0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$\left(\frac{-3}{p}\right)=(-1)^{p-1}\left(\frac{p}{3}\right)$, 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 $3$.

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.