<s>[INST] 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!
 [/INST] We have $(-6/p)=1$ if (i) $(2/p)=1$ and $(3/p)=1$ or (ii) $(2/p)=-1$ and $(3/p)=-1$.
You seem to have taken care of (i). We get the solutions $p\equiv 1\pmod{24}$ and $p\equiv 7\pmod{24}$.
For (ii) we want a) $p\equiv 3\pmod{8}$ and $p\equiv 2\pmod{3}$ or b) $p\equiv 5\pmod{8}$ and $p\equiv 2\pmod{3}$.
a) has the solution $p\equiv 11\pmod{24}$ and b) has the solution $p\equiv 5\pmod{24}$.
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