Let $p>3$ be a prime, and $a,b \in \Bbb F_p$ such that $4a^3+27b^2 \neq 0$.
Can I find some $x \in \Bbb F_p$ such that $x^3+ax+b$ is a square mod $p$ without using Hasse's bound? I know that $\Bbb F_p$ has $(p+1)/2$ squares (including $0$), but I don't see why $x^3+ax+b$ should take at least one square value.