TITLE: Given that $n > 3$ and $z$ is a Gaussian integer, when can $z^n \pm z$ be a rational integer?
QUESTION [7 upvotes]: I came across the following conjecture. If you have any thoughts on how to approach it, let me know.
Conjecture. For any integer $n > 3$ and any Gaussian integer $z$ that is not a unit, if $z^n - z$ or $z^n + z$ is a rational integer, then $z$ is a rational integer.

REPLY [2 votes]: The question can be rephrased: Find the integers $n > 3$ for which there exist Gaussian integers $z$ such that
$z^n \pm z = \overline{z}^n \pm \overline{z}$. Rewriting the equation as $(z^n -\overline{z}^n)/(z-\overline{z}) = \mp 1$, one recognizes an instance of the problem of finding the  terms of a Lucas sequence with no  primitive divisors. The complete answer to an even more general problem (Lehmer sequences instead of Lucas sequences) was given by Bilu, Hanrot, Voutier  J. Reine Angew. Math.
539, 75-122 (2001)  and Abouzaid  J. Théor. Nombres Bordx.
18, No. 2, 299--313 (2006) .  To settle the conjecture, it suffices to  look up the  tables in these papers .