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Integers then Equals

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  • 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 1 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 2 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 3 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 4 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 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.

              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 6 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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