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 $p \equiv 19 \mod 24$?
Best Answer
$3$ 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 $p\equiv19\bmod24$.