I am working on a mathematical statistics problem. Here is an intermediate step of my proof, but I am stuck now. My question is if we have $x_1, x_2, x_3, y_1, y_2, y_3$ and all values are positive, the system of nonlinear equations is

\begin{align*} \begin{cases} x_1 + x_2 + x_3 = y_1 + y_2 + y_3 \\ x_1^2 + x_2^2 + x_3^2 = y_1^2 + y_2^2 + y_3^2 \\ x_1^3 + x_2^3+ x_3^3 = y_1^3 + y_2^3 + y_3^3 . \end{cases} \end{align*}

I suspect that the solution should be $x_1 = y_1, x_2 = y_2, x_3 = y_3$ (or other permutation). But I can't find a way to justify my guess. In fact, I want to show that in general if $k=1,2, \cdots, n$, \begin{align*} x_1^k + x_2^k + \cdots + x_n^k = y_1^k + y_2^k + \cdots + y_n^k, \end{align*} then the solution is (my guess) $x_1 = y_1, x_2 = y_2, \cdots ,x_n = y_n$ (or other permutation). Thanks in advance!

  • $\begingroup$ Obviously, setting the $y$ values to any permutation of the $x$ values is a solution. So am I correct in guessing that what you really mean is you want to show that these are only solutions? $\endgroup$ – Paul Sinclair Jan 27 '17 at 0:05
  • $\begingroup$ Yes. I want to know whether they are the only solutions to this system. $\endgroup$ – jwyao Jan 28 '17 at 5:27

Hint: show, using Newton's relations, that the elementary symmetric polynomials are equal for $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)\,$. Therefore, each triplet is the set of roots to the same $3^{rd}$ degree equation, so they must be permutations of each other.

The argument generalizes to $n \gt 3$ in a straightforward way.

  • 1
    $\begingroup$ Very clear and concise explanation. Thank you so much! $\endgroup$ – jwyao Jan 30 '17 at 0:49

If I remember correctly, given a set of $n$ distinct real numbers $\mathcal Y :=\{y_1, y_2, \dots, y_n\}$, the following system of $n$ polynomial equations

$$\begin{array}{rl} x_1 + x_2 + \dots + x_n &= c_1\\ x_1^2 + x_2^2 + \dots + x_n^2 &= c_2\\ \vdots \\ x_1^n + x_2^n + \dots + x_n^n &= c_n\end{array}$$


$$c_k := y_1^k + y_2^k + \dots + y_n^k$$

has $n!$ solutions, namely, all $n!$ permutations of the $n$ distinct elements of $\mathcal Y$. One can use algebraic geometry to prove this. I vaguely recall this being related to moment problems.


Let $\mathcal Y := \{1,3,6\}$. Intersecting the plane with the sphere,

$$\begin{array}{rl} x_1 + x_2 + x_3 &= 10\\ x_1^2 + x_2^2 + x_3^2 &= 46\end{array}$$

we obtain a circle on the plane, as depicted below

enter image description here

However, intersecting these two with the cubic surface,

$$\begin{array}{rl} x_1 + x_2 + x_3 &= 10\\ x_1^2 + x_2^2 + x_3^2 &= 46\\ x_1^3 + x_2^3 + x_3^3 &= 244\end{array}$$

we obtain the $3!=6$ permutations of the elements of $\mathcal Y$, which are colored in red

enter image description here

If we plot the quadratic and the cubic surfaces, but not the plane,

enter image description here

  • $\begingroup$ This is a great way to visualize my question. I tried to find a geometrical representation of the system but I failed when I moved to higher dimensional cases. Graphs seem to be a good way to summarize the solutions and provide easy interpretations. Thank you so much! BTW, would you mind telling me what software you used to plot these equations? $\endgroup$ – jwyao Jan 30 '17 at 0:55
  • $\begingroup$ @jwyao I used Grapher. $\endgroup$ – Rodrigo de Azevedo Jan 30 '17 at 1:01
  • $\begingroup$ Cool!Thanks a lot! $\endgroup$ – jwyao Jan 30 '17 at 2:00

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