Find the set of primes $p$ for which $-3$ is quadratic residue $\text{mod } p$.

I have started my solution like this:

$1= \left(\dfrac{-3}{p}\right) = \left(\dfrac{-1}{p}\right)\left(\dfrac{3}{p}\right) = (-1)^\frac{p-1}{2}\left(\dfrac{3}{p}\right)$

Using quadratic reciprocity $\left(\dfrac{-3}{p}\right)$ becomes $(-1)^\frac{p-1}{2}\left(\dfrac{3}{p}\right)$

So up to here I have $1 = (-1)^\frac{p-1}{2}\cdot (-1)^\frac{p-1}{2}\left(\dfrac{p}{3}\right) = (-1)^{p-1}\left(\dfrac{p}{3}\right)$

Where $\left(\dfrac{a}{b}\right)$ stands for the Legendre symbol. What is my next step? I can not seem to see how to break down $p/3$ further. My solution should be when $p\equiv 1 \pmod 3$ but I cant seem to get there.


1 Answer 1


$\left(\frac{-3}{p}\right)=(-1)^{p-1}\left(\frac{p}{3}\right)$, as you've found. But $(-1)^{p-1}=1$, since $p$ is odd.

So $\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$. $\ 0^2\equiv \color{#0bc}{0},\ 1^2\equiv \color{#0bc}{1},\ 2^2\equiv \color{#0bc}{1}$ mod $3$, so $\color{#0bc}{0}$, $\color{#0bc}{1}$ are all quadratic residues mod $3$.

$\left(\frac{\color{#0bc}{0}}{3}\right)=0,\ \left(\frac{\color{#0bc}{1}}{3}\right)=1,\ \left(\frac{2}{3}\right)=\left(\frac{-1}{3}\right)=-1$.

$\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)=\left\{\begin{array}{}1,&p\equiv 1\pmod{\!3}\\-1, &p\equiv -1\pmod{\!3}\\0, &p=3\end{array}\right\}\equiv p\pmod{\!3}$.


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