Proof that there are infinitely many primes congruent to 3 modulo 4

Question

I'm having difficult proving this.

As a hint the exercise to prove first, that if $a\lneqq \pm 1$ satisfies $a \equiv 3 \pmod4$, then exist $p$ prime, $p \equiv 3 \pmod 4$ such $p\mid4$. But I'm not really getting for what purpose can this be used.

Answer 1

If there are only finitely many primes $\equiv 3 \pmod 4$, take the product of them and denote that product by $a$. Now look at $2a + 1$, and try to deduce a contradiction.

Answer 2

Use Euclid's proof showing that there are infinitely many primes, i.e., find an Euclidean polynomial you can use for your arithmetic progression $l \mod k$. Since $l^2\equiv 1 \mod k$ such an Euclidean polynomial exists - see http://www.mast.queensu.ca/~murty/murty-thain2.pdf how to do it (in particular, on page one, the case $4n+3$ is given, see [5]). For $8n+1$ see Infinitely many primes of the form $8n+1$.

Answer 3

Does this work?

This is a modification of Euclid’s proof. Let $p_1,p_2,…,p_k$ be the only 3 mod 4 primes. Consider two cases : (I) $k$ is odd Then notice $p_1p_2…p_k+4$ is a 3 mod 4 number, and in particular not divisible by any of the $p_i$s we have. Then it must be a product of some 1 mod 4 primes for it to be not a 3 mod 4 prime itself. But product of 1 mod 4 primes is always 1 mod 4, so not possible. (II) $k$ is even This is slightly trickier. Consider $p_1p_2…p_{k-1}+2^x$ for some $x \geq 2$. Again, this is 3 mod 4. I claim we can choose $x$ so that this is not divisible by $p_k$. Suppose not, then $2^{a+1} \equiv{2^{a}} \pmod{p_k}$,but theb 2 is congruent 1 mod p so 1 is divisible by p_k, which is absurd.