The Problem
I have a sum of $N$ known functions $f_k(x)$, $$ y(x) = \sum_{0<k\leq N} k^2f_k(x) \tag{1}\label{eq1} $$ I also have a constraint which is valid at the minima and maxima points $x_m$ $$ 0 = \sum_{0<k\leq N} k^3f_k(x) \bigg|_{x\in x_m} \tag{2}\label{eq2} $$ My question is, is there any general form(s) of $y(x)|_{x\in x_m}$?
My Attempts
Writing out the sums \eqref{eq1} and \eqref{eq2} in full, $$ y(x) = 1^2 f_1(x) + 2^2 f_2(x) + \dots + N^2 f_N(x) \tag{1} $$ $$ 0 = 1^3 f_1(x) + 2^3 f_2(x) + \dots + N^3 f_N(x)\bigg|_{x\in x_m} \tag{2} $$
From here I can choose a term to eliminate. If I choose the $n$th term, $$ n^2f_n(x)\bigg|_{x\in x_m} = -\frac{1^3}{n} f_1(x) - \dots - \frac{(n-1)^3}{n} f_{n-1}(x) - \frac{(n+1)^3}{n} f_{n+1}(x) - \dots - \frac{N^3}{n} f_N(x)\bigg|_{x\in x_m} $$ Inserting this into \eqref{eq1} $$ y(x)\bigg|_{x\in x_m} = \left(1^2-\frac{1^3}{n}\right) f_1(x) + \dots + \left((n-1)^2-\frac{(n-1)^3}{n}\right)f_{n-1}(x) \\ + \left((n+1)^2-\frac{(n+1)^3}{n}\right)f_{n+1}(x) + \dots + \left(N^2-\frac{N^3}{n}\right) f_N(x)\bigg|_{x\in x_m} $$ and refactoring into a sum $$ \require{cancel} y(x)|_{x\in x_m}= \sum_{0<k\leq N}\left[\left(k^2-\frac{k^3}{n}\right) f_k(x)\right] - \cancelto{0}{\left(n^2-\frac{n^3}{n}\right)}f_n(x) \bigg|_{x\in x_m} $$ I would conclude that the "general form" for $y(x)|_{x\in x_m}$ is $$ y(x)|_{x\in x_m}= \sum_{0<k\leq N}\left(k^2-\frac{k^3}{n}\right) f_k(x)\bigg|_{x\in x_m} $$ where $n$ is any integer in $[1,N]$.
Discussion
Is my working here correct? Are these $N$ different general solutions really the same solution? If this is the result of some theorem or basic linear algebra concept I'd love to find out about it. I tried deriving this result using matrix algebra but couldn't quite figure it out. Thank you :)