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Prize Puzzle : Primality Conjecture

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  • aldrich617
    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,
    Message 1 of 5 , Sep 8, 2008
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      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
    • Adam
      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
      Message 2 of 5 , Sep 10, 2008
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        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
        >
      • Bernardo Boncompagni
        ... That s his point. There must be NO square in the selected interval, but T(10) is a square. By the way, this is just a restatement of Euler s factorization
        Message 3 of 5 , Sep 10, 2008
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          Adam dice:

          > 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.

          That's his point. There must be NO square in the selected interval, but
          T(10) is a square. By the way, this is just a restatement of Euler's
          factorization method. If you put N = A(x) = m*n = (a-b)*(a+b) = a^2-b^2
          and do some algebraic tricks, it wouldn't be too difficult to prove the
          conjecture and claim your 50 bucks.

          Bernardo Boncompagni

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        • Jack Brennen
          You haven t given a counterexample. Read his conjecture again. Hint... his conjecture says nothing about cases where a square value DOES exist.
          Message 4 of 5 , 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
            >>
            >
            >
            >
            > ------------------------------------
            >
            > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
            > The Prime Pages : http://www.primepages.org/
            >
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            >
            >
            >
            >
            >
          • Peter Kosinar
            Hi Adam, ... Umm... pardon me? How does this contradict Aldrich s conjecture? As far as I understood the claim, A(x) is claimed to -be a prime- under the
            Message 5 of 5 , Sep 10, 2008
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              Hi Adam,

              Aldrich's conjecture:
              >> 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.

              Adam's proposed counterexample:
              > 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.

              Umm... pardon me? How does this contradict Aldrich's conjecture? As far as
              I understood the claim, A(x) is claimed to -be a prime- under the
              assumption that -no- value of T(k) between 1 and B(x) is a square.

              Peter

              --
              [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
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