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Update on CONTEST! and CONTEST++ $$

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  • aldrich617
    Not much action on two of the conjectures, and time grows short, so I will do the obvious and increase the purse. Maybe even DB himself will swoop out of the
    Message 1 of 3 , Dec 28, 2007
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      Not much action on two of the conjectures,
      and time grows short, so I will do the
      obvious and increase the purse. Maybe even
      DB himself will swoop out of the clouds to
      pounce on this one! Of course, it just may
      be possible that there really are no
      counterexamples and the durn things are
      impossible to prove. That would be unfortunate.

      I offer a $100 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)

      I offer a $101 prize to the first person who
      can submit a verifiable counterexample or proof
      by New Year's day for the following 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 exists a value of k that satisfies the
      conditions of the test, then A is composite,
      and there will be at least one value of k such
      that 2*k - 1 will have a factor in A.
      (issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

      Aldrich Stevens
    • Payam Samidoost
      Happy new year to all ... I provide counterexamples to the contrapositive: If A is composite then there exists a k in the interval that satisfies the test. The
      Message 2 of 3 , Dec 28, 2007
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        Happy new year to all

        > I offer a $100 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)
        I provide counterexamples to the contrapositive:
        If A is composite then there exists a k in the interval that satisfies the
        test.

        The maple code:

        *> for x from 1 to 9 do
        > A:=20*x^2+10*x+1;
        > if not isprime(A) then
        > B:=10*x^2+4*x+1;
        > c:=floor(A/sqrt(5))-1;
        > printf("%A %A-%A ", x, c, B-1);
        > for k from c to B-1 do
        > y := 5*(2^k-1)^2-4*A^2;
        > if y<=0 then next; end if;
        > if issqr(y) then printf("(%A)", k); break; end if;
        > end do;
        > printf("\n");
        > end if;
        > end do:
        *
        The output:

        4 160-176
        5 245-270
        6 348-384
        9 764-846
        Each row begins with a value x for which A is composite.
        It is followed by the range of k. That is from c to B.
        The code is written such that by finding a k value it is printed in ().
        But there is no (k)

        So we have at least 4 counterexamples trivially.
        Or I have misunderstood the problem

        Regards
        Payam


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      • Payam Samidoost
        Hello ... Thank you Aldrich and Peter Changing the code as Peter noted and adding a single dot after 5 in sqrt(5) to avoid maple s symbolic manipulations and
        Message 3 of 3 , Dec 29, 2007
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          Hello

          Aldrich Stevens noted offline that my result is not correct:

          > There must be an error in your program.
          > All of your proposed counterexamples
          > actually do have values of k that
          > satisfy the conditions of the test.
          > x = 4, k = 162
          > x = 5, k = 247
          > x = 6, k = 351, 358
          > x = 9, k = 769

          and Peter Kosinar again offline located the my mistake:

          >> (issquare test: 0 < 5*(2*k-1)^2 - 4*A^2 = f^2)
          >> y := 5*(2^k-1)^2 - 4*A^2;
          > It seems your code uses 2^k, while the original poster used 2*k.

          Thank you Aldrich and Peter

          Changing the code as Peter noted
          and adding a single dot after 5 in sqrt(5) to avoid maple's symbolic
          manipulations
          and extending the for x to higher values
          the output reads as follows:

          4 160-176 (162)
          5 245-270 (247)
          6 348-384 (351)
          9 764-864 (769)
          10 938-1040 (952)
          ...
          100 89889-100400 (89893)
          ...
          1000 8948743-10004000 (8959147)
          ...

          Yes, not the conjecture nor its inverse is trivial!

          Regards,
          Payam


          [Non-text portions of this message have been removed]
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