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15546Re: [PrimeNumbers] Extension of Cunningham chains

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  • Jack Brennen
    Nov 13, 2004
      Mike wrote:
      > With 3 divisors, each must be the square of a prime, of course.
      > Are there any such chains of length > 1? (I haven't searched.)

      Yes, there are a few easy examples. There are no examples of
      such chains with length > 2; this can probably be proven in
      several ways, but a simple analysis of the solutions to the
      Pell equations (x^2 = 2*y^2 +/-1) suffices.

      Of length 2, the following very small examples exist:

      k=5, n=(0,1), k*2^n-1 == (4,9) (Only example of form k*2^n-1)

      k=3, n=(3,4), k*2^n+1 == (25,49)
      k=105, n=(3,4), k*2^n+1 == (841,1681)

      And some bigger ones:

      k=248204571168918457275, n=(3,4), k*2^n+1
      k=22980046115755920139039424449209532289506425, n=(3,4), k*2^n+1

      I'm not sure if there are any more. An heuristic argument would
      indicate that the total number of such chains is finite.
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