It is proved [here](https://mathoverflow.net/questions/447134/can-the-equation-1zz2-zn-have-multiple-complex-roots) that the equation $1+z+z^2=z^n$ have no multiple complex roots.



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> Q. Let us consider the equation $1+z+z^q=z^n$ where $q$ and $n$ are natural numbers with $1<q<n$. Any characterization for the pair $(q,n)$ such that the equation $1+z+z^q=z^n$ have no multiple complex roots?
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Can the equation $1+z+z^q=z^n$ have multiple complex roots $z$?
The roots are always simple, this follows without further reasoning from the old paper [On the irreducibility of certain trinomials and quadrinomials](https://doi.org/10.7146/math.scand.a-10593) by Ljunggren. Set $f(z)=z^n-1-z^p-z^q$. Then Ljunggren proves (Theorem 1) that $f(z)=Q(z)g(z)$, where the roots of $Q$ are roots of unity (so $Q=1$ if there are no such roots), and where $g(z)\in\mathbb Q[z]$ is irreducible none of its roots is a root of unity. As an irreducible polynomial, $g$ has no multiple roots. Furthermore (Theorem 2), Ljunggren shows that $Q$ has simple roots either, which settles the question.


Remark: In case Ljunggren's paper is not accessible, [this review at zbMATH](https://zbmath.org/scans/095/013.gif) contains the full statement of Ljunggren's results needed here.


