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CONTEST!

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  • aldrich617
    I offer a $50 prize to the first person who can submit a verifiable counterexample or proof by New Year s day for the following primality conjecture:
    Message 1 of 3 , Dec 20, 2007
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      I offer a $50 prize to the first person who can submit
      a verifiable counterexample or proof by New Year's day
      for the following primality conjecture:

      (x,A,B,c,k,f : integers)
      Let A = 20x^2 + 10x + 1;
      Let B = 10x^2 + 4x + 1;
      Let c = trunc(A/sqrt(5)) - 1;

      For any x > 0, apply the issquare test to each k
      in the interval c <= k < B. If there is no value
      of k that satisfies the conditions of the test,
      then A is Prime.
      (issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)
    • Alan Eliasen
      ... Wasn t much of a challenge. Is it a homework problem? The first counterexample is at x=5 (which we can all agree is prime.) Here, A=551 B=271 c=245
      Message 2 of 3 , Dec 20, 2007
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        aldrich617 wrote:

        > I offer a $50 prize to the first person who can submit
        > a verifiable counterexample or proof by New Year's day
        > for the following primality conjecture:
        >
        > (x,A,B,c,k,f : integers)
        > Let A = 20x^2 + 10x + 1;
        > Let B = 10x^2 + 4x + 1;
        > Let c = trunc(A/sqrt(5)) - 1;
        >
        > For any x > 0, apply the issquare test to each k
        > in the interval c <= k < B. If there is no value
        > of k that satisfies the conditions of the test,
        > then A is Prime.
        > (issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

        Wasn't much of a challenge. Is it a homework problem? The first
        counterexample is at x=5 (which we can all agree is prime.)

        Here, A=551 B=271 c=245

        In this case, when k=247, the value of the expression
        5*(2*k -1)^2 - 4*A^2

        is 841, which equals 29^2. Thus, your algorithm would declare 5 not
        prime, which is incorrect.

        I can provide other examples where it fails in the opposite way, that
        is, declaring a composite number to be prime. The first occurs at x=8
        (which we can all agree is composite.)

        In this case, A=1361 B=673 c=607

        In this case, none of the positive values for 5*(2*k -1)^2 - 4*A^2
        from 607 to 672 produce a square.

        Aldrich Stevens, please transfer the money to my PayPal account by
        clicking the "donate" button here:

        http://futureboy.us/frinkdocs/donate.html

        --
        Alan Eliasen | "Furious activity is no substitute
        eliasen@... | for understanding."
        http://futureboy.us/ | --H.H. Williams
      • Jacques Tramu
        You have to read , Alan. (Furious activity is not substitue from carefully reading). The prime test is on A, not x . (I checked Aldrich conjecture for large
        Message 3 of 3 , Dec 25, 2007
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          You have to read , Alan. (Furious activity is not substitue from carefully reading).

          The prime test is on A, not x .

          (I checked Aldrich conjecture for large values of x, and found no counter example)

          Regards,
          JT

          ------------------------------------------------------
          http://www.echolalie.com
          http://www.echolalie.org/wiki

          ---------------------------------------------------------
          ----- Original Message -----
          From: Alan Eliasen
          To: aldrich617
          Cc: primenumbers@yahoogroups.com
          Sent: Friday, December 21, 2007 12:23 AM
          Subject: Re: [PrimeNumbers] CONTEST!


          aldrich617 wrote:

          > I offer a $50 prize to the first person who can submit
          > a verifiable counterexample or proof by New Year's day
          > for the following primality conjecture:
          >
          > (x,A,B,c,k,f : integers)
          > Let A = 20x^2 + 10x + 1;
          > Let B = 10x^2 + 4x + 1;
          > Let c = trunc(A/sqrt(5)) - 1;
          >
          > For any x > 0, apply the issquare test to each k
          > in the interval c <= k < B. If there is no value
          > of k that satisfies the conditions of the test,
          > then A is Prime.
          > (issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

          Wasn't much of a challenge. Is it a homework problem? The first
          counterexample is at x=5 (which we can all agree is prime.)

          Here, A=551 B=271 c=245

          In this case, when k=247, the value of the expression
          5*(2*k -1)^2 - 4*A^2

          is 841, which equals 29^2. Thus, your algorithm would declare 5 not
          prime, which is incorrect.

          I can provide other examples where it fails in the opposite way, that
          is, declaring a composite number to be prime. The first occurs at x=8
          (which we can all agree is composite.)

          In this case, A=1361 B=673 c=607

          In this case, none of the positive values for 5*(2*k -1)^2 - 4*A^2
          from 607 to 672 produce a square.

          Aldrich Stevens, please transfer the money to my PayPal account by
          clicking the "donate" button here:

          http://futureboy.us/frinkdocs/donate.html

          --
          Alan Eliasen | "Furious activity is no substitute
          eliasen@... | for understanding."
          http://futureboy.us/ | --H.H. Williams




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