Proof of infinitude of the number of primes of the form $4k+1$

Question

Would this work?

If the number of primes of a given form is finite, then the number $$M = (4k_1 + 1)(4k_2 + 1) \ldots (4k_n + 1) + 4$$ should be composite. But the product of numbers of the form $4k + 1$ also gives $4k + 1$, so $M = 4K + 1 + 4 = 4(K + 1) + 1$, and $M$ is prime because it leaves a remainder of $4$ after dividing by any prime of the given form; also $M$ is not divisible by $4$. So that should be contradiction.

Answer 1

All right, the argument doesn't work as it stands. But then, can we fix it? Do we have a way to construct numbers that definitely aren't divisible by any primes congruent to $3$ mod $4$?

The trick:

A sum of squares $a^2+b^2$ is divisible by a prime $p$ congruent to $3$ mod $4$ only if both $a$ and $b$ are divisible by $p$. So, then, we just need to replace that product of primes with a perfect square that fills the same role.

Also, as noted in the comment, you're skipping steps in the logic. What exactly are those $4k_i+1$? Why should the number you construct be composite? An experienced mathematician will likely know how these gaps should be filled, but you can't always assume that your audience will, or that the audience will give you the benefit of the doubt.

Answer 2

No, it is not correct. You cannot deduce that $M$ is prime. Note that $M$ could be the product of an even number of primes of the form $4k-1$.

Answer 3

You can make it work with a little adjustment.

Let $q$ be the largest prime of the form $q=4k+1$. Then let

$$x = 2\prod_{p \in \mathbb{P}; \ p \le q} p,$$

where here $\mathbb{P}$ denotes the set of odd primes. and let us set $M = x^2+1$. Now let $q'$ be a prime that divides $M$. Then as $M$ is odd, it follows that $q'$ must be odd, and as $q$ and no odd prime smaller than $q$ divides $M$ it follows that $q' > q$. Now if $q'$ is $M$ itself then we are done [why]. We note the following otherwise:

$$M - 1 = x^2 \equiv -1 \mod q'$$

This implies that

$x^4 \equiv 1$ mod $q'$ and so (as $x^2 \not \equiv 1$ mod $q'$) that the order of $x$ in $\left(\mathbb{F}_{q'}\right)^{\times}$ is precisely 4, which implies that $|\left(\mathbb{F}_{q'}\right)^{\times}| = q'-1$ is divisible by 4.

Can you finish the rest?