Claim: Let $\{p_{1},p_{2},\cdots,p_{K}\}$ be a probability distribution, that is, $p_{k}\in (0,1)$ and $\sum_\limits{k=1}^{K}p_{k}=1$.
Then the following system of equations has AT MOST finite number of solutions. (The system may not have any solution at all.) $$\begin{cases} p_{1}+p_{2}+\cdots+p_{k}=1=c_{1} \\ p^2_{1}+p^2_{2}+\cdots+p^2_{k}=c_{2} \\ \vdots \\ p_{1}^K+p_{2}^{K}+\cdots+p^{K}_{k}=c_{K} \\ \end{cases} $$ where $c_{1},\cdots,c_{K}$ are constants.
I believe the statement is true, in fact, I believe the number of solutions is upper bounded by K!, but I cannot prove it. Help!
You can recast these equations in terms of the elementary symmetric functions. For example, $(p_1+...+p_k)^2=\sum{p_k^2}+2\sum_{i<j}p_ip_j$ allows you to solve for $e_2(p_1,\dots,p_k)$
The elementary symmetric functions functions give the coefficients of a polynomial whose roots are $p_1,\dots,p_k$. For example $-e_1(p_1,\dots,p_k)$ is the coefficient of $x^{n-1}.$ Now the roots of the polynomial are unique, up to order, so there are at most $k!$ solutions, if all the roots turn out to be distinct.
Of course, some of the roots may not satisfy $0\le p_j \le 1,$ so there may not be any admissible solutions to the original system.