elementary number theory - Modular Arithmetic & Congruences

Show that if $p$ is an odd prime and $a \in \mathbb Z$ such that $p$ doesn't divide $a$ then $x^2\equiv a(mod p)$ has no solutions or exactly 2 incongruent solutions.

The only theorem that I thought could help was:

Let $a, b$ and $m$ be integers with $m > 0$ and $gcd(a,m)=d$. If $d$ doesn't divides $b$ then $ax\equiv b(mod p)$ has no solutions. If $d$ divides $b$ then $ax\equiv b(mod p)$ has exactly $d$ incongruent solutions modulo $m$.

But I feel that this would be an invalid theorem for this proof since $ax$ and $x^2$ are of different degrees.

Thoughts? Other methods to approach this?