Easiest system of equations to solve

Question

Suppose we need to solve the following system of equations:

$$ \left\{\begin{align*} &f_1(x_1) + f_1(x_2) + \ldots + f_1(x_k) = a_1,\\ &f_2(x_1) + f_2(x_2) + \ldots + f_2(x_k) = a_2,\\ &\vdots\\ &f_k(x_1) + f_k(x_2) + \ldots + f_k(x_k) = a_k,\\ &1 \le x_1 < x_2 < \ldots < x_k \le n\\ &x_1,x_2,\ldots,x_k\in \mathbb{N} \end{align*}\right. $$ However, the functions $f_i$ can be chosen freely before the system is formed.

Obviously, for $k = 1$ choosing the identity function $f_1(x)=x$ is the easiest. For $k = 2$ we can have $f_1$ as an identity and $f_2$ as a quadratic function, which would give a quadratic equation.

In general, we could have chosen $f_k(x)=x^k$, and then use some numeric methods to solve the resulting system, but I wonder if there's perhaps a better choice for the functions $f_i$ which would make solving the system easier?

With $f_k(x)=x^k$ it is straightforward to use Newton's identities to determine the $n^{th}$ degree polynomial whose roots are $x_1, x_2,\dots,x_n$. Given the roots are known to be integers, they can be then be found with the rational root theorem. — dxiv, Jun 16 '22 at 18:36

Easiest system of equations to solve

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