Browse Groups

• 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 6:59 AM
View Source
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 3:23 PM
View Source
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 9:42 AM
View Source
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

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.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.