If we consider $p$ be a prime number such that $p \pmod4 =3.$ How many elements of order $4$ are in $GF(p^n)^*?$
If modulo were $p \pmod 4 = 1.$ Then I could easily solve the question. Since $GF(p^n)^*$ is isomorphic to $Z_{p^{n}-1}$ which is cyclic. And $4$ divides $p-1$ and can easily show that $4$ divides $p^n-1.$
And using the Theorem: If $d$ is the divisor of $n$, then number of elements of order $d$ in a cyclic group of order $n$ is $\phi(d)$.
So I can say that there are $2$ elements of order $4.$
But above question please provide your hint.
The multiplicative group of the field is cyclic of order $p^n-1$. So since $p\equiv 3 \mod 4$, if $n$ is odd then it contains no subgroups of order $4$ by Lagrange and if $n$ is even it contains one subgroup of order $4$ hence $2$ elements of order $4$.