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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