Let $X=\{x_{1}, \cdots , x_{n}\}$ be a set of $n$ positive integers and integer $i \ge 1$. Let’s say that the set $X$ is $i$-sum-avoiding if for any nonnegative integers $c_{1}, \cdots, c_{n}$ such that $\sum_{j=1}^{n}c_{j} = n$ and $(c_{1},\cdots, c_{n}) \ne (1,\cdots, 1)$, it holds that
$\displaystyle \sum_{j=1}^{n}c_{j}x_{j}^{i} \ne \displaystyle \sum_{j=1}^{n}x_{j}^{i}$
Let $f(n,i)$ be the minimum value for a given $i \ge 1$ such that there exists an $i$-sum-avoiding set $X$ consisting of $n$ positive integers at most $f(n,i)$. Does there exist a constant $k_{i}$ for every $i$ such that $\forall n \in \mathbb N$, it holds $f(n,i) \le n^{k_{i}}$? If it does, what is the minimum of such $k_{i}$ for every $i$?
Showing such a set would help solve hard problems in computer science given some space relaxations. It seems that the hardness of such problems is directly related to non-existence of such sets. I could only show such sets when $k=n$ that is $k$ is not a constant. My example for $X$ is $X = \{n^{1}, n^{2},\cdots, n^{n}\}$.
Conjecture$\underline{Conjecture}$: $k_{i} = \infty$ $\forall i \ge 1$.