Infinitely many primes $p$ with $p^3\equiv5\mod6$

Question

Show that there are infinitely many primes $p$ with $p^3\equiv5\mod6$.

I've so far tried to look at proofs for 'simpler' statements like showing there are infinitely many primes $p\equiv3\mod8$ but it isn't helping. Any hints would be greatly appreciated.

Answer 1

Here is a proof that doesn't need Dirichlet's theorem. It is sufficient to show there are infinitely many primes $p \equiv 5 \mod 6$, since then $p^3 \equiv 5^3 \equiv 5 \mod 6$. Suppose there are only finitely many primes $p_1,\ldots, p_k$ greater than 5 that are congruent to 5 mod 6. Let $N=6p_1p_2\cdots p_k+5$. Then $p_i \nmid N$, and so all of prime factors of $N$ are congruent to 1 mod 6. But then $N$, being a product of numbers congruent to 1 mod 6, must be itself congruent to 1 mod 6. This is a contradiction, and so there are infinitely many primes $p \equiv 5 \mod 6$.

Answer 2

There are infinitely many primes congruent to $5\bmod6$ (Dirichlet's theorem on arithmetic progressions), so there are infinitely many primes whose cubes are congruent to $5^3\equiv5\bmod6$.