$A\in \mathbb{F}_q$, then there exists $n\in \mathbb{Z}_{\geq 0}$ such that $A \in \mathbb{F}_{q^n}$ a perfect square.

Question

I was wondering whether the statement in my title holds. I think it does, but I am not sure. I have managed to prove it for the case $A = -3$, but not in the general case. Any ideas or tips?

Answer 1

You are asking for the solvability of $X^2-A = 0$ over a finite extension of $\mathbb{F}_{q}$, which are of the form $\mathbb{F}_{q^n}$.

By adjoining all the roots of $X^2-A = 0$ (in an algebraic closure of $\mathbb{F}_q$) to $\mathbb{F}_q$, you get the desired finite extension. In fact, for $A$ such that $X^2-A = 0$ is irreducible, you may choose any even $n$.