First time calculating this and I apologize if I result a little bit confused; so I want to find p primes odds for Legendre symbol $(-3|p)=+1$ or $(-3|p)=-1$.
I know from Legendre's original definition that $(-3)^{{p-1} \over {2}}=\pm 1$ and that, to be this Legendre symbol valid, must be $-3\not |p$, so $p\not=3$ right?
If $(-3|p)=+1$ so $-3$ must be a quadratic residue $\mod p$ (in Gauss notation $-3Rp$).
Is this equivalent to find $p$ : $q$ be a square and $q \equiv -3 \mod p$, trying $q = 1,4,9,16,15…$ and solving by trying p?
Best Answer
If you work out that $\left(\frac{-1}{p}\right)=1$ when $p\equiv 1\pmod{4}$ and $\left(\frac{3}{p}\right) =1$ when $p\equiv \pm 1 \pmod{12},$ then the multiplicativity gives you that $\left(\frac{-3}{p}\right)=1$ when $p\equiv 1 \pmod{12}.$ One continues doing cases. If $p\equiv -1 \pmod{12}$ then $p\equiv 3 \pmod{4}$ so one has $\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = (-1)(-1)=1.$ Etc.
To work out $\left(\frac{3}{p}\right)$ we use quadratic reciprocity. If $p$ is prime and not $3$, then it is congruent to one of $1, 5, 7, 11 \pmod{12}$. Say $p\equiv 5 \pmod{12}$ then $\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right) =\left(\frac{2}{3}\right) = -1.$ Etc.