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21355Re: F = 20z^2 - n and (5r^2 -F)/n

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  • djbroadhurst
    Feb 28, 2010
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      --- In primenumbers@yahoogroups.com,
      Kevin Acres <research@...> wrote:

      > Was the hint that generous or did I miss the point entirely?

      The hint was indeed very generous and my point was this:

      > I conjecture that this generous hint gives the only answer.

      Explanation: I had solved Aldrich's "issquare problem"
      (5*r^2 - F)/11 = a^2, with F = 20*z^2 - 11,
      by setting r = 3*p, a = 2*p, where
      p = 63304192701203884454712276208334615126937940298786013924413
      is, most crucially, a prime.

      Then F = p^2 is the square of a prime. Thus I conjecture that
      (5*r^2 - F)/11 is never a square for 2*z < r < 3*p,
      though it is certainly a square for r = 2*z and for r = 3*p.
      As Kevin noticed, 3*p is very close to 6*sqrt(5)*z, since
      z^2 = (p^2 + 11)/20.

      Puzzle: (p^2 + 11)/20 is a square for the primes
      p = 3, 13, 67, 4253, 21587, 6950947 ...
      Find the 17th prime in this sequence.

      Comment: Extra credit will be gained for proving
      that your answer is correct.

      David
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