Infinitely many primes of the form $4n+3$ I've found at least 3 other posts$^*$ regarding this theorem, but the posts don't address the issues that I have.
Below is a proof that for infinitely many primes of the form $4n+3$, there's a few questions I have in the proof which I'll mark accordingly.
Proof: Suppose there were only finitely many primes $p_1,\dots, p_k$, which are of the form $4n+3$. Let $N = 4p_1\cdots p_k - 1$. This number is of the form $4n+3$ and is also not prime as it is larger than all the possible primes of the same form. Therefore, it is divisible by a prime $ \color{green}{ \text{(How did they get to this conclusion?)}}$. However, none of the $p_1,\dots, p_k$ divide $N$. So every prime which divides $N$ must be of the form $4n+1$ $ \color{green}{ \text{(Why must it be of this form?)}}$. But notice any two numbers of the form $4n+1$ form a product of the same form, which contradicts the definition of $N$. Contradiction. $\square$
Then as a follow-up question, the text asks "Why does a proof of this flavor fail for primes of the form $4n+1$? $ \color{green}{ \text{(This is my last question.)}}$

$^*$One involves congruences, which I haven't learned yet. The other is a solution-verification type question. The last one makes use of a lemma that is actually one of my questions, but wasn't a question in that post.


Therefore, it is divisible by a prime (How did they get to this conclusion?).

All integers are divisible by some prime!

So every prime which divides N must be of the form 4n+1 (Why must it be of this form?).

Because we've assumed that $p_1, \dots, p_k$ are the only primes of the form 4n+3. If none of those divide N, and 2 doesn't divide N, then all its prime factors must be of the form 4n+1.

"Why does a proof of this flavor fail for primes of the form 4n+1? (This is my last question.)

Can you do this yourself now? (Do you understand how the contradiction works in the proof you have? What happens if you multiply together two numbers of the form 4n+3?)
