How to solve system: x_1+x_2+...+x_n=a

Question

How to solve this system:

$$\left\{\begin{matrix} x_1 & + &x_2 & + & \ldots & + & x_n &= & a \\ x^2_1& + &x^2_2 &+ & \ldots & + &x^2_n &= & a^2 \\ \vdots && \vdots && \vdots &&\vdots && \vdots\\ x^n_1 & +&x^n_2 &+ & \ldots & + &x^n_n &= & a^n \end{matrix}\right.$$

Thank you very much!

Answer 1

Using Newton's identities we can prove by induction that $e_1=a,e_k=0$ for $1<k\leq n$ (see the article for relevant notation). Indeed, if we know that for $1<k<l\leq n$, $e_k=0$, then $le_l=e_{l-1}p_1+\cdots\pm e_1p_{l-1}\mp p_l=\pm a\cdot a^{l-1}\mp a^l=0$ (all terms in $...$ evaluate to $0$).

By Viete's formulas, $x_1,...,x_n$ form a complete set of roots of the polynomial $x^n-e_1x^{n-1}+e_2x^{n-2}-\cdots\pm e_n=x^n-ax^{n-1}=(x-a)x^{n-1}$. Clearly the roots of this polynomial are $a$ and $n-1$ times $0$, hence all the solutions are indeed of the form $(0,...,0,a,0,...,0)$.

Answer 2

There are trivial solutions $e_i =(0,0, ...0,\underbrace{a}_i ,0,...,0).$ Maybe there are no more solutions.