Equation involving the Jacobi symbol: $\left( \frac {-6} p \right) = 1$? I have to determine the values of $p \in \{0, \dots, 23 \}$ such that $\left( \frac {-6} p \right) = 1$.
I have that:
$$\left( \frac {-6} p \right) = \left( \frac 2 p \right) \left( \frac {-3} p \right)$$
and I know that $\left( \frac 2 p \right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left( \frac 2 p \right) = -1$ if $p \equiv 3,5 \pmod 8$, and also that $\left( \frac {-3} p \right) = 1$ if $p\equiv 1 \pmod 3$.
Using the Chinese Remainder Theorem, I find that $p\equiv 1,?,7,?\pmod{24}$.
Thanks in advance!

I saw this the other day as well. Call this a side note, the primes $1,7 \pmod {24}$ are all represented by $x^2 + 6 y^2.$ It is easy to see that if $p = x^2 + 6 y^2$ in integers, then $(-6|p) = 1.$ 
$$   1,      7,     31,     73,     79,     97,    103,    127,    151,    193,
    199,
 $$  
the primes $5,11 \pmod {24}$ are all represented by $2x^2 + 3 y^2.$ 
$$       2,      3,      5,     11,     29,     53,     59,     83,    101,    107,
    131,    149,    173,    179,    197,
$$
Of course, $2x^2 + 3 y^2$ also represents $2$ and $3,$ but those divide $6.$
Meanwhile, IF $(-6|p)= 1,$ it is easy enough to construct a (positive) binary quadratic form $f(x,y) = p x^2 + Bxy + C y^2$ of discriminant $-24,$ and this form reduces to one of the two indicated forms above; that process tells us how to represent $p.$ For part of that, prove if n - natural number divide number $34x^2-42xy+13y^2$ then n is sum of two square number
Why not: if $(-6|p)= 1,$ then $(-24|p)= 1,$ we have some integer solution to $\beta^2 \equiv -24 \pmod p.$ If the original $\beta$ we found was odd, replace it by $B= p - \beta$ so that $B$ is even, and $B^2 \equiv -24 \pmod {4p}.$ That is, $B^2 = -24 + 4pC.$ Well, $B^2 - 4pC = -24,$ and the positive binary quadratic form $\langle p, B, C \rangle$ has discriminant $-24.$ The notation $\langle p, B, C \rangle$ means the form $f(x,y) = p x^2 + Bxy + C y^2$
