Two questions about the infinitude of prime Euclid numbers: $E_{n}=p_{1}\cdot p_{2}\cdot \ldots \cdot p_{n} + 1$ First of all I was wondering what results there are out there about the prime Euclid numbers? I would like to read up on any results that are out there.
Secondly I wanted someone to look at the following argument about the infinitude of the prime Euclid numbers.
According to Dirichlets theorem: If $a,b\in \mathbb{Z^{+}}$ such that $(a,b)=1$, then there are infinitely many prime numbers of the form $am+b$ where $m$ is a non-negative integer. Now, if we denote the $n$th primorial (i.e the product of the first $n$ primes) as $\displaystyle \prod_{i=1}^{n} p_{i}$ we define the Euclid numbers as $$E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1.$$ Note that $$E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1=2\cdot {\displaystyle \prod_{i=2}^{n} p_{i}}+1$$ where the first term $$2\cdot {\displaystyle \prod_{i=2}^{n} p_{i}}\equiv 2\pmod 4.$$ Thus $$E_{n}\equiv 3\pmod 4.$$ Meaning in particular that $$\boxed{E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1=4m+3}$$ for some non-negative integer $m$.
Since $(4,3)=1$, then by Dirichlets theorem we have that $4m+3$, where $m$ is a non-negative integer is a prime number infinitely often. And since the $RHS$ of the boxed equation above has this form, then does this imply that whatever is on the $LHS$ must also be a prime number infinitely often? In particular, does this mean that $E_{n}$ is prime infinitely often?
 A: I will try to elaborate what the comments are trying to say with a little bit more formality.
It is true that there are infinite primes of the form $4m + 3$. But you should note that you will get this infinitude of primes if you run $m$ all over the set of natural numbers. To see this let's say, take $t \in \{2n^3 - 1| n \in \mathbb{N}\} \subset \mathbb{N}$. You will  see that  you will never get a prime of the form $4t + 3$ except $7$. My point is that you can't guarantee that there will be infinite list primes of the form $4m+3$ if you run $m$ through arbitrary subset of $\mathbb{N}$.
Now, let's go back to the boxed equation and see the set through which you are running $m$:
$$E_n = 2\prod_{i = 2}^n p_i + 1 = 4m_n + 3$$
$$2\prod_{i=2}^n p_i - 2 = 4m_n$$
$$ \prod_{i=2}^n p_i = 2m_n + 1$$
This means if $m_n$ runs through $\mathbb{N},$ then $2m_n+ 1$ should run through all odd numbers greater than $2$(or $0 $ if you include $m_1 = 0$). But the product doesn't run through $\{2k + 1 | k \in \mathbb{N},\ \mu(2k+1) = 0\}$. This implies $m_n$ runs through a set that is considerably "smaller" (not smaller per se, but you get the idea) than $\mathbb{N}$. This shows the boxed argument is not sufficient to imply the infinitude of Euclid primes.
A: No, you can't arguee that. You only know that at least one prime factor of $E_n$ is of the form $4m+3$. But it is not clear at all how many of the $E_n$ are prime.
The set
$$\{ E_n : n \in \mathbb N\}$$
is a proper subset of
$$\{ 4m+1 : m \in \mathbb N_0\}.$$
We only know that in the latter are infinitely many primes.
