When x=9, k=10, then A(9)=361 is not prime, T(10)=361 is a square,

10=k > x=9, 9=x >3, T(k) = 361 is between 1 and B(x)=1681.

Please contact me off list for mailing address for my $50 prize.

There are 644 other counterexamples for x<=1000.

Adam

--- In

primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>

wrote:

>

> I offer a $50 prize to the first person who can submit

> a verifiable counterexample or proof by 10/1/8

> for the following primality conjecture:

>

> x, A(x), B(x), k, T(k) : integers;

>

> Let A(x) = 5x^2 - 5x +1;

> Let B(x) = 25x^2 - 40x + 16;

> Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;

>

> If x > 3 and k > x , then A(x) will be prime if no value

> T(k) exists such that 1 < T(k) < B(x) is a square.

>

> Aldrich Stevens

>