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.


1.How did they get to this conclusion?

Well, it already assumed that prime numbers of $4n+3$ are finite. If you multiply all of them and $+3$ (or $-1$, if you like), by assumption, $N$ is not in the prime set of $4n+3$ form. So according to the finite assumption, $N$ is not a prime number. 


2.Why must it be of this form $(4n+1)$?

We classify all prime numbers into four sets of the form $4n,4n+1,4n+2,4n+3$. from $1$ we already see that prime number in $4n+3$ can't divide $N$ by our assumption. Prime numbers of $4n$ and $4n+2$ contain only $2$, however, $N=4k+3$, so $N$ is odd which can't be divided by $2$. The only result is that prime numbers of the form $4n+1$ could divide $N$.


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

$(4a+1)(4b+1) = 4(4ab+a+b)+1$ which is still of the form $4n+1$. And that's why our assumption of $N$ is not a prime number fails. So we see $N$ is a new prime number of the form $4n+3$. Repeat this process, we will get the infinite numbers of elements with the form $4n+3$.
