For which primes $p$ is $-3\in Q_p$?

Question

I'm really bad at doing these for some reason, just need some help, this is not a hw question. I just need to do these smaller problems to gain some understanding.

For which primes $p$ is $-3\in Q_p$ (quadratic residues of p)?

I know that $(\frac{-3}{p})=(\frac{-1}{p})(\frac{3}{p})$

For $(\frac{-1}{p})$ = 1 if $p\equiv 1\pmod{4}$ and -1 if $p\equiv 3\pmod{4}$

Also for $(\frac{-3}{p})$ = 1 if $p\equiv \pm 1\pmod{12}$ and -1 if $p\equiv \pm 5\pmod{12}$

If I CRT these it's wrong.

Answer 1

I'll assume that $p>3$. Then by quadratic reciprocity, $$ \Big(\frac{3}{p}\Big)=(-1)^{\frac{p-1}{2}\frac{3-1}{2}}\Big(\frac{p}{3}\Big)=(-1)^{\frac{p-1}{2}}\Big(\frac{p}{3}\Big)$$ Since $$ \Big(\frac{-1}{p}\Big)=(-1)^{\frac{p-1}{2}}$$ it follows that $$ \Big(\frac{-3}{p}\Big)=\Big(\frac{-1}{p}\Big)\Big(\frac{3}{p}\Big)=\Big(\frac{p}{3}\Big)$$ which is $1$ when $p\equiv 1$ (mod $3$) and is $-1$ when $p\equiv -1$ (mod $3$).

Answer 2

Here is a more elegant proof without quadratic reciprocity. Suppose, $-3$ is a quadratic residue modulo $p$, i.e. there exists an $x$ such that $$ x^2 \equiv -3 \pmod{p}. $$ Let, $x =2y+1$ (note that $\frac{x-1}{2}$ is a well-defined quantity modulo $p$). We have, $$ 4y^2+4y+4\equiv 0 \pmod{p}\implies p\mid (y^2+y+1)\implies p\mid y^3 - 1. $$ Let $d$ be the smallest positive integer for which $y^d\equiv 1\pmod{p}$. Clearly, $d\mid p-1$. Also, $d\mid 3$. We know that $d$ can't be $1$ (otherwise, $p$ would be $3$), hence, $d = 3$. Therefore, $$ 3|p-1 \implies p\equiv 1\pmod{3}. $$