To find the number of Elements of order 4 in Galois Field $GF(p^n)^*$, 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 $4$.
It comes down to whether $4|p^n-1$.
Say $p=7$.  Then since $4|p^2-1$, and the multiplicative group of a finite field is cyclic, we get $\varphi(4)=2$ for $\Bbb F_{49}$.
But for $\Bbb F_7$ there are none, since $4\not|7-1=6$

Now, when $n=2k$ is even, we have $p^n-1=p^{2k}-1=(p^k+1)(p^k-1)$, which is divisible by $4$ as the product of two even numbers.
On the other hand, when $n=2k+1$ is odd, we have $p^n=(4l+3)^n\cong (4l+3)^{2k}(4l+3)\cong 4l+3\cong3\bmod4$, by Euler's theorem.  So $p^n-1\cong2\bmod4$. Thus $4\not|p^n-1$.