- 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

> - Adam dice:

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

That's his point. There must be NO square in the selected interval, but

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

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.

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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> The Prime Pages : http://www.primepages.org/

>

> Yahoo! Groups Links

>

>

>

>

> - Hi Adam,

Aldrich's conjecture:>> If x > 3 and k > x , then A(x) will be prime if no value

Adam's proposed counterexample:

>> 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,

Umm... pardon me? How does this contradict Aldrich's conjecture? As far as

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

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