Number of certain elements in a finite field having zero trace I have a question concerning certain elements having zero trace in a finite field extension and I do have the feeling that additive characters should play a role, but I am not sure how. I am stating the problem slightly more generally than what I am really interested in.
Let $q$ be prime power, let $\mathbb{F}_{q}$ the finite field of cardinality $q$ and let $n$ be a prime with $n$ dividing $q+1$. I denote with $\mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}:\mathbb{F}_{q^{2n}}\to \mathbb{F}_{q^2}$ the trace map.
I am interested in the cardinality of the set $$\{x\in \mathbb{F}_{q^{2n}}|\mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}(x)=0 \hbox{ and } x=y^n \textrm{ for some }y\in \mathbb{F}_{q^{2n}}\}.$$
For instance when $n=3$, I have a bit of evidence that that this set has cardinality $(q+1)^2(q^2-1)/3$.
 A: We have
$$ \# \{x\in \mathbb{F}_{q^{2n}}|\mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}(x)=0 \hbox{ and } x=y^n \textrm{ for some }y\in \mathbb{F}_{q^{2n}}\} = \frac{\#\{y\in \mathbb{F}_{q^{2n}}|\mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}(y^n)=0 \} +n-1}{ n}$$
since $\mathbb F_{q^n}$ contains all the $n$th roots of unity, so it suffices to calculate.
\begin{align*} 
\# \{y\in \mathbb{F}_{q^{2n}}|\mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}(y^n)=0 \} &= \frac{1}{q^2}  \sum_{\lambda \in \mathbb F_{q^2} } \sum_{ y \in \mathbb F_{q^{2n}} }\psi_2 (\lambda \mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}(y^n) ) \\ &=  \frac{1}{q^2}  \sum_{\lambda \in \mathbb F_{q^2} } \sum_{ y \in \mathbb F_{q^{2n}} }\psi_{2n}  (\lambda y^n) ) \\& = q^{2n-2} +  \frac{1}{q^2} \sum_{\lambda \in \mathbb F_{q^2}^\times  } \sum_{ y \in \mathbb F_{q^{2n}} }\psi_{2n}  (\lambda y^n) )  \\ &=q^{2n-2} + \frac{1}{q^2}  \sum_{\lambda \in \mathbb F_{q^2}^\times  } \sum_{ x \in \mathbb F_{q^{2n}} } \sum_{\substack{  \chi \colon \mathbb F_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \psi_{2n}  (\lambda x)  \chi(x) \\ &=q^{2n-2} + \frac{1}{q^2}  \sum_{\lambda \in \mathbb F_{q^2}^\times  } \sum_{ x \in \mathbb F_{q^{2n}} } \sum_{\substack{  \chi \colon \mathbb F_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \psi_{2n}  (x)  \chi(\lambda^{-1} x) \end{align*}
where $\psi_2 \colon \mathbb F_{q^2} \to \mathbb C^\times$ is a nondegenerate additive character and $\psi_{2n} = \psi_2 \circ \mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}$
Now since $\chi$ has order $n$ and the quotient group $\mathbb F_{q^{2n}}^\times / \mathbb F_{q^2}^\times $ has order $q^{2n-2} + q^{2n-4} + \dots + 1$ which is divisible by $n$, $\chi$ must factor through this quotient group so $\chi(\lambda^{-1})=1$, giving
$$\#\{y\in \mathbb{F}_{q^{2n}}|\mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}(y^n)=0 \} = q^{2n-2} + \frac{q^2-1}{q^2}  \sum_{\substack{  \chi \colon \mathbb F_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } }  \sum_{ x \in \mathbb F_{q^{2n}} }  \psi_{2n}  (x)  \chi( x) .$$
Here the inner sum $\sum_{ x \in \mathbb F_{q^{2n}} }  \psi_{2n}  (x)  \chi( x)$ vanishes for $\chi$ trivial and is a Gauss sum of absolute value exactly $q^{n}$ for $\chi$ trivial. This gives an estimate with main term and error term
$$ \left| \#\{y\in \mathbb{F}_{q^{2n}}|\mathrm{Tr}_{\mathbb{F}_{q^{2n}}/\mathbb{F}_{q^2}}(y^n)=0 \} -q^{2n-2} \right| \leq (n-1)  (q^2-1) q^{n-2} $$
