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

a verifiable counterexample or proof by New Year's day

for the following primality conjecture:

(x,A,B,c,k,f : integers)

Let A = 20x^2 + 10x + 1;

Let B = 10x^2 + 4x + 1;

Let c = trunc(A/sqrt(5)) - 1;

For any x > 0, apply the issquare test to each k

in the interval c <= k < B. If there is no value

of k that satisfies the conditions of the test,

then A is Prime.

(issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2) - aldrich617 wrote:

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

Wasn't much of a challenge. Is it a homework problem? The first

> a verifiable counterexample or proof by New Year's day

> for the following primality conjecture:

>

> (x,A,B,c,k,f : integers)

> Let A = 20x^2 + 10x + 1;

> Let B = 10x^2 + 4x + 1;

> Let c = trunc(A/sqrt(5)) - 1;

>

> For any x > 0, apply the issquare test to each k

> in the interval c <= k < B. If there is no value

> of k that satisfies the conditions of the test,

> then A is Prime.

> (issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

counterexample is at x=5 (which we can all agree is prime.)

Here, A=551 B=271 c=245

In this case, when k=247, the value of the expression

5*(2*k -1)^2 - 4*A^2

is 841, which equals 29^2. Thus, your algorithm would declare 5 not

prime, which is incorrect.

I can provide other examples where it fails in the opposite way, that

is, declaring a composite number to be prime. The first occurs at x=8

(which we can all agree is composite.)

In this case, A=1361 B=673 c=607

In this case, none of the positive values for 5*(2*k -1)^2 - 4*A^2

from 607 to 672 produce a square.

Aldrich Stevens, please transfer the money to my PayPal account by

clicking the "donate" button here:

http://futureboy.us/frinkdocs/donate.html

--

Alan Eliasen | "Furious activity is no substitute

eliasen@... | for understanding."

http://futureboy.us/ | --H.H. Williams - You have to read , Alan. (Furious activity is not substitue from carefully reading).

The prime test is on A, not x .

(I checked Aldrich conjecture for large values of x, and found no counter example)

Regards,

JT

------------------------------------------------------

http://www.echolalie.com

http://www.echolalie.org/wiki

-------------------------------------------------------------- Original Message -----

From: Alan Eliasen

To: aldrich617

Cc: primenumbers@yahoogroups.com

Sent: Friday, December 21, 2007 12:23 AM

Subject: Re: [PrimeNumbers] CONTEST!

aldrich617 wrote:

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

> a verifiable counterexample or proof by New Year's day

> for the following primality conjecture:

>

> (x,A,B,c,k,f : integers)

> Let A = 20x^2 + 10x + 1;

> Let B = 10x^2 + 4x + 1;

> Let c = trunc(A/sqrt(5)) - 1;

>

> For any x > 0, apply the issquare test to each k

> in the interval c <= k < B. If there is no value

> of k that satisfies the conditions of the test,

> then A is Prime.

> (issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

Wasn't much of a challenge. Is it a homework problem? The first

counterexample is at x=5 (which we can all agree is prime.)

Here, A=551 B=271 c=245

In this case, when k=247, the value of the expression

5*(2*k -1)^2 - 4*A^2

is 841, which equals 29^2. Thus, your algorithm would declare 5 not

prime, which is incorrect.

I can provide other examples where it fails in the opposite way, that

is, declaring a composite number to be prime. The first occurs at x=8

(which we can all agree is composite.)

In this case, A=1361 B=673 c=607

In this case, none of the positive values for 5*(2*k -1)^2 - 4*A^2

from 607 to 672 produce a square.

Aldrich Stevens, please transfer the money to my PayPal account by

clicking the "donate" button here:

http://futureboy.us/frinkdocs/donate.html

--

Alan Eliasen | "Furious activity is no substitute

eliasen@... | for understanding."

http://futureboy.us/ | --H.H. Williams

[Non-text portions of this message have been removed]