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.
Best Answer
(1) The way I see it, we are doing a proof similar to Euclid's proof that there are infinitely many primes that most are familiar with, where we get a contradiction because every integer has a prime divisor (Euclid's case say $p_1,\dots,p_n$ are all the primes, let $N=p_1\cdots p_n+1$. $N$ has a prime divisor, but then it must be one of the $p_i$ but then as $p_i \mid (p_1\cdots p_n +1)$ this implies $p_i \mid 1$, a contradiction). Our goal here is basically to construct an integer $N$ that is of the form $4k+3$ so then it will be divisible by some prime of the form $4m+3$, and then we want that to lead to a contradiction. In this case, it leads to a prime dividing $-1$, a contradiction.
(3) The product of an even number integers of the form $4k+3$ is of the form $4k+1$. A product of an odd number integers of the form $4k+3$ is again of the form $4k+3$.
(4) Every integer of the form $4k+3$ has a prime factor of the form $4k+3$ however not every integer of the form $4k+1$ has a prime factor of the form $4k+1$.
Could consider $21=3\cdot7$, $21$ is of the form $4k+1$ but $3,7$ are both of the form $4k+3$ so it has no prime factors of the form $4k+1$.