Why the sum of the squares of the roots of the $n$th Hermite polynomial is equal to $n(n-1)/2$? 
How to prove that the sum of the squares of the roots of the $n$th Hermite polynomial is $\frac{n(n-1)}{2}$?

I tried with Vieta formulas, but it's hard. I appreciate a proof or reference to it.
An idea is to use the definition of sum of Hermite polynomials, but do not know.
 A: Let us write 
\begin{align}H_n(x)&=A_n(x-x_1)\ldots(x-x_n)=\\
&=A_n\left(x^n-e_1(x_1,\ldots,x_n)x^{n-1}+e_2(x_1,\ldots,x_n)x^{n-2}+\mathrm{poly}_{n-3}(x)\right),\tag{1}
\end{align}
where $e_k(x_1,\ldots,x_n)$ denote elementary symmetric polynomials:
\begin{align}
& e_1(x_1,\ldots,x_n)=\sum_{k=1}^nx_k,\\
& e_2(x_1,\ldots,x_n)=\sum_{1\leq i<j\leq n}^n x_ix_j.
\end{align} 
We want to find
$$\sum_{k=1}^{n}x_k^2=e_1^2(x_1,\ldots,x_n)-2e_2(x_1,\ldots,x_n),\tag{2}$$
and therefore it will suffice to know the coefficients of $x^n$, $x^{n-1}$ and $x^{n-2}$ in $H_n(x)$. But they can be determined from the series representations of Hermite polynomials:
$$H_n(x)=2^n\left(x^n-\frac{n(n-2)}{4}x^{n-2}+\mathrm{poly}_{n-4}(x)\right).\tag{3}$$
Together with (1) and (2), this gives the result:
$$\sum_{k=1}^{n}x_k^2=\frac{n(n-1)}{2}.$$
A: I would suggest to use Statisticians Hermite polynomials instead.  Of course it does not make the answer different, just imho easier not having to deal with all the 2s.  Also monic polynomials have a simpler Vieta formula.
The answer given is correct - with one minor typo
$$\operatorname{He}_n(x) = n!\sum_{m=0}^n \frac{(-1)^m}{2^m m!(n-2m)!}x^{n-2m}$$
Giving the coefficient in front of the $x^{n-2}$ term
$-\frac{n(n-1)}{2}$ (the 2 is just due to the statisticians Hermite, it is the 1 in the numerator that is the typo).
The fact that there is no $n-1$ coefficient means that the sum of the roots are zero.  Implying that they always come in symmetric pairs, for every root $x_k$ there is a value that is $-x_k$.  For odd Hermite polynomials, this implies zero is always a root.
This also means that $e_2$ is explicitly the negative sum of all squares itself, no need for the subtraction step in the earlier response.
$$e_2(\{x_k\}) = x_0x_1 + x_1x_2 + ... x_{n-1}x_n$$
Since every $x_k$ has a negative pair, you can pair them up as you wish and these become negative squares of $x_k$.  If $n$ is even you have accounted for all of them, if odd, the unpaired root is zero, so its entries do not contribute to the sum.
However, the sum $\sum_k x_k^2$ actually counts each value twice - since each root has a symmetric (negative) partner, the value will appear twice in this sum.  So, the sum of squares is $-2$ times the coefficient of the $n-2$ polynomial expansion of $He_n(x)$.
Now all you have to do is read off the coefficient in front of $x^{n-2}$ as in the original response.
$$\sum_k x_k^2=-2e_2(\{x_k\})$$
$$ \sum_k x_k^2= n(n-1)$$
Therefore the sum of the squares is not $n \choose 2$, but twice this amount.  Look at the first few for confirmation.  $n=2, x_k\in\{-1,1\}\implies\sum_k x_k^2=2$, $n=3,x_k\in\{-\sqrt{3},0,\sqrt{3}\}\implies\sum_k x_k^2=6$.
