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Re: [PrimeNumbers] Interesting Sets of Three Consecutive Primes

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  • 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 1 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 2 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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