Dirichlet's theorem for modulus eight

Statement

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

$p\equiv a{\pmod {8}}$.

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 constant term of absolute value one takes infinitely many pairwise relatively prime values

Proof

We need to consider four congruence classes modulo $8$: $-3,-1,1,3$. We do these case by case.

Congruence class of $-1$ modulo $8$

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

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 $\pm 1{\pmod {8}}$. 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 $p\equiv 1,3{\pmod {8}}$. In particular, a prime $p$ can divide $n^{2}+2$ for some natural number $n$ if and only if $p\equiv \pm 1{\pmod {8}}$.

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 $1,3{\pmod {8}}$. 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 $8$.