Dirichlet's theorem for modulus four

Statement

Suppose $a$ is an odd natural number. Then, there exist infinitely many prime numbers $p$ such that:

$p\equiv a(\mathrm{mod}4)$.

Facts used

1. Congruence condition for two to be a quadratic residue
2. Congruence condition for minus one to be a quadratic residue
3. Nonconstant polynomial with integer coefficients and nonzero constant term takes infinitely many pairwise relatively prime values

Proof

We need to check two congruence classes modulo $4$: the congruence class of $1$ and the congruence class of $-1$.

The congruence class of $1$ modulo $4$

By fact (2), $-1$ is a quadratic residue modulo $p$ if and only if $p\equiv 1(\mathrm{mod}4)$. In particular, a prime $p$ can divide $n^2+1$ for some natural number $n$ if and only if $p\equiv 1(\mathrm{mod}4)$.

Consider now the polynomial $f(x)=x^2+1$. For any natural number $n$, all prime divisors of $f(n)$ are congruent to $1$ modulo $4$. By fact (3), there are infinitely many pairwise relatively prime values of $f(n)$, so we get infinitely many primes that are congruent to $1$ modulo $4$.

Congruence class of $-1$ modulo $4$

By fact (1), $2$ is a quadratic residue modulo $p$ if and only if $p\equiv \pm 1(\mathrm{mod}8)$. In particular, a prime $p$ can divide $n^2-2$ for some natural number $n$ if and only if $p\equiv \pm 1(\mathrm{mod}8)$.

Consider now the polynomial $f(x)=(2x+1{)}^2-2=4x^2+4x-1$. For any natural number $n$, all prime divisors of $f(n)$ are congruent to $\pm 1(\mathrm{mod}8)$. However, $f(n)$ itself is $-1$ modulo $8$, so $f(n)$ must have at least one prime divisor that is $-1$ modulo $8$. By fact (3), there are infinitely many pairwise relatively prime values of $f(n)$, yielding infinitely many distinct primes that are $-1$ modulo $8$. In particular, all these primes are $-1$ modulo $4$.