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19184Re: [PrimeNumbers] Update on CONTEST! and CONTEST++ $$

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  • Payam Samidoost
    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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