Prime divisors of $4k - 1$ In this question, I have to prove that there exists infinitely many primes for $n = 4k - 1$. We can consider this sequence $3, 5, 7, 11, 19, 23, 31, \ldots$, which are all a form of $4k - 1$.
So, we start by considering the positive integers of the form $n=4k-1$ and their possible prime divisors.


*

*Show that every prime divisor of $4k - 1$ is odd.

*Show that an odd number is either of the form $4k - 1$ or $4k + 1$.

*Show that a product of two numbers of the form $4k + 1$ again has that form (more precisely, given two numbers $4k + 1$, $4l + 1$, show that their product has the form $4m + 1$ for some integer $m$).
I need help up to this point to solve the rest of the question. Any pointers will be greatly appreciated!
 A: *

*Every divisor of an odd number is odd. As $4k$ is always even, $4k-1$ is therefore odd.
2.
Every number leaves a remainder of $0,1,2,$ or $3$ when divided by 4. If the number was originally odd, then it leaves a remainder of $1$ or $3$ when divided by 4.
If the number leaves a remainder of $1$ when it's divided by $4$, then it's clearly of the form $4k + 1$
If the number leaves a remainder of 3 when it's divided by 4, then it's of the form $4x + 3=4(x+1)-1$, which is of the form $4k-1$
3.
$$\begin{align}(4k+1)(4l+1)&=16kl+4k+4l+1 \\ &=4(4kl+k+l)+1 \end{align}$$
which is of the form $4m+1$
A: Assume the first three points have been proved. Then if $n=4k-1$ is not prime, it has only odd divisors (point 1), which must therefore all be of the form $4k'\pm 1$ (point 2). If all of these are of the form $4k+1,$ then $n$ itself would have to take this form (by point 3). Thus, if $n=4k-1$ is not prime, then it has a proper divisor of the form $4k'-1.$
Suppose by way of contradiction that there are only finitely many primes of the form $4k-1$, $p_{1},p_{2},\ldots,p_{n}.$ If $n$ is odd, then $\prod_{i=1}^{n}p_{i}$ is of the form $4k-1,$ and if $n$ is even, this product is of the form $4k+1,$ so let $p=\prod_{i=1}^{n}p_{i}+4$ if $n$ is odd, and otherwise let $p=\prod_{i=1}^{n}p_{i}+2.$ In either case, $p$ is of the form $4k-1.$ If $p$ were composite, it would have a proper divisor of the form $4k'-1,$ as we showed above. If this divisor is not prime, then it too has a divisor of the form $4k''-1,$ and we may repeat this argument to say that $p$ has a prime divisor $p'$ of the form $4k-1.$ But $p'$ must divide $\prod_{i=1}^{n}p_{i},$ since this is the product of all primes of the form $4k-1,$ and it cannot be the case that $p'|4$ or $p'|2,$ so in fact $p'\not|\,p,$ and thus $p$ is prime. Since $p$ is of the form $4k-1$, and strictly larger than any prime in the list $p_{1},\ldots,p_{n},$ this contradicts our assumption that this was the list of all of the primes of the form $4k-1.$ This contradiction completes the proof.
