Describing the primes $p$ for which the Legendre symbol $(\frac{-6}{p})=1$ I would love your help with describing the primes $p$ for which the Legendre symbol $(\frac{-6}{p})=1$.
From the properties of the Legendre symbol I know that $(\frac{-6}{p})=(\frac{-1}{p})(\frac{2}{p})(\frac{3}{p})$,$(\frac{-1}{p})=(-1)^{\frac{p-1}{2}}$, and $(\frac{2}{p})=(-1)^{\frac{p^2-1}{8}}$. If I knew about a similar formula for $3$, it would help me, but I'm afraid there isn't one.
What should I do for solving this one?
Thanks!
 A: Actually, the main trick is that
$$ (-3 | p) = (p | 3)    $$ and depends on the fact that $$ 3 \equiv 3 \pmod 4.  $$ Quite a time saver. I use the horizontal typesetting of the symbol, which was  introduced by L. E. Dickson. The vertical style always makes me think of fractions.
So, why? If $p \equiv 1 \pmod 4,$ then
$$  (-3|p) = (-1|p ) \cdot (3 |p) = 1 \cdot (p|3) = (p|3).    $$
Switching to another letter, if prime $q \equiv 3 \pmod 4,$ then
$$  (-3|q) = (-1|q ) \cdot (3 |q) = -1 \cdot -(q|3) = (q|3).    $$ 
What does a value of 1 tell us? if $(-24 | p) = (-6|p)=1$ for a prime $p \neq 2,3,$ we get either an expression
$$ p = u^2 + 6 v^2, \; \mbox{as in} \; \; \{  7, 31, 73, 79, 97, 103, 127, \ldots \},    $$ all of which are $\equiv 1 \; \mbox{or} \; 7 \pmod {24},$ or
 $$ p = 2 x^2 + 3 y^2, \; \mbox{as in} \; \; \{ 5,11,29,53,59,83,101,107,131,149, \ldots \},    $$ all of which are $\equiv 5 \; \mbox{or} \; 11 \pmod {24}.$ As I said, the primes $2,3$ are to be considered separately. 
The same thing works with the Jacobi symbol, which is just a product of Legendre symbols. Just an example, $ (-35 | p) = (p | 35).$ Note the required $35 \equiv 3 \pmod 4.$
A: Hint: Use the quadratic reciprocity law to express $\left(\frac{3}{p}\right)$ in terms of $\left( \frac{p}{3}\right)$.
A: Continuing with Marlu's answer and comment:
(1) $\,p=1\pmod 4\Longrightarrow \left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)=1\Longleftrightarrow p=0,1\pmod 3$
2) $\,p=3 \pmod 4\Longrightarrow \left(\frac{3}{p}\right)=-\left(\frac{p}{3}\right)=1\Longleftrightarrow p=2\pmod 3$
