Inifinitely many primes $p\equiv -1 \mod12$

Question

I haven't been able to prove this statement from my Elementary Number course:

There are infinitely many primes $p$ such that $p\equiv -1 \mod12$.

From here I know that there exists a "Eulcidean style" proof for this problem. I think it should be related to quadratic residues.

Thanks,

EDIT: I know it follows from Dirichlet's theorem, but I am expected to find an elemntary proof.

Answer 1

Suppose there are only finitely many primes of this form, say $\{p_1,\dots,p_n\}$. Let $P = p_1\dots p_n$ denote their product. Consider now the following expression: $$Q = 12P^2-1$$ Then we observe the following:

Every prime divisor of $Q$ is $\pm 1 \pmod{12}$

Proof. Let $q$ be a (necessarily) odd prime such that $q \mid 12P^2-1$. Then $$3(2P)^2 \equiv 1 \pmod{q}$$ So $3$ is a quadratic residue. But as an application of the quadratic reciprocity law, this only happens when $q \equiv \pm 1 \pmod{12}$.

So every prime divisor is of the form $12n+1$ or $12n-1$. But they cannot all be of the form $12n+1$, otherwise the product would also be of this form. We conclude that there is a prime of the form $12n-1$ dividing $Q$ and this prime cannot divide $P$, so we're done.

Answer 2

The answer comes directly from Dirichlet's theorem: Let us have the sequence $12k+11$. The members of the sequences are all $\equiv -1$ (modulo $12$), and according to Dirichlet's theorem, since $gcd(11,12)=1$, there are infinite amount of primes in our sequence.