# Prime Numbers

• Feb 9th 2007, 01:26 AM
Ideasman
Prime Numbers
Prove that if p, a prime number, is in the arithmetic progression of 3n + 1, n = 1, 2, 3, ..., then its also in the arithmetic progression of 6n + 1, n = 1, 2, 3, ....
• Feb 9th 2007, 03:30 AM
topsquark
Quote:

Originally Posted by Ideasman
Prove that if p, a prime number, is in the arithmetic progression of 3n + 1, n = 1, 2, 3, ..., then its also in the arithmetic progression of 6n + 1, n = 1, 2, 3, ....

It doesn't matter whether the number is prime or not. The entire series 6n + 1 is contained within 3n + 1. To see this suppose we have a number k = 6n + 1. What is n' for k = 3n' + 1?

k = 6n + 1 = 3n' + 1

6n + 1 = 3n' + 1

6n = 3n'

n' = 2n.

Since there is an n' for every n, thus every number 6n + 1 can be written in the form 3n' + 1.

-Dan
• Feb 9th 2007, 12:53 PM
AfterShock
Quote:

Originally Posted by topsquark
It doesn't matter whether the number is prime or not. The entire series 6n + 1 is contained within 3n + 1. To see this suppose we have a number k = 6n + 1. What is n' for k = 3n' + 1?

k = 6n + 1 = 3n' + 1

6n + 1 = 3n' + 1

6n = 3n'

n' = 2n.

Since there is an n' for every n, thus every number 6n + 1 can be written in the form 3n' + 1.

-Dan

You have to be careful, Dan.

You have:

{6n + 1 : n is a pos integer} "is a subset of (although not equal) " {3n + 1 : n is a pos int}

Yes I will agree with that;

Looking at the 3n + 1 set;

Given a prime in the largest set, we want to know if it a prime in the smaller set, too.

We know p > 2

3n + 1 is odd; therefore, 3n must be even

If the above is prime, then n has to be even = 2k

3n + 1 = 3(2k) + 1 = 3k + 1