Uniqueness of the nodes for Gauss-Legendre quadrature Gauss-Legendre quadrature approximates $\int_{~1}^{1}f(x)dx$ by $\sum_{i=1}^nw_if(x_i)$.
Wikipedia says that

This choice of quadrature weights $w_i$ and quadrature nodes $x_i$ is the unique choice that allows the quadrature rule to integrate degree $2n − 1$ polynomials exactly.

https://en.wikipedia.org/wiki/Gauss–Legendre_quadrature
The uniqueness of $w_i$ satisfying the condition for fixed $x_i$ comes from the invertibility of the Vandermonde's matrix.
How to prove the uniqueness of $x_i$?
 A: It is known that there are nodes $x_1<x_2<\ldots < x_n$ and the weights $w_i>0$ such that
$$\int\limits_{-1}^1p(x)\,dx =\sum_{i=1}^n w_ip(x_i),\qquad \deg p\le 2n-1$$
Assume
there are other nodes $x_1'<x_2'<\ldots <x_n'$ and quantities $w_i'$ (not necessarily nonnegative) such that
$$\sum_{i=1}^n w_ip(x_i)=\sum_{i=1}^n w_i'p(x_i'),\qquad \deg p\le 2n-1\quad (*)$$
Assume there exists a polynomial $q,$ $\deg q\le 2n-1,$ $q(x_i)\ge 0$ for $i=1,2,\ldots, n$  and
$q(x)=0$ iff $x\in \{x_1',x_2',\ldots, x_n'\}.$
Then the formula $(*)$ implies $q(x_i)=0$ for $i=1,2,\ldots, n.$ Therefore $x_i'=x_i$ for $i=1,2,\ldots, n.$
Now we are going to construct a polynomial $q$ with the properties described above. Assume one of the intervals $[x_k,x_{k+1}]$ contains more than one element of $\{x_j'\}_{j=1}^n$
$$x_k\le x'_{l} <x'_{l+1}\le x_{k+1}$$
Then
$$q(x)=(x-x'_l)(x-x'_{l+1})\prod_{j\neq l,l+1}^n(x-x'_j)^2$$
In the opposite case every interval $[x_i,x_{i+1}]$ contains at most one element $x_j'.$ By the pigeonhole principle either $x_1'<x_1$ or $x_n'>x_n.$
In the first case let
$$q(x)=(x-x_1')\prod_{j=2}^n(x-x'_j)^2$$ and in the second case
$$q(x)=(x_n'-x)\prod_{j=1}^{n-1}(x-x'_j)^2$$
In all three cases we have $\deg q\le 2n-1,$ $q(x_i')=0$ and $q(x_i)\ge 0.$
