# How to show that power sums determines ${x_1,…,x_n}$

How can I prove that for real numbers $x_k$ and $y_k$,

$x_1^k+x_2^k+ \dots + x_n^k=y_1^k+y_2^k+ \dots + y_n^k$ $(k=1,2,\dots,n)$,

where $x_1\le x_2 \le \dots \le x_n$ and $y_1\le y_2 \le \dots \le y_n$

implies $x_k=y_k (k=1,2,\dots,n)$?

I considered the polynomial with roots $x_k$, $(t-x_1)(t-x_2)\dots(t-x_n)$and its coefficients, but could not proceed.

• Are the numbers supposed to be nonnegative? Otherwise $1^2=(-1)^2$ is a counterexample. – Jose Brox Jun 19 '18 at 14:17
• If you have a single number, the only possible value for $k$ is $1$. – José Carlos Santos Jun 19 '18 at 14:20
• @JoséCarlosSantos Oh, I see now that the equation is meant for all $k$ up to $n$. – Jose Brox Jun 19 '18 at 14:26

It follows from Newton's identities and from your hypothesis that, if $\sigma_1,\ldots,\sigma_n$ are the elementary symmetric polynomials, then$$(\forall k\in\{1,\ldots,n\}):\sigma_k(y_1,\ldots,x_n)=\sigma_k(y_1,\ldots,y_n).$$Therefore$$(X-x_1)\ldots(X-x_n)=(X-y_1)\ldots(X-y_n),$$and this means that $\{x_1,\ldots,x_n\}=\{y_1,\ldots,y_n\}$.