Square roots in $GF(2^k)$

Question

In my lecture slides it is stated that :

In $GF(2^k)$ for $k > 1$, $a^{2^{k - 1}}$ is the unique square root of $a$.

I can't see why.

So, if $a^{2^k - 1}$ is a sq. root of $a$ then it must be $a^{2^k} = a$. How does this hold?

Answer 1

The multiplicative group of $GF(2^k)$ has order $n=2^k-1$.

Therefore, $a^n=1$ implies $(a^{(n+1)/2})^2=a^{n+1}=a$, and so $b=a^{2^{k-1}}$ is a square root of $a$.

This square root is unique because $x^2=a=b^2$ implies $0=x^2-b^2=(x-b)^2$, since $GF(2^k)$ has characteristic $2$.