Dirichlet's theorem for modulus eight

Statement

For any odd natural number $a$, there are infinitely many primes $p$ such that:

$p\equiv a(\mathrm{mod}8)$.

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 consider four congruence classes modulo $8$: $1,3,5,7$. We do these case by case.

Congruence class of $-1$ modulo $8$

By fact (2), $2$ 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 $8$.

Congruence class of $3$ modulo $8$

By facts (1) and (2), we can deduce that $-2$ is a quadratic residue modulo $p$ if and only if $p\equiv 1,3(\mathrm{mod}8)$. 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 $1,3(\mathrm{mod}8)$. However, $f(n)$ itself is $3$ modulo $8$, so $f(n)$ must have at least one prime divisor that is $3$ modulo $8$. By fact (3), there are infinitely many pairwise relatively prime values of $f(n)$, yielding infinitely many distinct primes that are $3$ modulo $8$.