# Elements of finite fields with many powers of trace zero

Let $p$ be an odd prime number, $n>1$ be an integer, and $\mathrm{tr}$ be the trace map of the field extension $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$. For which pair $(p,n)$ does there exists $x\in\mathrm{GF}(p^{2n})$ such that $x^{2(1+p^n)}=1$ and $\mathrm{tr}(x^{1+p^j})=0$ for $j=0,1,\dots,n-1$?

After computing the gcd of the corresponding $n+1$ polynomials over $\mathrm{GF}(p)$ for small $p$ and $n$, I believe that the answer should be if and only if $p$ divides $n$" and moreover, when $p$ divides $n$, all such $x$ would form a subgroup of $\mathrm{GF}(p^{2n})^\times$ of index $2(1+p^{n/p})$.

Update. If $(p,n)$ is such a pair, then denoting by $X$ the $(2n)\times(2n)$ matrix with $(i,j)$th entry $x^{p^{i+j-2}}$ we deduce from the conditions that $$X^2= \begin{pmatrix} 0&2naI_n\\ 2naI_n&0 \end{pmatrix},$$ where $a=x^{1+p^n}=\pm1$. This yields two necessary conditions:

(1). If $n$ is not divisible by $p$ then $n$ is odd.

Proof. Suppose on the contrary that $n$ is even. Then $$|X|^2= \begin{vmatrix} 0&2naI_n\\ 2naI_n&0 \end{vmatrix} =(-1)^n(2na)^{2n}=(2na)^{2n}$$ is a square in $\mathrm{GF}(p)$, which implies that $|X|\in\mathrm{GF}(p)$. However, changing columns shows that $-|X|=(-1)^{2n-1}|X|$ is the determinant of the matrix with $(i,j)$th entry $x^{p^{i+j-1}}$. Hence $-|X|=|X|^p$ and so $|X|\notin\mathrm{GF}(p)$, a contradiction.

(2). If $n$ is not divisible by $p$ then $x,x^p,\dots,x^{p^{2n-1}}$ is a normal basis of $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$.

Proof. $|X|^2=(-1)^n(2na)^{2n}\neq0$ implies that $|X|\neq0$ and so $x,x^p,\dots,x^{p^{2n-1}}$ are $\mathrm{GF}(p)$-linearly independent.