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 constant term of absolute value one 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 $4$.