If $p$ mod $4=1$, how many elements of order $4$ are in $GF(p^n)^*$

Question

If $p$ mod $4 = 1$ then how many elements of order $4$ are in $GF(p^n)^*$

I know it has something to do with $GF(p^n)^*$ being isomorphic to $Z_4$, thus having $2$ elements of order $4$, but I am not sure the steps in between, any help would be appreciated.

Answer 1

Since we're in a field, the equation $x^4=1$ has at most $4$ solutions.

$-1$ and $1$ are two solutions but they don't have order $4$.

When $p \equiv 1 \bmod 4$, we have two other solutions: $\pm \left(\dfrac{p-1}{2}\right)!$ . $\quad$(*)

These are the two elements of order $4$ in $GF(p^n)$, and they live in the prime field $GF(p)$.

(*) This follows from Wilson's theorem.

Answer 2

$\newcommand{\GF}{\mathrm{GF}}$You should know that the multiplicative group $\GF(p^{n})^{*}$ is cyclic, hence it has a unique subgroup of order $m$ for each divisor $m$ of its order $p^{n} -1$, and each of these subgroups is cyclic.

Now you know that $4$ divides $p-1$, and thus $m = 4$ divides $p^{n} - 1 = (p-1) (p^{n-1} + p^{n-2} + \dots + p + 1)$.

So $\GF(p^{n})^{*}$ has exactly one subgroup of order $4$, which is cyclic, and thus has exactly two elements of order $4$.