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
$Y\subset X$.)



Perfectly balanced sets of complex numbers
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](https://en.wikipedia.org/wiki/Skolem%E2%80%93Mahler%E2%80%93Lech_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'](https://www.sciencedirect.com/science/article/pii/S0021869399980894?via%3Dihub)


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.


