Proof - There're infinitely many primes of the form 3k + 2 — origin of $3q_1..q_n + 2$ Origin — Elementary Number Theory — Jones — p28 — Exercise 2.6
To instigate a contradiction, postulate $q_1,q_2,\dots,q_n$ as all the primes $\neq 2 (=$ the only even prime) of the form $3k+2$. Consider  $N=3q_1q_2\dots q_n+2.$ None of the $q_i$ divides $N$, and $3 \not | N$.
$N$ = 3(the product of odd numbers) + 2 = odd number + even number = odd. Because $N \ge 2 $,
$\color{brown}{♯}$ because $N$ is odd $\implies 2 \not | N$,
thence by $\color{brown}{♯}$ and the Fundamental Theorem of Arithmetic, N =  a product of one or more $\color{brown}{odd}$ primes.
The prime divisors of $N$ cannot be all of the shape $3k+1$. At least one of these primes is of the form $3k+2$ — Why? $(3a + 1)(3b + 1) = 3(...) + 1$, thence 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 $N$. Overhead in first paragraph, we proved $q_i \not|N$ for all $i$. Therefore $p \notin$  $\{q_1,\dots,q_n\}$ of primes of the form $3k+2$, contradiction. 


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*The general proof just starts with primes. Therefore how can you prefigure this proof's different start with odd primes ? 

*Where did this choice of $N$ hail from — feels uncanny?

*I don't understand how none of the $q_i$ divides $N$? 

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*Purport that you start with all primes of the form $3k+2$ including $\color{red}2$, thence you start with $M = 3q_1 ... q_n \color{red}2 + 2$ You will quickly realize that the argument don't work, because $\color{red}2$ always divides $M$ and we want a number outside the list to divide $M$. You would not be able to conclude that there are primes of the form $3k+2$ NOT on the list.
So remove $\color{red}2$ from the list is just a natural thing to do. 

*You want a number that doesn't divide any primes on the list and is of the form $3k+2$. You know that the number have to be of the form $3k+2$. So start out by thinking what is the possible value of $k$. If $k$ is not divisible by a prime $q_{i}$ on the list, you will quickly find that $q_{i}$ divides $N$. Because you want all $q_{i}|k$, their product $\Pi \, q_i$ is just one possible choice.

*$q_{i}$ divide $N-2$ by definition. So if $q_{i}$ divide $N$ too, you end up with $q_{i}$ divide $2$, which is not possible because $q_{i}$ is odd and is a prime.

*$N$ is odd so it is not divisible by any even number, prime or not.
