# infinitely many primes of the form 6k + 5 and 6K + 1

• Oct 21st 2009, 08:51 AM
MariaJJ
infinitely many primes of the form 6k + 5 and 6K + 1
hey the question is

show that there are infinitely many primes of the form 6k + 5. does the method work for 6k + 1.

suppose there are finite primes of the form 6k + 5
order them such: p(1) < p(2) <....< p(n)
let R = 6(p(1)p(2)...p(n)) + 5
R can't be prime, if it is R > p(n)
R can't be composite as any division will give a remainder of 5
therefore there are infinitely many primes :)

i think there's something not quite right with it and i can't use Dirichlet's Theorem

kudos for any help! :)
• Oct 22nd 2009, 05:25 AM
Media_Man
good work
You have hit the nail on the head. Your proof is correct. Here are some clarifications:

Let $p_1 be a finite list of primes of the form $6k+5$. Let $R=6p_1p_2p_3...p_n+5$. Since $R>p_n$, $R$ cannot be prime, so it is composite. When divided by any $p_i$, it leaves a remainder $5$, therefore $R$ is not divisible by any prime of the form $6k+5$. So $R$ must be the product of primes only of the form $6k+1$. But if you take the product of numbers of the form $6k+1$, the product is also of the form $6k+1$. So $R$ must leave a remainder $1$ when divided by $6$. But by construction, $R$ leaves a remainder $5$ when divided by $6$. We have reached a contradiction, therefore our premise is false: there must be an infinite number of primes of the form $6k+5$.

This argument does not work for the $6k+1$ case. Can you figure out why?