Existence of certain cubes in finite fields

Question

Consider $F := GF(q)$ where $q = p^e$ and $E := GF(q^2)$ where w is a primitive element of $E$.
Fix $\theta := w^{q - 2}$.

Starting point: can I always write $1 + \theta$ as a power of $\theta$?

If $Gcd (q^2 - 1, q - 2) = 1$, $\theta$ is a primitive element for $E$ and the answer is yes.

So we can restrict to the case $q \mod 3 = 2$ where all cubes of elements of $E$ are powers of $\theta$.

Is $1 + \theta$ a cube in $E$? No, when $e$ is odd; otherwise yes.

So in the remaining odd case we want to replace the role of $1 + \theta$ by $1 + \theta^j$ for some $j$, leading to the following conjecture.

Conjecture: Consider $q = p^e$ where $p = 2 \mod 3$ and $e$ is odd. There exists $j$ in $\{1, \dots, q^2 - 1\}$ such that $Gcd (q^2 - 1, j) = 1$ and $1 + \theta^j$ is a cube in $E$.

Computational evidence in support of this conjecture is strong. While it does not hold for $q = 5$ and $q = 8$, such $j$ exists for all relevant $q$ in the range $[11\dots 10^8]$. Always $j$ is at most $q - 1$.

The context for the query is the study of presentations for the classical groups $SU(3, q) \leq GL(d, q^2)$ where a generator has action described by $\theta$.

Pointers towards a proof would be much appreciated.

Answer 1

I think what you want is a solution of $1+x^3=y^3$ such that $x$ is a primitive element of $E$ by writing $x^3=\theta^j$. This is known to exist for $q$ large and is not very difficult to extract a bound. One uses the Weil bound to count the number of solutions of $1+x^3=y^3$ as well as the number of solutions of $1+x^{3d}=y^3$, for divisors $d$ of $q^2-1$ (and for large $d$ replace the Weil bound by a direct estimate), then apply inclusion exclusion to get the desired result. The technique goes back to the work of Bilharz (Math Ann 1937) on the function field analogue of Artin's conjecture.