Proving that there are infinitely many primes with remainder of 2 when divided by 3 I need to prove that there are infinitely many primes with remainder of 2 when divided by 3. I started out similarly to Euclid's classic proof of an infinite number of prime numbers:
Suppose there is only a finite set of prime numbers with remainder of 2 when divided by 3, then we can write their product as:
$$ P = q_1 \cdot q_2 \cdots q_r, \qquad \text {for some integer } r, = $$ $$ (3q_1+2)\cdot(3q_2+2)\cdots(3q_r+2), $$ for integers $q_r$.
This is where I am stuck. I do not know how to get to a similar contradiction as Euclid did when he considered $P$+$1$ and how the $q_i$'s could not divide $P$+$1$ since they divided $P$. (If they divided $P$+$1$ then they would divide $P$ and $1$, where dividing $1$ is the contradiction). Any ideas on how I can get to a similar contradiction?
Origin —  Elementary Number Theory — Jones — p28 — Exercise 2.6
 A: $3P - 1$ is either a prime itself or has at least one prime factor $q \equiv 2 \pmod 3.$
The reason for this is that the product of any number of primes that are $\equiv 1 \pmod 3$ is again $\equiv 1 \pmod 3.$ So, since $3P - 1 \equiv 2 \pmod 3,$ we know there is some $q$ as described.
Note that $$  \gcd(3P-1,P) = 1, $$ so the new $q$ cannot divide $P,$ and therefore is not equal to any of your original list of primes. 
A: Let $q_1,q_2,\dots,q_n$ be odd primes of the form $3k+2$. Consider the number $N$, where 
$$N=3q_1q_2\dots q_n+2.$$
It is clear that none of the $q_i$ divides $N$, and that $3$ does not divide $N$.
Since $N$ is odd and greater than $1$, it is a product of one or more odd primes. We will show that at least one of these primes is of the form $3k+2$. 
The prime divisors of $N$ cannot be all of the shape $3k+1$. For the product of any number of (not necessarily distinct) primes of the form $3k+1$  is itself of the form $3k+1$. But $N$ is not of the form $3k+1$. So some prime $p$ of the form $3k+2$ divides $N$. We already saw that $p$ cannot be one of $q_,\dots,q_n$.  It follows that given any collection $\{q_1,\dots,q_n\}$ of primes of the form $3k+2$, there is a prime $p$ of the same form which is not in the collection. Thus the number of primes of the form $3k+2$ cannot be finite.
A: Let $N > 2$ be an integer.  Early on in Chapter 10 of these notes, on p120 as Theorem 121, I explain how Euclid's proof of the infinitude of primes can be very slightly modified to prove the following generalization:
There are infinitely many prime numbers $p$ such that $p$ is not of the form $kN+1$.
When $N = 3$, this means that there are infinitely primes which are either of the form $3k$ or $3k+2$.  Since there is only one prime of the form $3k$, this answers your question.
(The notes go on to mention that essentially the same argument proves that for any proper subgroup $H$ of $(\mathbb{Z}/N\mathbb{Z})^{\times}$, there are infinitely many primes $p$
such that the reduction of $p$ modulo $N$ does not lie in $H$.)
A: Break it into two cases depending on the parity of $r$. Start by showing that if $r$ is even, $P$ has a remainder of $1$ when divided by $3$, while if $r$ is odd, it has a remainder of $2$. In the first case show that $P+1$ must have a prime factor of the right form; in the second case, consider $P+3$ instead.
A: Try squaring P and adding 1. (The square of a number of form 3n+2 is of the form 3n+1)
A: An intuitive proof is simply this: Note that least prime factors are governed by combinatorial units, the smallest of which is $6$ elements.
$3, 2, n_{1}, 2, n_{2}, 2, 3$
Apart from multiples of $3$, all odd numbers are $n_{1}$, and $n_{2}$. These are potentially prime. If either one is prime, the odd composite less than it or greater than it is a multiple of $3$.
