There is so much literature on the relation between the multiplicative structure of a finite field and elements having zero trace, that I am hoping that the following is known.


Let $q$ be a prime power, let $n$ be an odd prime number, let $\mathbb{F}\_{q^{2n}}$ be the field of cardinality $q^{2n}$ and let $\mathrm{Tr}:\mathbb{F}\_{q^{2n}}\to \mathbb{F}\_{q^2}$ be the trace map. Let $\mathcal{A}$ be the subgroup of $\mathbb{F}\_{q^{2n}}^\times$ having order $(q^n+1)/(q+1)$.


Is there an element $a\in \mathcal{A}$ and an element $y$ in the multiplicative group $\mathbb{F}\_{q^n}^\ast$ of the field $\mathbb{F}\_{q^n}$ with
$\mathrm{Tr}(y)\ne 0$ and $\mathrm{Tr}(ay)=0$?


It seems to me that $(n,q)=(3,2)$ is the only exception.



Zero trace elements in finite fields
I can show that exceptions occur at most for $n=3$. (Primality of $n$ is never used.)


Since $n$ is odd, $\mathbb F\_{q^{2n}} = \mathbb F\_{q^2} \otimes\_{\mathbb F\_q} \mathbb F\_{q^n}$. The trace map $\operatorname{Tr}:\mathbb F\_{q^{2n}} \to \mathbb F\_{q^2}$ is obtained by tensoring the identity map $\mathbb F\_{q^2} \to \mathbb F\_{q^2}$ with the trace map $\operatorname{tr} : \mathbb F\_{q^n} \to \mathbb F\_q$.


Thus, choosing an arbitrary basis of $\mathbb F\_{q^2}$, we can write any $a$ as a pair of elements $a\_1,a\_2 \in \mathbb F\_{q^n}$, and your condition that $y \in \mathbb F\_{q^n}$ satisfies $\operatorname{Tr}(y)\neq 0$ but $\operatorname{Tr}(ay)=0$ is equivalent to the condition that $\operatorname{tr} (y) \neq 0$ but $\operatorname{tr} (a\_1 y ) =\operatorname{tr}(a\_2 y)=0$. (We can ignore the condition that $y\neq 0$ as it is implied by the condition that $y$ has trace zero.)


Since the trace map of a product is a perfect $\mathbb F\_q$-linear pairing on $\mathbb F\_q^n$, such a $y$ exists unless $1$ is an $\mathbb F\_q$-linear combination of $a\_1$ and $a\_2$.


I will show there must exist a member of $\mathcal A$ that has this unusual property by bounding the number of members of $\mathcal A$ that do have this unusual property.


Note that every member of $\mathcal A$ is in the subgroup of order $q^n+1$, thus has norm to $\mathbb F\_{q^n}$ equal to $1$. This is a nonsingular quadratic equation in $a\_1,a\_2$. For each $\lambda\_1,\lambda\_2$ in $\mathbb F\_q$, not both zero, $\lambda\_1 a\_1 + \lambda\_2 a\_2 =1$ is a linear equation. There can be at most two solutions to a linear equation together with an nonsingular quadratic equation in two variables, since it gives a nontrivial quadratic equation in one variable.


Summing over possible choices of $\lambda\_1,\lambda\_2$, the number of members of $\mathcal A $ with this unusual property is at most $2 (q^2-1)$. So we can only have all members of $\mathcal A$ with this property if


$$ \frac{q^n+1}{q+1} \geq 2 (q^2-1)$$
i.e.


$$q^n+1 \geq 2 (q^2-1) (q+1).$$


For $n\geq 5$, the left side dominates the right side for any $q$.


