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• ## Re: Prize Puzzle : Primality Conjecture

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• 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
Message 1 of 5 , Sep 10, 2008
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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.

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
>
• ... That s his point. There must be NO square in the selected interval, but T(10) is a square. By the way, this is just a restatement of Euler s factorization
Message 1 of 5 , Sep 10, 2008
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> 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.

That's his point. There must be NO square in the selected interval, but
T(10) is a square. By the way, this is just a restatement of Euler's
factorization method. If you put N = A(x) = m*n = (a-b)*(a+b) = a^2-b^2
and do some algebraic tricks, it wouldn't be too difficult to prove the
conjecture and claim your 50 bucks.

Bernardo Boncompagni

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• You haven t given a counterexample. Read his conjecture again. Hint... his conjecture says nothing about cases where a square value DOES exist.
Message 1 of 5 , Sep 10, 2008
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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
>>
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
>
>
• Hi Adam, ... Umm... pardon me? How does this contradict Aldrich s conjecture? As far as I understood the claim, A(x) is claimed to -be a prime- under the
Message 1 of 5 , Sep 10, 2008
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Aldrich's conjecture:
>> 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.

> 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.

Umm... pardon me? How does this contradict Aldrich's conjecture? As far as
I understood the claim, A(x) is claimed to -be a prime- under the
assumption that -no- value of T(k) between 1 and B(x) is a square.

Peter

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
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