Can this system of polynomial equations be solved in polynomial time?

Question

I have these $n$ equations, with $n$ variables. Variables are first $n$ positive integers, constants can be any rational number including zero. Given that there is always a solution, how do we find a solution to the system:

\begin{align} ax^{1}+by^{1}+cz^{1}+\cdots&=k_{1}\\ ax^{2}+by^{2}+cz^{2}+\cdots&=k_{2}\\ ax^{3}+by^{3}+cz^{3}+\cdots&=k_{3} \end{align}
so on ... till $$ax^{n}+by^{n}+cz^{n}+\cdots=k_{n}$$

for variables $x,y,z\in \{1,2\ldots, n\}$.

The value of constants $a$,$b$,$c$... remain same in all these equations.

And these constants can also be equal to each other. For e.g: $a = b$ so that values of $x$ and $y$ can become interchangeable. In such cases one working solution is suffice.

Can we find a solution or determine none exist in polynomial time?

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Edit: For the sake of clarity here is a simple example.

If $n=3$, I know $x,y,z$ can only take 1, 2 or 3 as their values. So if I have $x + y + z = 9$, I clearly know that $x, y$ and $z$ are all 3. But if I have $x + y + z = 6$, I don't know whether they are $2,2,2$ or $1,2,3$. Then I can make use of $x^2 + y^2 + z^2 = 14$. Then I will know they are $1,2,3$.

Answer 1

Let us denote the unknowns by $x_1,\ldots,x_n$, and the coefficients by $c_1,\ldots,c_n$. For a polynomial $P(x)$, we denote $$ [P(x)] = \sum_{i=1}^n c_i P(x_i). $$ We are given the values of $[1],[x],\ldots,[x^n]$, from which we can calculate $[P(x)]$ for every polynomial of degree at most $n$. In particular, we can compute, for each $j \in \{1,\ldots,n\}$, the value $$ \left[ \prod_{k \neq j} \frac{x-k}{j-k} \right] = \sum_{i\colon x_i=j} c_i. $$ To complete the solution, we need to solve several SUBSET-SUM instances, which possibly interact.

Conversely, we can encode SUBSET-SUM using your equations. Given a set of positive coefficients $c_1,\ldots,c_n$ summing to $C$ and a target $S$, we encode a solution using $\{1,2\}$-valued variables (where $1$ represents a value in the subset summing to $S$), and can compute $[x^k] = S + (C-S)2^k$. These numbers satisfy $$ [x^k(x-1)(x-2)] = [x^{k+2}-3x^{k+1}+2x^k] = \\ (S + (C-S)2^{k+2}) - 3(S+(C-S)2^{k+1}) + 2(S+(C-S)2^k) = \\ (1-3+2)S + (4-6+2)(C-S)2^k = 0. $$ Therefore the argument above shows that every $\{1,\ldots,n\}$-valued solution is actually $\{1,2\}$-valued. Therefore a solution to your problem exists if and only if the SUBSET-SUM instance is a Yes instance.

When all coefficients $c_i$ are equal, the argument above allows us to calculate directly the number of $x_i$'s equal to $j$. Let us take your example: $[1] = 3$, $[x] = 6$ and $[x^2] = 14$. Denoting by $y_j$ the number of $x_i$'s equal to $j$, we compute $$ \begin{align*} y_1 &= \left[ \frac{(x-2)(x-3)}{(1-2)(1-3)} \right] = \frac{[x^2-5x+6]}{2} = \frac{14-5\cdot6+6\cdot3}{2} = 1, \\ y_2 &= \left[ \frac{(x-1)(x-3)}{(2-1)(2-3)} \right] = [-x^2+4x-3] = -14+4\cdot6-3\cdot3 = 1, \\ y_3 &= \left[ \frac{(x-1)(x-2)}{(3-1)(3-2)} \right] = \frac{[x^2-3x+2]}{2} = \frac{14-3\cdot6+2\cdot3}{2} = 1. \end{align*} $$

Answer 2

My algorithm is not an exact one but actually an approximation algorithm.

Algorithm

1. Define three matrices

$$ X = \left[ \begin{matrix} x & y & z & \cdots \\ x^2 & y^2 & z^2 & \cdots \\ \vdots & \vdots & \vdots & \ddots \\ x^n & y^n & z^n & \cdots \end{matrix} \right]_{n \times n} $$

$$ C = \left[ \begin{matrix} a \\ b \\ c \\ \vdots \end{matrix} \right]_{ n \times 1} $$

$$ K = \left[ \begin{matrix} k_1 \\ k_2 \\ k_3 \\ \vdots \end{matrix} \right]_{ n \times 1} $$

We can see that $X \cdot C = K$

2. Find the Inverse Matrix for C. (Yes, inverse of a column matrix) $$ C^{-1} = \frac{C^T}{C_{sq}} \\ C_{sq} = \sum_{i=1}^n c_i^2 = a^2 + b^2 + c^2 + \cdots $$

$C^{-1}$ satisfies all properties that a normal matrix inverse does $$ C^{-1} \cdot C = I_{1 \times 1} \\ C \cdot C^{-1} \cdot C = C $$

3. Now comes the tricky part. Calculate $C^* = C \cdot C^{-1}$. We can now use this to obtain

$$ X \cdot C \cdot C^{-1} = K \cdot C^{-1} \\ X \cdot C^* = K \cdot C^{-1} \\ X = K \cdot C^{-1} \cdot C^{*{-1}} $$

The tricky part is that $C^*$ is always singular i.e. $|C^*| = 0$. So it does not have a unique inverse, but has many pseudo-inverses. (Refer: Moore-Penrose Inverse ).

So we have to keep searching for all pseudo-inverses of $C^*$ i.e. $C^{*{-1}}$ such that the result $K \cdot C^{-1} \cdot C^{*{-1}}$ we obtain, matches the pattern observed in $X$.