Sorry, an error occurred while loading the content.

CONTEST!

Expand Messages
• 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:
Message 1 of 3 , Dec 20, 2007
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
Message 2 of 3 , Dec 20, 2007
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
• 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
Message 3 of 3 , Dec 25, 2007
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
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]
Your message has been successfully submitted and would be delivered to recipients shortly.