Consider F := GF (q)$F := GF(q)$ where q = p^e and E := GF$q = p^e$ and (q^2)$E := GF(q^2)$ where w is a primitive element of E$E$.
Fix theta := w^(q - 2)$\theta := w^{q - 2}$.
Starting point: can I always write 1 + theta$1 + \theta$ as a power of theta$\theta$?
If Gcd (q^2 - 1$Gcd (q^2 - 1, q - 2) = 1$, q - 2) = 1, theta$\theta$ is a primitive element for E$E$ and the answer is yes.
So we can restrict to the case q mod 3 = 2$q \mod 3 = 2$ where all cubes of elements of E$E$ are powers of theta of $\theta$.
Is 1 + theta$1 + \theta$ a cube in E$E$? No, when e$e$ is odd; otherwise yes.
So in the remaining odd case we want to replace the role of 1 + theta$1 + \theta$ by 1 + theta^j $1 + \theta^j$ for some j some $j$, leading to the following conjecture.
Conjecture: Consider q = p^e$q = p^e$ where p = 2 mod 3$p = 2 \mod 3$ and e$e$ is odd. There exists j in {1, ..., q^2$j$ in - 1}$\{1, \dots, q^2 - 1\}$ such that Gcd (q^2 - 1, j) = 1$Gcd (q^2 - 1, j) = 1$ and 1 + theta^j$1 + \theta^j$ is a cube in E$E$.
Computational evidence in support of this conjecture is strong. While it does not hold for q = 5$q = 5$ and q = 8$q = 8$, such j$j$ exists for all relevant q$q$ in the range [11..10^8]$[11\dots 10^8]$. Always j$j$ is at most q - 1$q - 1$.
The context for the query is the study of presentations for the classical groups SU(3, q) \leq GL(d, q^2)$SU(3, q) \leq GL(d, q^2)$ where a generator has action described by theta$\theta$.
Pointers towards a proof would be much appreciated.
