# Perfectly balanced sets of complex numbers

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 $$Y\subset X$$.)

• For question $2$: consider $X := \{1, \zeta_{2i}\}$ for any integer $i$. Then for any $i|n, 2i\not|n$, we have $\sum_{x \in X} x^n = 1 + (-1) = 0$. – user44191 Feb 3 at 18:33
• 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? – user44191 Feb 3 at 18:40
• Thanks! Actually, we can take even $\{1,i\}$ (where $i^2=-1$). I have edited the question. – Anton Klyachko Feb 3 at 18:49
• @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. – Peter Mueller Feb 4 at 15:00
• @AntonKlyachko That set doesn't work; $\{1, e^{i\pi k/17}\}$ satisfies the equation for $n \equiv 17 (\mod 34)$. – user44191 Feb 15 at 1:24