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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!

hardmath
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Riju
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    When you say "smaller subfield", do you mean "smaller field extension of $\mathbb{F}_q$" or literally "smaller subfield"? Because that makes a difference. – darij grinberg Mar 27 '19 at 17:57
  • no literally smaller subfield!! Will smaller subfield extension make sense in the question? Because not all finite fields are contained in $\mathbb{F}{q^n}$. $\mathbb{F}{q^d}$ is contained in $\mathbb{F}_{q^n}$ iff $d\mid n$. – Riju Mar 28 '19 at 06:34
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    If you mean literally "smaller subfield", then your computation of $\left|S\right|$ doesn't work already (because your Möbius inversion only accounts for the subfields that contain $\mathbb{F}_q$). – darij grinberg Mar 28 '19 at 06:53
  • Ok I know sww what you are saying! I am taking subfields containing $\mathbb{F}_{q}$. – Riju Mar 28 '19 at 07:21
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    Care to edit your question to correct it? – darij grinberg Mar 29 '19 at 03:51
  • I have edited it now!! – Riju Mar 29 '19 at 15:33
  • @grinberg Where does the solution go wrong? Can you tell me? – Riju Mar 29 '19 at 15:35
  • Some details on "$a\in S \implies \text{either } a^2\in S \text{ or } a^2\in \mathbb{F}_{q^{n/2}}^{*}$" would be helpful. It's not something I immediately see in a hurry. – darij grinberg Mar 29 '19 at 16:06

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