In a finite field $x^2 = a$ has at most two roots.

Question

Given that in a finite field $K$ the equation $x^2 = 1$ has zero or two solutions depending if $1 \neq -1$, is it true that $x^2 = a$ has at most two roots for a given $a \in K$?

Answer 1

That's right. If $a$ has some square root $b$, we can write $$ x^2=b^2\implies (x-b)(x+b)=0 $$ Recall that fields do not have zero divisors.

Answer 2

It follows from a more general statement: in any integral domain, a polynomial of degree $n$ has at most $n$ roots. Note that fields are, by definition, integral domains.

You can easily find a proof of this fact, for example, here. It relies on Bézout's theorem and lack of zero divisors.