I know there are sundry questions — like this pdf — and this
Because $4k + 3 = 2(2k + 1) + 1$, any number of the form $4k + 3$ must be odd.
It can't have any factors of the form $4k = 2(2k) $ or $4k + 2 = 2(2k + 1)$ which are even
— so they must have forms $4k + 1$ and $4k + 3$.
Suppose that they were all of the form $4k + 1$. Multiplying two such forms yields $(4k+1)(4m+1) = 4(4km+k+m) +1$, another $4k+1.$
Thus $\Pi$ (factors of the form $4k + 1$) must be another $4... + 1.$
Thus $\Pi$ (factors of the form $4k + 3$) must have a prime factor of the form $4k + 3\quad (♯)$.
I still don't understand Elementary Number Theory — Jones — p28 — Theorem 2.9.
Prove by contradiction. Suppose that there are only finitely many primes of this form $4k + 3$, say $p_1, ... , p_k$. Let $\color{red}{m = 4(p_1 ... p_k - 1) + 3}$. Since $m$ is odd, and the only even prime is $2,$ so each prime $p$ dividing m is odd.
(1 — Red) Where did this choice of $m$ hail from — feels uncanny?
By reason of $(♯)$ overhead, $m$ must be divisible by at least one prime of the form $4k + 3$ - name it $p_i$. Thence $p_i$ divides $(4p_l ... p_k -m = 1) \implies p_i = \pm 1$, a contradiction because $p_i$ is prime.
(3) How can I prefigure to consider $4p_1...p_k - m = 1$, in order to instigate a contradiction?
(4) Why does the method of proof here fail for proving infinitely many primes of the form $4k + 1$? I tried https://math.stackexchange.com/a/391103/85100.