If $p \equiv 3 \mod 8$ is prime and 3 is non-residue of $p$, then $p\equiv 19 \mod 24$.

Question

I'm reading Stark's paper "a complete determination of the complex quadratic fields of class number one".

He argues that if $p \equiv 3 \mod 8$ is prime and 3 is non-residue of $p$, then $p\equiv 19 \mod 24$, which I don't follow.

Suppose $p\equiv 11 \mod 24$. Since $$\left(\frac{24k+11}{3}\right)=\left(\frac{2}{3}\right)=-1,$$ if I want to use the Quadratic Reciprocity so that I can have $\left(\frac{3}{24k+11}\right)=-\left(\frac{24k+11}{3}\right)=1$, then I need to show $p=24k+11 \equiv 3 \mod 14$. I don't know how to show this condition holds whenever $p$ is prime. Is there a way to show $p \not \equiv 11 \mod 24$ other than using Quadratic Reciprocity? Also, how do I show that $p \equiv 19 \mod 24$?

Answer 1

$3$ is a non-residue of $p$ means $p\equiv\pm5\bmod12$.

If $p\equiv5\bmod12$ then $p\equiv1\bmod4$, so we can't have $p\equiv3\bmod8$.

So $p\equiv7\bmod12$ and $p\equiv3\bmod8$; i.e., $p\equiv1\bmod3$ and $p\equiv3\bmod8$.

By the Chinese remainder theorem, that means $p\equiv19\bmod24$.

Answer 2

We just have to show $p\equiv1\bmod3$ and apply the Chinese remainder theorem. By quadratic reciprocity $$\left(\frac3p\right)\left(\frac p3\right)=(-1)^{(p-1)/2\cdot(3-1)/2}=(-1)^{(p-1)/2}$$ Since $3$ is a non-residue modulo $p$: $$\left(\frac p3\right)=(-1)^{(p+1)/2}$$ The RHS must be $+1$ since $p+1\equiv4\bmod8$. The LHS is $+1$ iff $p\equiv1\bmod3$, which completes the proof.