# Finding a Polynominal

Let $$x_1 + x_2 + \dots+ x_n = A$$

The value of $x_1$ to $x_n$ is not given. Suppose we have $n$ variables $y_1, y_2,\dots, y_n$. Is there any way to find $x_1y_1 + x_2y_2 + x_3y_3 + \dots+ x_ny_n$, based on the value of $A$ and $y_1$ to $y_n$?

• Very unlikely. Any permutation of $x_1, \ldots, x_n$ preserves the identity $x_1 + \ldots + x_n = A$, but may change the value of $x_1 y_1 + \ldots + x_n y_n$ – Martin R Dec 26 '17 at 9:39
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Dec 26 '17 at 9:41
• You have 1 equation for $n$ unknown variables. So in general it cant work. Maybe you missed to mention that it is a polynomial function $f_m(x) = y$ of degree $m$. Then you could solve it for the case of $m=0$. – Rudi_Birnbaum Dec 26 '17 at 9:54
• And likewise for the case of $n=1$ the system is solvable. – Rudi_Birnbaum Dec 26 '17 at 10:02

In most cases no, the answer is yes iff $y_i=y_j\forall i,j\in\{1,2,\cdots,n\}$
Consider $x_1 + x_2 + \dots+ x_n = A$ and that the condition I stated are false.
In this case I can rewrite the $y_1x_1 +y_n x_2 + \dots+ y_nx_n=B$ it as $w_1u_1+w_{2}u_{2}+\cdots+w_\ell u_\ell=B$ where $1<\ell\le n$, $w_k=y_h$ for some $h$ and $u_k$ is combination of some values of $x_s$.
This is equation is the most reduce form, i.e. has the least amount of variables we can get, so unless I have only one variable I can't solve it, but one of my conditions is that not all of $y_i$ are equal, so I have at least 3 variables thus I can't solve this.
• It does not work even if $A= 0$: $(+1)1 + (-1)2 = -1$, $(-1)1 + (+1)2 =+1$. – Martin R Dec 26 '17 at 12:26