Let $p$ and $q$ be distinct odd primes for which $(2/p)$ and $(3/q)$ are both $1$. Prove that there exists a number $x$ such that $x^2 ≡ 2$ (mod $p$) and $x^2 ≡ 3$ (mod $q$).
This is my attempt to solve it:
$x^2=2$ (mod $p$) and $y^2=3$ (mod $q$)
$p \equiv 1$ or $7$ (mod $8$)
$q \equiv 1$ or $11$ (mod $12$)
I'm totally lost after this.