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

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• 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

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