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 $n$.


*

*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
$Y\subset X$.)
 A: First of all, thank you Gerry Myerson for bringing this problem to my attention at West Coast Number Theory 2019. 
The answer to this question 1 is No.
Edit on 12/26 : With a few more detailed analysis of the paper by Lam and Leung, it is possible to complete the proof, a crucial point is indicated by bold face italic.
We prove that under the assumptions of this problem, $|X|$ must be prime. 
Step 1 : Reduction
Let $|X|>2$ and $\zeta_N=\exp(2\pi i (1/N))$. 
The sequence $s_n=\sum_{x\in X} x^n$ satisfies a linear recurrence relation. By Skolem-Mahler-Lech theorem, there is an arithmetic progression $\{an+b\}_{n\geq 0}\subseteq \mathbb{N}$ with $a>0$, $b\in \mathbb{N}$ such that 
$$
s_{an+b}=\sum_{x\in X} x^{an+b} = 0 \ \textrm{for all} \ n\in\mathbb{N}.
$$
Step 2 : Vandermonde
Let $X=\{x_1,\ldots,x_k\}$. The result from Step 1 forms a Vandermonde system
$$
\begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_1^a & x_2^a & \cdots & x_k^a \\ x_1^{2a} & x_2^{2a} & \cdots & x_k^{2a} \\
\vdots& \ddots& \cdots& \vdots \\
x_1^{(k-1)a} & x_2^{(k-1)a} & \cdots & x_k^{(k-1)a}\end{pmatrix}\begin{pmatrix} x_1^b \\ x_2^b \\ \vdots \\ x_k^b \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}.
$$
Since $x_i\neq 0$, the Vandermonde matrix must be singular. This yields $x_i^a = x_j^a$ for some $i\neq j$. Without loss of generality, assume that $x_1^a=x_2^a$. Then we may rewrite the system as 
$$
\begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_2^a & x_3^a & \cdots & x_k^a \\ x_2^{2a} & x_3^{2a} & \cdots & x_k^{2a} \\
\vdots& \ddots& \cdots& \vdots \\
x_2^{(k-2)a} & x_3^{(k-2)a} & \cdots & x_k^{(k-2)a}\end{pmatrix}\begin{pmatrix} x_1^b+x_2^b \\ x_3^b \\ \vdots \\ x_k^b \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}.
$$
If $x_1^b+x_2^b=0$, it yields a shorter vanishing sums in the expression of $s_{an+b}=0$. So, we must have $x_1^b+x_2^b\neq 0$. Thus the above Vandermonde matrix is also singular, and we obtain another $i\neq j$ (both $\geq 2$) with $x_i^a=x_j^a$. Repeating this process, we obtain 
$$
x_1^a = x_2^a = \cdots = x_k^a. 
$$
Dividing by $x_1$, we may assume that all members of $X$ are roots of unity. 
Step 3 : Vanishing sums of roots of unity
We refer to the results of this paper: T. Y. Lam, K. H. Leung, 'On the vanishing sums of roots of unity'
Let $s_b=0$ from Step 1 be written as a vanishing sum of roots of unity. In this paper, the vanishing sums of roots of unity $\alpha_1+\cdots+\alpha_k=0$ is called minimal if no proper sums vanish. It should be noted here that repetition is allowed. By the inclusion-minimal assumption, $s_b=0$ should be a minimal vanishing sum of roots of unity. 
By Corollary 3.2 of the paper, we may assume (up to a rotation) all the roots of unity are $\zeta_N^s$ where $N=p_1p_2\cdots p_r$ is square free, $p_1<p_2<\cdots <p_r$ are primes. Let $G=\langle g\rangle$ be a cyclic group of order $N$  generated by $g$. 
Let $\varphi: \mathbb{Z}G \rightarrow \mathbb{Z}[\zeta_N]$ be defined by the $\mathbb{Z}$-linear group-ring homomorphism such that $\varphi(g)=\zeta_N$. Let $G=H\times P_r$ where $H=\langle g_{N/p_r} \rangle \simeq\mathbb{Z}/p_1\cdots p_{r-1}\mathbb{Z}$ and $P_r=\langle g_{p_r} \rangle\simeq\mathbb{Z}/ p_r \mathbb{Z}$. 
We write the vanishing sum $s_b=0$ as an element $v=\sum_{i\leq N-1} v_i g^i$, $v_i\in\mathbb{Z}$ of the group ring $\mathbb{Z}G$ such that $\varphi(v)=0$. Note that the coefficients of $v$ must be nonnegative. The element $v$ can also be written as
$$
v=h_0 + h_1 g_{p_r} + \cdots + h_{p_r-1} g_{p_r}^{p_r-1},
$$
with $h_i\in \mathbb{Z}H$. Here the coefficeients of $h_i$ are also nonnegative.  
We have $\varphi(h_i)\in \mathbb{Q}(\zeta_{N/p_r})$ for all $i\leq p_r-1$. By linear disjointness of $\mathbb{Q}(\zeta_{N/p_r})$ and $\mathbb{Q}(\zeta_{p_r})$, we must have
$$
\varphi(h_0)=\varphi(h_1)=\cdots=\varphi(h_{p_r-1}).
$$
Case 1 
If $\varphi(h_0)\neq 0$, this means that each $h_i=\sum_{j\leq N/p_r -1} h_{ij} g_{N/p_r}^j$, $h_{ij}\in\mathbb{Z}$, contains at least one positive coefficient. Let $h_{ij_i}$ be such positive coefficient of $h_i$. Then a sub-sum $$
h_{0j_0} g_{N/p_r}^{j_0} +h_{1j_1} g_{N/p_r}^{j_1} g_{p_r}  + \cdots +  h_{(N/p_r-1)j_{N/p_r -1}} g_{N/p_r}^{j_{N/p_r-1}}g_{p_r}^{p_r-1}
$$
has positive coeffients. 
Taking $N/p_r$-th powers of each group element and applying $\varphi$, we obtain a vanishing sub-sum of powers
$$
1+\zeta_{p_r}^{N/p_r} + \cdots + \zeta_{p_r}^{(N/p_r)(p_r-1)}=0.
$$
Case 2 
If $\varphi(h_0)=0$, then we may repeat the above argument with fewer prime factors, e. g. $N$ is replaced by $N/p_r$.  
Therefore, under the inclusion-minimality assumption, $s_b=0$ yields a minimal vanishing sum of roots of unity 
$$
1+\zeta_p+\cdots + \zeta_p^{p-1}
$$
for some prime $p$ up to a rotation.
