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Suppose that $X$ is an inclusion-minimal finite set of non-zero complex numbers such that $\sum\limits_{x\in X}x^n=0\ $ for infinitely many integers $n$.

  1. Can the cardinality of $X$ be a composite number?

2. Can $X$ be something different from $\root^p\of c$ (for some $c\in\mathbb C$ and prime $p$)?

(Inclusion-minimal means that the number of $n\in\mathbb Z$ such that $\sum\limits_{x\in Y}x^n=0$ is finite for any proper subset $Y\subset X$.)

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    $\begingroup$ For question $2$: consider $X := \{1, \zeta_{2i}\}$ for any integer $i$. Then for any $i|n, 2i\not|n$, we have $\sum_{x \in X} x^n = 1 + (-1) = 0$. $\endgroup$ – user44191 Feb 3 at 18:33
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    $\begingroup$ When you say "inclusion-minimal", what do you mean precisely? Do you mean that any subset doesn't satisfy the same equation for the same $n$? That for any subset, the corresponding set of $n$ must be finite? $\endgroup$ – user44191 Feb 3 at 18:40
  • $\begingroup$ Thanks! Actually, we can take even $\{1,i\}$ (where $i^2=-1$). I have edited the question. $\endgroup$ – Anton Klyachko Feb 3 at 18:49
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    $\begingroup$ @IlyaBogdanov: Doesn't work: $X=\{\zeta_7,\zeta_{15},\zeta_{15}^{-1}\}$, where $\zeta_i$ is a primitive $i$-th root of unity. Then $\sum_{x\in X}x^n=0$ whenever $n\equiv35\pmod{7\cdot15}$, so your example isn't inclusion-minimal. $\endgroup$ – Peter Mueller Feb 4 at 15:00
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    $\begingroup$ @AntonKlyachko That set doesn't work; $\{1, e^{i\pi k/17}\}$ satisfies the equation for $n \equiv 17 (\mod 34)$. $\endgroup$ – user44191 Feb 15 at 1:24

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