Solving a simple quadratic in modular arithmetic

Question

I've been trying to solve this congruence> $$x^2 \equiv 9\pmod{13}$$

I keep coming back to $x =3$ but it seems to be wrong, any help on what's the correct technique to find the answer?

Answer 1

One way to see this is solving $x^2=9$ over the integers yields $x = \pm 3$. Clearly, $3$ is one solution, and similarly, $-3 \equiv 10 \pmod{13}$ so $10$ is another solution. But if $3$ is a solution, so must be $3+13=16$, etc. Can you find a general form?

Answer 2

Because $13$ is a prime number. Then $\Bbb Z_{13}$ is a finite field. That means that
$x^2 \equiv 0 \pmod{13}$ has exactly two solutions $x \in \{3, -3\} = \{3, 10\}$