- Hi, All!

Is the next statement true?

Let n is prime, b and c are integer, c>b>0, GCD(c,b)=1,GCD(c-b,n)=1.

Then

all prime factors of the number (c^n-b^n)/(c-b) have the form 2kn+1,

where

k>0 is integer.

Where can I find it?

Thanks.

--

Best regards. Someone.

http://www.someoneltd.boom.ru/ http://home.tula.net/frazez/ - From: someoneltd

Date: 01/22/06 11:41:50

To: primenumbers@yahoogroups.com

Subject: [PrimeNumbers] Prime factors of (c^n-b^n)/(c-b)

Hi, All!

Is the next statement true?

Let n is prime, b and c are integer, c>b>0, GCD(c,b)=1,GCD(c-b,n)=1.

Then

all prime factors of the number (c^n-b^n)/(c-b) have the form 2kn+1,

where

k>0 is integer.

Where can I find it?

Thanks.

Kermit says:

p divides (c^n - b^n)/(c-b)

p divides (c^n - b^n)

C^n - b^n = 0 mod p

c^n = b^n mod p

(c/b)^n = 1 mod p

n divides (p-1)

kn = p-1

kn+1 = p

p = kn + 1

[Non-text portions of this message have been removed] - In an email dated Sun, 22 1 2006 4:38:44 pm GMT, "someoneltd" <umax@...> writes:

>Is the next statement true?

You statement as written is false.

>

>Let n is prime, b and c are integer, c>b>0, GCD(c,b)=1,GCD(c-b,n)=1.

>Then

>all prime factors of the number (c^n-b^n)/(c-b) have the form 2kn+1,

>where

>k>0 is integer.

Counterexample: c = 4, b = 3, n = 2.

For n = 2, you can't have the "2" in "2kn+1".

For n > 2 (and so odd) your statement is true.

-Mike Oakes