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Re: [PrimeNumbers] Update on CONTEST! and CONTEST++ $$

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  • 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 1 of 3 , Dec 28, 2007
      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


      [Non-text portions of this message have been removed]
    • 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 2 of 3 , Dec 29, 2007
        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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