Let $\mathbb{F}_{q}$ be the finite field of order $q$, where $q$ is odd. Now, Let $S$ be the set of elements $x$ in $\mathbb{F}_{q^n}^*$, such that $x$ doesnot belong to any smaller subfield(containing $\mathbb{F}_{q}$) of $\mathbb{F}_{q^n}$. Let $S^2=\{\ x^2 | x\in S \}\ $. I want to find out $|S\cap S^2|$.
I have made some calculations and found out the answer, but I am not sure whether it is perfectly correct, or whether there is easier way to see this!
Here is my idea: $|S|=\sum_{d|n}\mu(d)q^{n/d}$. Here $\mu$ denotes the Mobius Function. Now I divide this in two case.
Case 1: Suppose $n$ is odd.
Then observe that $a\in S \implies a^2\in S$. This is because there is no intermediate extension of even order. Now, since exactly two element of $S$ gives the same squared element, we have $|S\cap S^2|=\frac{|S|}{2}= \frac{1}{2}\sum_{d|n}\mu(d)q^{n/d}$.
Case 2: Suppose $n$ is even
In this case: $a\in S \implies \text{either } a^2\in S \text{ or } a^2\in \mathbb{F}_{q^{n/2}}^{*}$. Now, there are exactly $\frac{q^{n/2}-1}{2}$ non-squares in $\mathbb{F}_{q^{n/2}}^{*}$. These non-squares are squares of elements in $S$. So, in this way there are $q^{n/2}-1$ elements in $S$, whose squares lie in $\mathbb{F}_{q^{n/2}}^{*}$. Hence there are $|S|- (q^{n/2}-1)$ number of elements in $S$ whose squares are in $S$. Hence $|S\cap S^2|=\frac{1}{2}[|S|-(q^{n/2}-1)]= \frac{1}{2}\Big[\sum_{d|n}\mu(d)q^{n/d} - (q^{n/2}-1)\Big]$.
This completes the solution. Please let me know whether this is a fine solution, or am I missing something! Let me know, if one can approach this problem any easier way.
Thank you!