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?
Best Answer
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 $n=(p_1p_2\cdots p_k)p_k+1$.