# How to prove that the sum of the squares of the roots of the nth Hermite polynomial is (n(n−1))/2?

How to prove that the sum of the squares of the roots of the nth Hermite polynomial is $\frac{n\left(n-1\right)}{2}$?
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Lisa Acevedo
Let us write
$\begin{array}{rl}{H}_{n}\left(x\right)& ={A}_{n}\left(x-{x}_{1}\right)\dots \left(x-{x}_{n}\right)=\\ \text{(1)}& & ={A}_{n}\left({x}^{n}-{e}_{1}\left({x}_{1},\dots ,{x}_{n}\right){x}^{n-1}+{e}_{2}\left({x}_{1},\dots ,{x}_{n}\right){x}^{n-2}+{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}}_{n-3}\left(x\right)\right),\end{array}$
where ${e}_{k}\left({x}_{1},\dots ,{x}_{n}\right)$ denote elementary symmetric polynomials:
$\begin{array}{rl}& {e}_{1}\left({x}_{1},\dots ,{x}_{n}\right)=\sum _{k=1}^{n}{x}_{k},\\ & {e}_{2}\left({x}_{1},\dots ,{x}_{n}\right)=\sum _{1\le i
We want to find
$\begin{array}{}\text{(2)}& \sum _{k=1}^{n}{x}_{k}^{2}={e}_{1}^{2}\left({x}_{1},\dots ,{x}_{n}\right)-2{e}_{2}\left({x}_{1},\dots ,{x}_{n}\right),\end{array}$
and therefore it will suffice to know the coefficients of ${x}^{n}$, ${x}^{n-1}$ and ${x}^{n-2}$ in ${H}_{n}\left(x\right)$. But they can be determined from the series representations of Hermite polynomials:
$\begin{array}{}\text{(3)}& {H}_{n}\left(x\right)={2}^{n}\left({x}^{n}-\frac{n\left(n-2\right)}{4}{x}^{n-2}+{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}}_{n-4}\left(x\right)\right).\end{array}$
Together with (1) and (2), this gives the result:
$\sum _{k=1}^{n}{x}_{k}^{2}=\frac{n\left(n-1\right)}{2}.$