Can this system of equations about Newton's formula have concrete result?

Question

Try to solve this system of equations:

$$ S_1=x_1+\dots+x_n=a;\\ S_2=x_1^2+\dots+x_n^2=a;\\ {}\cdots\\ S_n=x_1^n+\dots+x_n^n=a; $$ And find the value of $S_{n+1}=x_1^{n+1}+\dots+x_n^{n+1},a\in\mathbb{C}.$

Now, by using Newton's identities I have find that $$\sigma_1=\sum x_i= a,\quad \sigma_2=\sum _{1\le j<k\le n}x_{j}x_{k}=\frac{a(a-1)}{2},\dots,\\\sigma_n=x_1x_2\dots x_n=\frac{(a-n+1)(a-n+2)\dots21}{n!}$$ So $x_1,x_2,\dots,x_n$ satisfy this equation:$$x^n-\sigma_1x^{n-1}+\sigma_2x^{n-1}+\dots+(-1)^n\sigma_n=0$$ But I can't figure out what each $x_i$ is, so can I find explicit solutions? I'm able to find $S_{n+1}$ for the second part of the question. The only difficulty is to find the explicit of each $x_i(i=1,2 ,\dots , n).$

Answer 1

Consider $a$ as a transcendental over $\mathbb C$. Then each $x_i$ generates an extension of $\mathbb C(a)$ of degree $n$ whose Galois closure has Galois group the symmetric group $S_n$. Thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.

Answer 2

Hilarious little detail: the OP assumes that $a\in\mathbb{C}$.

One may wonder if the system is solvable by radicals; then it is easier to assume that $a\in\mathbb{Q}$.

Always there exists a unique -up to order- solution $(x_i)_i$ . Let $p$ be the required polynomial of degree $n$ whose root are the $x_i$'s.

$\textbf{Question.}$ Is there a rational $a$ such that $p$ is irreducible over $\mathbb{Q}$ and solvable by radicals?

The answer is not obvious.

For example, assume that $n=5$. Then, if $a$ is an integer in $[[-10^5,10^5]]$, then $galois(p)=S_5$ except when $a\in [[-1,6]]$, where $p$ is reducible.