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19584Re: Prize Puzzle : Primality Conjecture

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  • Adam
    Sep 10, 2008
      When x=9, k=10, then A(9)=361 is not prime, T(10)=361 is a square,
      10=k > x=9, 9=x >3, T(k) = 361 is between 1 and B(x)=1681.

      Please contact me off list for mailing address for my $50 prize.

      There are 644 other counterexamples for x<=1000.

      Adam

      --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>
      wrote:
      >
      > I offer a $50 prize to the first person who can submit
      > a verifiable counterexample or proof by 10/1/8
      > for the following primality conjecture:
      >
      > x, A(x), B(x), k, T(k) : integers;
      >
      > Let A(x) = 5x^2 - 5x +1;
      > Let B(x) = 25x^2 - 40x + 16;
      > Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;
      >
      > If x > 3 and k > x , then A(x) will be prime if no value
      > T(k) exists such that 1 < T(k) < B(x) is a square.
      >
      > Aldrich Stevens
      >
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