Square root for Galois fields $GF(2^m)$ Can we define a function similar to square root for $G = GF(2^m)$ (Galois field with $2^m$ elements) as $\sqrt{x} = y$ if $y^2 = y \cdot y = x$ ? For which elements $x \in G : \exists y \in G : y^2 = x$ this function would be defined?
Can I approach this question like this:
If we can generate all elements of $G$ except $0$ from another element $a$ as $a^k : k = 1 \ldots (2^m-1)$, then any $x \neq 0$ can be expressed as $x=a^r$ for some $r$ and $y$ would be also $y = a^s$ for some $s$. $y^2 = x$ means that $$r = 2s \mod 2^m-1$$ $$r,s \in 0 \ldots 2^m-2$$
It looks like I can find $s$ for any $r$ to satisfy equation. That would mean that there is a "square root" for any element in $G$, right?
PS: I'm looking into options to analyze streams of data (bytes, 16 bit or 32 bit integers) as part of another computational task, therefore only specific Galois fields are interesting for me: $GF(2^m)$. Be prepared that I can be way off in the field theory, that was very long time since I touched it - but any comments are welcome!

Another way to look at it:
If we are working in a field of characteristic $2$, the derivative of $X^2-a$ is $0$.  This implies that the two roots of $X^2-a$ coincide, so square roots, when they exist, are unique.
In other words, the map $b\mapsto b^2$ is injective.  If our field is finite, this map must be surjective as well.
