I need to solve this question:
Find the primes $p$ such that the equation: $x^{2} + 6x + 15 = 0 $ has a solution modulo $ p $.
My approach was: I checked for $p = 2$ and there is no solution.
Now if $p \neq 2 $ so the equation has a solution $\iff 6^{2} -4 \times 15 = -24$ is a square modulo $p$.
Now $-24 $ is a square modulo $p \iff \left(\frac{-24}{p}\right) = 1$ (Legendre symbol) $ \iff \left(\frac{-1}{p}\right) \left(\frac{2}{p}\right) \left(\frac{3}{p}\right) = 1.$
And now there are many cases to check and I'm not sure how to do it... For example - case 1:
$ \left(\frac{-1}{p}\right) =1 $ and $ \left(\frac{2}{p}\right) =1 $ and $ \left(\frac{3}{p}\right) = 1$ :
$ \left(\frac{-1}{p}\right) =1 \iff p \equiv 1 (4) $ ,
$ \left(\frac{2}{p}\right) =1 \iff p \equiv 1, -1 (8) $ ,
$ \left(\frac{3}{p}\right) = 1 \iff p \equiv 1, -1 (12) $
so how do I combine these results for case 1? and after that, do I really need now to check all the other cases - that 2 of the Legendre symbols are $(-1)$ and one is $1$ ?