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Can the equation $1+z+z^2=z^n$ for natural $n$ have multiple complex roots $z$?
The question is stated in the title of this post.
It is easy to see that, if $z$ is a multiple root of $p_n(z):=1+z+z^2-z^n$, then $(n-2)z^2+(n-1)z+n=0$, so that we can successively express $z^2,\dots,...