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

Every number $n>1$ is divisible by some prime $p$ (which includes the case $n=p$). Assume otherwise and let $n$ be the smallest such number. As this $n$ is not prime, it has a nontrivivial divisor $d$ with $1<d<n$. By minimality of $n$, $d$ is divisible by some prime $p$. But then $p$ also divides $n$.

All numbers are of the form $4n$, $4n+1$, $4n+2$, or $4n+3$. This is also true for primes $p$, but $p=4n$ is not possible and $p=2n$ only for $p=2$. Here, we have excluded $p=2$ as well as $p=4n+3$ by construction, which leaves only primes $p=4n+1$.

This proof fails for $p=4n+1$ because a number of the form $4n+1$ may well be the product of two numbers of the form $4n-1$. For example $3\cdot 7=21$. Therefore the step that at least one divisor must be of form $4n+1$ fails.

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?)

Question 1: Any integer is divisible by a prime (you actually don't even need the full strength of the unique factorization theorem; this is provable by induction).

Question 2: It's not divisible by any of the primes of the form $4n+3$ (namely, $p_1,\ldots,p_k$, which we're assuming is all of them), and it's not divisible by $2$ because it's odd. Thus, any prime dividing it must be of the form $4n+1$.

Question 3: Multiplying an even number of things of the form $4n+3$ together gives something of the form $4n+1$.