Quite surprisingly the following question appears while studying the billiard dynamics.

Assume we have $2n$ real numbers: $ x_1, x_2,..., x_{2n}$.

Assume also that $S_k=0$ for any odd positive integer $k$, where $S_k$ is a sum of $k$-th powers of these $2n$ numbers.

Conjecture: these $2n$ numbers can be split on $n$ pairs of type $(x, -x)$.

  • $\begingroup$ By "a sum of $k$-th powers" I assume you mean "the sum of the $k$-th powers". In other words, you want to add together the $k$-th powers of all of them. $\endgroup$ Commented Jun 13 at 15:36
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    $\begingroup$ @Dave, yes,exactly $\endgroup$ Commented Jun 13 at 19:15

2 Answers 2


The conjecture is true. Without loss of generality, $$x_1\geq\dotsb\geq x_n.$$

  • If $x_1+x_n>0$, then $\lim S_k/x_1^k=|\{i:x_i=x_1\}|$, so $S_k>0$ for $k$ large and odd.
  • If $x_1+x_n<0$, then $\lim S_k/x_n^k=|\{i:x_i=x_n\}|$, so $S_k<0$ for $k$ large and odd.

Hence $x_1+x_n=0$, and by omitting $x_1$ and $x_n$, we reduce the problem to $2n-2$ numbers.

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    $\begingroup$ That is a simple and nice idea, thanks! I was thinking in similar lines, some small refinement would even generalize for a field C $\endgroup$ Commented Jun 13 at 19:11

This is true over any field of characteristic zero, or characteristic larger than $2n$. It has nothing to do with the field being the real numbers. Rather, it has to do with symmetric functions.

The point is that in any characteristic, if you know the elementary symmetric functions $e_1,\dots,e_{2n}$ of the $x_i$ then you know the unordered set $\{x_1,\dots,x_{2n}\}$ (with multiplicities - i.e., you know the list up to permutations). The power sums $p_1,\dots,p_{2n}$ are expressible in terms of $e_1,\dots,e_{2n}$ in any characteristic, but provided the numbers $1,\dots,2n$ are invertible (i.e., the characteristic is zero or larger than $2n$) then $e_1,\dots,e_{2n}$ are expressible in terms of $p_1,\dots,p_{2n}$, and so if you know these power sums then you know $\{x_1,\dots,x_{2n}\}$. What goes wrong in small characteristic is that for example we have $p_2=e_1^2-2e_2$, so you can't recover $e_2$ from $p_2$ in characteristic two. Recovering the elementary symmetric functions from the power sums is via Newton's identities http://en.wikipedia.org/wiki/Newton's_identities.

Now suppose that $p_k=0$ for $k$ odd. Then $p_k(x_1,x_2\dots,x_{2n})=p_k(-x_1,-x_2,\dots,-x_{2n})$ for all $1\leqslant k\leqslant 2n$ and so $\{x_1,x_2\dots,x_{2n}\}=\{-x_1,-x_2,\dots,-x_{2n}\}$. So each $-x_i$ is some $x_j$, and an even number of them are equal to zero, which implies that you can pair them up this way.


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