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

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 $$p \equiv 19 \mod 24$$?

• Is mod $14$ a typo.? $\left(\frac{3}{p}\right)=-1\iff p\equiv\pm5\bmod12$; that and $p\equiv3\bmod8$ means $p\equiv19\bmod24$ Commented Apr 3, 2020 at 4:27
• I was using my old notes from number theory class: $\left(\frac{p}{q} \right)= \left(\frac{q}{p} \right)$ if $p \equiv 1 \mod 4$ or $q \equiv 1 \mod 11$ and $=-\left(\frac{q}{p} \right)$ if $p, q \equiv 3 \mod 14$. Maybe it's wrong or has some typo? Anyway, you answered my question.
– Andy
Commented Apr 3, 2020 at 4:42
• $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)/2}(-1)^{(q-1)/2}=1$ if $p$ or $q\equiv 1\bmod 4$ and $-1$ if $p$ and $q\equiv3\bmod4$ Commented Apr 3, 2020 at 4:45

$$3$$ 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 $$p\equiv19\bmod24$$.
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.