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suppose we have given positive real numbers $a_1,...,a_n>0$. Consider the following system of equations:

$$\sum_{i=1}^{n} (x_{i})^{2k-1} = a_k,\quad k= 1,.....,n$$

with $x_1,...,x_n>0$.

This system of equations does not have always solutions (see e.g. the answer of Leo163 below). But suppose $a_1,...,a_n$ are choosen in such a way that there exist a solution.

The question is: How many solutions can this system have? By solutions I mean any multi set $\{ x_1,…,x_n \}$ such that the above equations are satisfied. Are there conditions such that the solution becomes unique?

I would really appreciate any help.

Best wishes

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  • $\begingroup$ By symmetry, when there exists a solution, there are at least $n!$ of them, as the can be freely permuted. $\endgroup$
    – user65203
    Commented Dec 21, 2016 at 18:21
  • $\begingroup$ @YvesDaoust Right, unless $x_i=x_j$ which happens iff $\cfrac{a_k}{n}=\left(\cfrac{a_1}{n}\right)^{2k-1}$. $\endgroup$
    – dxiv
    Commented Dec 21, 2016 at 18:24
  • $\begingroup$ By solution I mean a multiset of n positive numbers which satisfy all equations. So your permutations are one and the same solution. Sorry that I did not make this clear in advance $\endgroup$
    – Denilson Orr
    Commented Dec 21, 2016 at 18:56

2 Answers 2

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I don't know if that's what you wanted to know, but it is not the case that a solution always exists. For instance, $\begin{cases} x_1+x_2=1\\ x_1^3+x_2^3=2 \end{cases}$ has no solutions: $3x_1x_2(x_1+x_2)=-1$, impossible since you are requiring that $x_1$ and $x_2$ are positive.

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The maximum number is $1\cdot3\cdot5\ldots\cdot2k-1$. You can see this in following example for k = 3. The first equation defines a hyperplane in $\Bbb C^3$, the second equation is algebraic equivalent with three hyperplanes, so the intersection with the first gives three lines and the third equation is equivalent to $5$ hyperplanes, so the intersectiono with the three lines gives $15$ points, maximal.

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