Infinitely many primes of the form $4n+3$ proof

Question

So I have seen this proof plastered everywhere, and here is a version of it from my textbook. No matter where I read, I don't understand one step of the proof, and I will highlight below. It surely has something to do with the Lemma that states for $x=y+z$, if $a\mid y$ and $a \mid z$, then $a\mid x$. Thank you.

Fact: There are infinitely many primes of the form $4n+3,$ where $n$ is a positive integer.

Proof: Let us assume that there are only a finite number of primes of the form $4n+3,$ say $p_0=3,p_1,p_2,...,p_r$.
Let $$Q = 4p_1p_2\cdot\cdot\cdot p_r+3.$$ Then, there is at least one prime in the factorization of $Q$ of the form $4n + 3$. Otherwise, all of these primes would be of the form $4n + 1$, and by Lemma 2.6, this would imply that $Q$ would also be of this form, which is a contradiction. However, none of the primes $p_0,p_1,...p_n$ divides $Q$. The prime $3$ does not divide $Q$, for if $3|Q$, then $3|(Q-3)=4p_1p_2...p_r,$ (How did they reach this? I don't understand why $3$ does not divide $4p_1p_2...!$ ) which is a contradiction.

Likewise, none of the primes $p_j$ can divide $Q$, because $p_j|Q$ implies $p_j|Q-(4p_1p_2...p_r)=3$ which is absurd. Hence, there are infinitely many primes of the form $4n + 3$.

Answer 1

It's actually a rearrangement ( or alteration) of that lemma : $$x=y+z\implies x-y=z\land x-z=y$$

So we get by factoring out that: $$a\mid x \land a\mid y \implies a\mid z$$ and $$a\mid x \land a\mid z\implies a\mid y$$

This is equivalent to $$a\nmid x \land a\mid y\implies a\nmid z$$ and $$a\nmid x \land a\mid z\implies a\nmid y$$

Answer 2

I prefer saying the primes in the list are $4n-1,$ then defining

$$Q = 4p_1p_2\cdot\cdot\cdot p_r-1.$$