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Interesting Sets of Three Consecutive Primes

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  • w_sindelar@juno.com
    This exercise stems from my previous post A Remarkable Property of Consecutive Integer Pairs; Odd, Even and Prime. Anyone care to carry it further? There
    Message 1 of 4 , Sep 8, 2005
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      This exercise stems from my previous post "A Remarkable Property of
      Consecutive Integer Pairs; Odd, Even and Prime." Anyone care to carry it
      further?
      There exist sets of 3 CONSECUTIVE primes P, Q and R, such that the
      PRODUCT (2*Q* R) is equal to the SUM of 2 distinct PRIMES S<T, whose
      DIFFERENCE (T-S) is divisible by the SUM (Q+R) and yields a quotient
      which equals P. Here is the smallest and largest examples I was able to
      find out of a total of 12:
      P=3, Q=5, R=7, (2*Q*R)=70, S=17, T=53, (T-S)=36, (Q+R)=12, quotient=3
      which equals P.
      P=1039, Q=1049, R=1051, (2*Q*R)=2204998, S=11549, T=2193449,
      (T-S)=2181900, (Q+R)=2100, quotient=1039 which equals P.
      I looked for sets of 3 consecutive primes in arithmetic progression but
      found only one so far, which is the first example shown above. Maybe more
      3-PAP's of this type cannot exist and that sets of 3 consecutive primes
      of this type are finite in number. I'm without a clue. I would appreciate
      comments. Thanks folks.
      Bill Sindelar
    • Jens Kruse Andersen
      ... In other words: There are cases of consecutive primes P, Q, R such that Q*R +/- P*(Q+R)/2 are both prime. If P, Q, R is a prime AP3 with P 3 then P, Q, R
      Message 2 of 4 , Sep 8, 2005
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        Bill Sindelar wrote:

        > There exist sets of 3 CONSECUTIVE primes P, Q and R, such that the
        > PRODUCT (2*Q* R) is equal to the SUM of 2 distinct PRIMES S<T, whose
        > DIFFERENCE (T-S) is divisible by the SUM (Q+R) and yields a quotient
        > which equals P.

        In other words:
        There are cases of consecutive primes P, Q, R such that Q*R +/- P*(Q+R)/2 are
        both prime.

        If P, Q, R is a prime AP3 with P>3 then P, Q, R are either all 1 or all 2
        modulo 3.
        In both cases 3 divides Q*R - P*(Q+R)/2.

        --
        Jens Kruse Andersen
      • w_sindelar@juno.com
        On Thu, 8 Sep 2005 23:28:10 +0200 Jens Kruse Andersen ... Jens thank you for your quick response. But you have me stumped. Can I interpret your response as
        Message 3 of 4 , Sep 10, 2005
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          On Thu, 8 Sep 2005 23:28:10 +0200 "Jens Kruse Andersen"
          <jens.k.a@...> writes:
          > Bill Sindelar wrote:
          >
          > > There exist sets of 3 CONSECUTIVE primes P, Q and R, such that
          > the
          > > PRODUCT (2*Q* R) is equal to the SUM of 2 distinct PRIMES S<T,
          > whose
          > > DIFFERENCE (T-S) is divisible by the SUM (Q+R) and yields a
          > quotient
          > > which equals P.
          >
          > In other words:
          > There are cases of consecutive primes P, Q, R such that Q*R +/-
          > P*(Q+R)/2 are
          > both prime.
          >
          > If P, Q, R is a prime AP3 with P>3 then P, Q, R are either all 1 or
          > all 2
          > modulo 3.
          > In both cases 3 divides Q*R - P*(Q+R)/2.
          >
          > --
          > Jens Kruse Andersen

          Jens thank you for your quick response. But you have me stumped. Can I
          interpret your response as follows?
          Yes, there exist sets of 3 CONSECUTIVE primes P, Q and R, such that the
          PRODUCT (2*Q* R) is equal to the SUM of 2 distinct PRIMES S<T, whose
          DIFFERENCE (T-S) is divisible by the SUM (Q+R) and yields a QUOTIENT
          WHICH EQUALS P. However, if the 3 consecutive primes are in arithmetical
          progression, with the sole exception of P=3, Q=5, R=7, then no such sets
          can exist.
          Thanks again and regards.
          Bill Sindelar
        • Jens Kruse Andersen
          ... Yes, that is correctly interpreted. My formulation: There are cases of consecutive primes P, Q, R such that Q*R +/- P*(Q+R)/2 are both prime is simply a
          Message 4 of 4 , Sep 10, 2005
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            Bill Sindelar wrote:

            > Jens thank you for your quick response. But you have me stumped. Can I
            > interpret your response as follows?
            > Yes, there exist sets of 3 CONSECUTIVE primes P, Q and R, such that the
            > PRODUCT (2*Q* R) is equal to the SUM of 2 distinct PRIMES S<T, whose
            > DIFFERENCE (T-S) is divisible by the SUM (Q+R) and yields a QUOTIENT
            > WHICH EQUALS P. However, if the 3 consecutive primes are in arithmetical
            > progression, with the sole exception of P=3, Q=5, R=7, then no such sets
            > can exist.

            Yes, that is correctly interpreted. My formulation:
            "There are cases of consecutive primes P, Q, R such that Q*R +/- P*(Q+R)/2 are
            both prime"
            is simply a rewording of your slightly convoluted:
            "There exist ..... which equals P".
            You can check this by setting S, T = Q*R +/- P*(Q+R)/2.

            P, Q, R = 10^100+8556049 + 0, 78, 464 is a solution found in a few minutes
            with PARI/GP.
            Heuristics support infinitely many solutions (reminds of twin prime
            conjecture) but I have no idea how to prove it.

            If P, Q, R, is an AP3 solution other than 3, 5, 7, then your S cannot be prime
            since 3 always divides it when S and T meet the other conditions.

            --
            Jens Kruse Andersen
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