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21353Re: F = 20z^2 - n and (5r^2 -F)/n

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  • djbroadhurst
    Feb 27, 2010
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      --- In primenumbers@yahoogroups.com,
      "aldrich617" <aldrich617@...> conjectured:

      > Let F = 20*z^2 - 7, for integer z > 0.
      > If there is at least one integer r where 2*z < r <= z*sqrt(14)
      > such that (5*r^2 - F)/7 = a perfect square, then F is composite.

      Thanks for writing sqrt(14), which indeed looks like
      the correct limit, in this case. You may write
      r < z*sqrt(14), since the issue of equality cannot arise.

      > Its companion conjecture that " otherwise F is prime " would
      > also probably be true.

      No. Here you need the caveat that for z = 0 mod 7
      it is F/7, and not F, that is conjectured to be prime.

      Your list of "n" values for which you have a conjecture was

      > n = 1,6,10,12,13,14,15,19,21,22, and 23

      Now consider the case n = 11, about which you were strangely
      silent. Here, the closely parallel conjecture is as follows:

      Let F = 20*z^2 - 11, for integer z > 0, and let
      G = F/11, if z = 0 mod 11, and G = F, otherwise.
      Then I conjecture that G is prime if and only if
      there is no integer r such that 2 < r/z < 6*sqrt(5)
      and (5*r^2 - F)/11 is the square of an integer.

      Exercise: Let
      z = 14155247814063791862875274459749906082392395640576933548527
      and let F = 20*z^2 - 11. Find an integer r such that
      2 < r/z < 6*sqrt(5) and (5*r^2 - F)/11 is the square of an integer.

      Hint: An answer can be found very easily, if you pay good
      attention to my carefully chosen upper limit for r/z. Moreover,
      I conjecture that this generous hint gives the only answer.

      David
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