You haven't given a counterexample. Read his conjecture again.

Hint... his conjecture says nothing about cases where a square

value DOES exist.

Adam wrote:

> 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

>>

>

>

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