## 19586Re: [PrimeNumbers] Re: Prize Puzzle : Primality Conjecture

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• Sep 10, 2008
You haven't given a counterexample. Read his conjecture again.

Hint... his conjecture says nothing about cases where a square
value DOES exist.

> 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.
>
>
> There are 644 other counterexamples for x<=1000.
>
>
> --- 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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