Evaluating quadratic residue $(6/p) $

Question

Evaluating quadratic residue $(6/p) $ I want to find for which numbers mod 6 this residue is equal to 1 and for which it's equal to -1. I've tried splitting it into $(3/p)(2/p)$ and expressing it as $(-1)^{(5 (p-1))/4} (p/6) $ both methods haven't led to me finding a solution. Is there an easier way?

Answer 1

For every prime $p\geq 7$ we know that: $$\left(\frac{-3}{p}\right)=1 \Longleftrightarrow p\equiv 1\pmod{3} $$ $$\left(\frac{-2}{p}\right)=1 \Longleftrightarrow p\equiv 1,3\pmod{8} $$ hence for every prime $p\geq 7$ we have: $$\left(\frac{6}{p}\right)=1 \Longleftrightarrow p\equiv \pm 1,\pm 5\pmod{24}. $$

Answer 2

I would simply go with your first method; the second is problematic as $6$ is not prime.

For $\left( \frac{2}{p} \right)$ use the second supplement law of quadratic reciprocity, to get it is $1$ for $\pm 1 \mod 8$.

For $\left( \frac{3}{p} \right)$ you can use quadratic reciprocity to reduce to $ (-1)^{(p-1)/2}\left( \frac{p}{3} \right)$.

Since $1$ is a quadratic residue $\mod 3$ while $2$ is not.

You get $\left( \frac{p}{3} \right) =1$ for $p \equiv 1 \mod 3$ and $-1$ otherwise.

Now, note that $(-1)^{(p-1)/2}$ is even for $p \equiv 1 \pmod 4$. So, $\left( \frac{3}{p} \right)$ is $1$ for $ p \equiv 1 \pmod{12}$ (as $1 \times 1$) and $ p \equiv 11 \pmod{12}$ as $(-1)(-1)$ and $-1$ for $ p$ congruent to $5,7 \pmod{12}$.

Now, combine the two and get a condition modulo $24$.