is indeed more or less trivial.

              As I see it, the nontrivial part of the assertion is:

              There is NO m such that
              * m²-7 is prime
              * (m+k)²+1 is prime for some k>0 (i.e. k>=2)
              * there is only one prime between m²-7 and (m+k)²+1

              I think that one would need to prove something like Cramer's conjecture to disprove this statement.


              > 3) is superfluous, because it follows
              > from 1) by simple transformation.

              Of course.


              > 3.) From p1=N-7 and p2=p1+6 follows p2=N-1=m²-1=(m+1)*(m-1),
              > not prime.

              As I wrote some days ago, cf. http://tech.groups.yahoo.com/group/primenumbers/message/19901/.


              > Thus 1.) is the only possibility.
              >
              > Btw: The final digits of p1,p2,p3 are 9,1,7. Who proves?

              m²-7 can't be 1 mod 10.

              Maximilian
            • Maximilian Hasler
              ... er... I read but you didn t write: ...and there is a prime between N-7 and N+1, ... else we have counter-examples for m =
              Message 6 of 7 , Mar 18, 2009
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                > --- In primenumbers@yahoogroups.com, "Werner D. Sand" wrote:
                > >
                > > In words: If N is a square (N=m²) and N+1 and N-7 are primes,
                > > then N-5 is a prime, too.
                >
                > Stated like this, it is indeed more or less trivial.

                er... I read but you didn't write:
                "...and there is a prime between N-7 and N+1, ..."

                else we have counter-examples for m =
                54,66,90,156,240,270,306,474,570,576,636,750,780,1080,1320,1350,2034,
                2154,2406,2700,2760,3204,3240,3306,3480,3516,3756,3774,3984,4056,4086,
                4140,4146,4176,4716,4734,4794,5154,5370,5424,5550,5664,5700,5850,5856,
                5970,6030,6060,6120,6366,6576,6714,6786,7050,7164,8196,8424,8454,8940,
                9180,9246,9486,9696,9804...
              • Werner D. Sand
                ... O.k., put it in.
                Message 7 of 7 , Mar 20, 2009
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                  --- In primenumbers@yahoogroups.com, "Maximilian Hasler" <maximilian.hasler@...> wrote:
                  >
                  > > --- In primenumbers@yahoogroups.com, "Werner D. Sand" wrote:
                  > > >
                  > > > In words: If N is a square (N=m²) and N+1 and N-7 are primes,
                  > > > then N-5 is a prime, too.
                  > >
                  > > Stated like this, it is indeed more or less trivial.
                  >
                  > er... I read but you didn't write:
                  > "...and there is a prime between N-7 and N+1, ..."
                  >
                  > else we have counter-examples for m =
                  > 54,66,90,156,240,270,306,474,570,576,636,750,780,1080,1320,1350,2034,
                  > 2154,2406,2700,2760,3204,3240,3306,3480,3516,3756,3774,3984,4056,4086,
                  > 4140,4146,4176,4716,4734,4794,5154,5370,5424,5550,5664,5700,5850,5856,
                  > 5970,6030,6060,6120,6366,6576,6714,6786,7050,7164,8196,8424,8454,8940,
                  > 9180,9246,9486,9696,9804...
                  >



                  O.k., put it in.
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