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
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