Show for any odd prime $p\geq 5,$ $(-3/p)=1$ or $ -1$

Question

Show for any odd prime $$p\geq 5,$$ $$\left ( \frac{-3}{p} \right ) =\begin{cases} 1 & \text{ if } p\equiv 1,-5\pmod{12} \\ -1& \text{ if } p\equiv -1,5\pmod{12} \end{cases}$$

So far I have that

(1) Let $$p\equiv 1\pmod{4}$$ then $$p\equiv 1\pmod{3}$$ to get $$\left ( \frac{-3}{p} \right )=-\left ( \frac{p}{3} \right )=-\left ( \frac{1}{3} \right )=-1$$

(2)Let $$p\equiv 1\pmod{4}$$ then $$p\equiv 2\pmod{3}$$ to get $$\left ( \frac{-3}{p} \right )=-\left ( \frac{p}{3} \right )=-\left ( \frac{2}{3} \right )=1$$

(3)Let $$p\equiv 3\pmod{4}$$ then $$p\equiv 1\pmod{3}$$ to get $$\left ( \frac{-3}{p} \right )=\left ( \frac{p}{3} \right )=\left ( \frac{1}{3} \right )=1$$

(4)Let $$p\equiv 3\pmod{4}$$ then $$p\equiv 2\pmod{3}$$ to get $$\left ( \frac{-3}{p} \right )=\left ( \frac{p}{3} \right )=\left ( \frac{2}{3} \right )=-1$$

After solving CRT systems I get, $$\left ( \frac{-3}{p} \right ) =\begin{cases} 1 & \text{ if } p\equiv 5,-5\pmod{12} \\ -1& \text{ if } p\equiv 1,-1\pmod{12} \end{cases}.$$

So I'm not sure where I'm messing up. Any help would be appreciated.

Answer 1

You messed up some calculations. When $p\equiv1\pmod4$, $\left(\dfrac{-3}p\right)=\left(\dfrac{-1}p\right)\left(\dfrac{3}p\right)=1\left(\dfrac p3\right)$,

so in those cases $\left(\dfrac{-3}p\right)=1$ when $p\equiv1\pmod3$ and $\left(\dfrac{-3}p\right)=-1$ when $p\equiv2\pmod3$ .