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19586Re: [PrimeNumbers] Re: Prize Puzzle : Primality Conjecture

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  • Jack Brennen
    Sep 10, 2008
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      You haven't given a counterexample. Read his conjecture again.

      Hint... his conjecture says nothing about cases where a square
      value DOES exist.

      Adam wrote:
      > 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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