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The math overflow post asks whether the equation $1+z^p+z^q=z^n$ can have multiple complex roots where $p<q<n$ (On the irreducibility of certain trinomials and quadrinomials).

Q. Let us consider the polynomial $\mathbf{P}(z)=1+z^p+z^q+z^r-z^n$ where $p,q,r$ and $n$ are natural numbers with $1<p<q<r<n$. Has $\mathbf{P}$ any multiple complex root?

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    $\begingroup$ Why does this have a functional analysis tag? $\endgroup$
    – KConrad
    May 25, 2023 at 19:13
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    $\begingroup$ I found no counter-example up to degree $n=50$ (even allowing $p=1$). There are counter-examples for sums of $6$ monomials, like $1+z+z^2+z^3+z^5-z^7$ which has $z=-1$ as double root. $\endgroup$
    – François Brunault
    May 25, 2023 at 21:10

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