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On a research problem of C&P

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  • Décio Luiz Gazzoni Filho
    Hello, ... 1.85. Note that 61 divides 67*71 + 1. Are there three larger consecutive primes p
    Message 1 of 2 , Dec 12, 2004
      Hello,

      I've been looking at this research problem in Crandall & Pomerance's book:

      -----
      1.85. Note that 61 divides 67*71 + 1. Are there three larger consecutive
      primes p < q < r such that p | qr + 1?
      -----

      I've also pondered the similar problem p | qr - 1, while upholding the
      restrictions on p,q,r. Then it dawned on me that it implies qr - 1 == 0 mod
      p, and thus qr == 1 mod p. In other words, q and r are modular inverses of
      each other.

      Now let q = p+s and r = p+t. Then (p+s)(p+t) == 1 mod p, so that finally we
      arrive at st == 1 mod p. Now s and t are prime gaps and it is known on the
      PNT that both are O(log p), so that their product is O(log^2 p). On the other
      hand, st == 1 mod p implies that st = 1 + kp. This condition means that we
      need, at best, st = O(p). This heuristic would suggest that no large
      solutions exist (the largest I've found is 83 | 89*97 - 1).

      This heuristic nicely carries over to the original statement. Now st == -1 mod
      p, so that st = -1 + kp = (p-1) + (k-1)p. Again the PNT suggests st = O(log^2
      p) while the congruence conditions require at least st = O(p).

      Any flaws in this reasoning?

      Décio
    • Décio Luiz Gazzoni Filho
      Further info about the argument below can be found in the NMBRTHRY posting http://listserv.nodak.edu/scripts/wa.exe?A2=ind0412&L=nmbrthry&F=&S=&P=2781 There s
      Message 2 of 2 , Dec 13, 2004
        Further info about the argument below can be found in the NMBRTHRY posting

        http://listserv.nodak.edu/scripts/wa.exe?A2=ind0412&L=nmbrthry&F=&S=&P=2781

        There's a little evidence, regarding what's known about first occurrence prime
        gaps, to support the conjecture.

        Décio

        On Monday 13 December 2004 00:52, you wrote:
        > Hello,
        >
        > I've been looking at this research problem in Crandall & Pomerance's book:
        >
        > -----
        > 1.85. Note that 61 divides 67*71 + 1. Are there three larger consecutive
        > primes p < q < r such that p | qr + 1?
        > -----
        >
        > I've also pondered the similar problem p | qr - 1, while upholding the
        > restrictions on p,q,r. Then it dawned on me that it implies qr - 1 == 0 mod
        > p, and thus qr == 1 mod p. In other words, q and r are modular inverses of
        > each other.
        >
        > Now let q = p+s and r = p+t. Then (p+s)(p+t) == 1 mod p, so that finally we
        > arrive at st == 1 mod p. Now s and t are prime gaps and it is known on the
        > PNT that both are O(log p), so that their product is O(log^2 p). On the
        > other hand, st == 1 mod p implies that st = 1 + kp. This condition means
        > that we need, at best, st = O(p). This heuristic would suggest that no
        > large solutions exist (the largest I've found is 83 | 89*97 - 1).
        >
        > This heuristic nicely carries over to the original statement. Now st == -1
        > mod p, so that st = -1 + kp = (p-1) + (k-1)p. Again the PNT suggests st =
        > O(log^2 p) while the congruence conditions require at least st = O(p).
        >
        > Any flaws in this reasoning?
        >
        > Décio


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