I saw similar questions related to this about the primes of the form $4k+3, 4k+1, \cdots$, but still I had some issues to ask.
To prove that there are infinitely many primes of the form $4k+3$, where $k$ is nonnegative integers, we can do that with proof by contradiction.
By Contradiction:
Let $m = 4q_1q_2\cdots q_n -1$, $m$ is of the form $4k+3$ because it's equivalent to $4k-1$. If we assume that primes $q_1, \cdots, q_n$ are primes of the form $4k+3$, which is equivalent to $4k-1$.
Let $m = 4q_1q_2\cdots q_n -1$, but this number is not of the form $4k+3$ as it's larger than all primes $q_1, \cdots, q_n$ of form $4k+3$. But since $m$ is not prime as it's not one of the primes of form $4k+3$, then we conclude that it's a multiple of primes.
None of the $q_i's$ of the form $4k+3$ divides $m$ as none divides $-1$, so we can conclude that $m$ needs to be the product of prime numbers of the form $4k+1$.
We have a contradiction that $m$ is of two forms $4k+1, 4k+3$.
Question: Why we concluded that since none of the $q_i's$ of the form $4k+3$ divides $m$ as none divides $-1$, so we can conclude that $m$ needs to be the product of prime numbers of the form $4k+1$? Why this form and not any other form specifically?
Best Answer
Rewriting this answer to make it constructive, as per @BillDubuque.
Let $S$ be a finite set of primes of the form $4k+3$. The following argument shows there is another such prime, not in $S$. It follows that there must be infinitely many such primes.
The product of two numbers of the form $4k+1$ is again of that form (easy to check).
There are only two kinds of odd primes when you work modulo $4$: some of the form $4k+1$, others of the form $4k+3$, since the numbers of the form $4k$ and $4k+2$ are even.
Let $Z$ be the product of all the primes in $S$ and let $m = 4Z+3$. Since when you divide $m$ by $4$ the remainder is $3$, $m$ cannot be the product only of primes of the form $4k+1$. It must have at least one prime factor of the form $4k+3$. Since none of the primes in $S$ divides $m$ you have found another prime of that form.
This kind of argument does not work to show that there are infinitely many primes of the form $4k+1$. That's true, but you have to work harder to show it.