An element of a $a\in \mathbb{F}_{q^2}$ such that $a^2 \in \mathbb{F}_q$.

Question

Suppose that $2\nmid q$ and $a\in \mathbb{F}_{q^2}$ (more precisely: $a \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$) such that $a^2 \in \mathbb{F}_q$. Why is it true that actually $a = -a^q$?

Answer 1

Since $a^2 \in \mathbb{F}_q$, we have $a^{2q} = a^2$. This means $(a^{q}-a)(a^q+a)=0$. The first bracket is nonzero since $a \not\in \mathbb{F}_q$, hence the result.