-3 is quadratic residue if and only if $p \equiv 1,7 \pmod {12}$

Question

I have to prove that -3 is quadratic residue if and only if $$p \equiv 1,7 \pmod {12}$$ I know one method (with symbol Legendre'a) but I don't get. If someone can explain me I will be happy or give another method.

Answer 1

The Legendre symbol is the easiest way to do it... Just use the properties: $$ \left( \frac{-3}{p} \right)=\left( \frac{3}{p} \right)\left( \frac{-1}{p} \right)$$

Now $\left( \frac{-1}{p} \right)=(-1)^{(p-1)/2}$ which can easily be evaluated.

Next for $\left( \frac{3}{p} \right)$ you could use the formula given in the wikipedia article or the quadratic reciprocity law $$ \left( \frac{p}{q} \right)\left( \frac{q}{p} \right)=(-1)^{(p-1)(q-1)/4}$$ which for $p=3$ gives $$ \left( \frac{p}{3} \right)\left( \frac{3}{p} \right)=(-1)^{(p-1)/2}$$

Note that $\left( \frac{p}{3} \right)$ is easily computable since a number is quadratic residue modulo $3$ if it has residue $0$ or $1$ modulo $3$.

Now the rest it's just casework modulo $12$.

Answer 2

We have

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

so $\;\left(\frac{-3}p\right)=1\iff\;$ either both $\;-1,3\;$ are quadratic residues modulo $\;p\;$ or both aren't.

If they both are: then

$$p=1\pmod 4\implies p=1,5,9\pmod{12}$$

but it also must be $\;p=1\pmod 3\; $ ( we're assuming $\;p>3\;$) , so the only options left is $\;p=1\pmod 3\;$

If the both aren't : then

$$p=3\mod 4\;\;\text{and also}\;\;p=1\pmod 3 -- \text{because}\;\;\left(\frac3p\right)=-\left(\frac p3\right)\;\;\text{by QR}\;--\implies$$

$$\implies p=7\pmod{12}$$