Infinite primes of the form 3n+2 Without recourse to Dirichlet's theorem, of course.
We're going to go over the problems in class but I'd prefer to know the answer today.
Let $S = \{3n+2 \in \mathbb P: n \in \mathbb N_{\ge 1}\}$
edit:
The original question is "the set of all primes of the form $3n + 2$, but I was only considering odd primes because of a reason I don't remember anymore.
How can I show $S$ is infinite? I start by assuming it's finite. 
I've tried: 
Assuming that the product $(3n_1+2)(3n_2+2)...(3n_m+2)$ is of the form "something" so I can show the product contains a prime factor not in the finite list of primes, which would be a contradiction, but came up with no useful "something".
I tried showing that the product would not be square-free, but couldn't show that.
I tried showing the product or sum was both even and odd, but couldn't show that.
What else should I try? Or was one of the above methods correct?
 A: The general strategy is to find a (large) number $n$ that is relatively prime to each of the existing list of such primes, and is also congruent to 2 modulo 3.  The prime factorization of $n$ cannot consist only of primes congruent to $1$ modulo $3$, since the product of any number of such is still $1$ modulo $3$.  Hence there must be some prime factor of $n$ that is congruent to $2$ modulo $3$, which must be not on our list by the construction of $n$.
Now, how to construct such an $n$?  Suppose the finite list is $\{p_1, p_2, \ldots, p_k\}$.  If $k$ is even, then take $n=p_1p_2\cdots p_k+1$.  If $k$ is odd, then take $n=(p_1p_2\cdots p_k)p_k+1$.
A: If $n$ in your question is even, the number can't be a prime. So it suffices to prove there are infinitely many primes of the form $6n-1$.
Assume there are only finitely many of them let $n_1\ldots,n_k$ be their representants. Then
$$6\bigl((6n_1-1)(6n_2-1)\ldots(6n_k-1)\bigr)-1$$
is a number of the form $6n-1$ which is not divisible by $2,3,$ and any of the primes of the form $6n-1$.
But all primes are either $2$, $3$, or of the form $6n\pm1$, so the number must be a product of primes of the form $6n+1$, but that's impossible because product of such numbers has again the form $6n+1$.
A: Hint: 
Assume there are only $m$ such primes. Consider
$$P=(3n_1+2)(3n_2+2)...(3n_m+2) = (-1)^m \mod 3$$
Now if $2|m, P+2 = -1\mod 3$.
If $2\nmid m$ then $P=1\mod 3$.
