## Prime factors of (c^n-b^n)/(c-b)

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• 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)
Message 1 of 3 , Jan 22, 2006
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
Message 2 of 3 , Jan 22, 2006
From: someoneltd
Date: 01/22/06 11:41:50
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]
• ... You statement as written is false. 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
Message 3 of 3 , Jan 23, 2006
In an email dated Sun, 22 1 2006 4:38:44 pm GMT, "someoneltd" <umax@...> writes:

>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.

You statement as written is false.
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
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