Let $x$ be an indeterminate and $n$ a non-negative integer. 



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> **Question.** The following seems to be true. Is it?
>  $$x\prod\_{k=1}^n(k^2-x^2)=\frac1{4^n}\sum\_{m=0}^n\binom{n-x}m\binom{n+x}{n-m}(x+2m-n)^{2n+1}.\tag1$$
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The problem came out of simplifying some work which reduced to the RHS of (1), but with a GOOD fortune this appeared to factor neatly into the LHS.



$\prod_k(x\pm k)$ in binomial basis?
It is sufficient to prove that the two sides agree when $x = n + \ell$ for $\ell \in \mathbb{N}$. As $k$ varies, the first term in $\prod\_{k=1}^n (k-n-\ell)(k+n+\ell)$ contributes 


$$-\ell(-\ell-1) \ldots (-\ell-n+1) = (-1)^n (\ell+n-1)\ldots (\ell+1)\ell$$ 


and the second term contributes $(n+\ell+1)\ldots (2n+\ell)$. Taking into account the factor $x$ before the product, the left-hand side is therefore 


$$(-1)^n \frac{(2n+\ell)!}{(\ell-1)!}.$$


The right-hand side can be rewritten as follows:


$$\begin{align\*} \frac{1}{4^n}{}&{} \sum\_{m=0}^n \binom{-\ell}{m}\binom{2n+\ell}{n-m}(2m+\ell)^{2n+1} \\
&= \frac{1}{4^n} \sum\_{m=0}^n (-1)^m \binom{m+\ell-1}{m}\binom{2n+\ell}{n-m}(2m+\ell)^{2n+1} \\
&= \frac{(-1)^n}{4^n} \sum\_{m=0}^n (-1)^m \binom{n-m+\ell-1}{\ell-1} \binom{2n+\ell}{m} (2n-2m+\ell)^{2n+1} \\
&= \frac{(-1)^n}{2.4^n} \sum\_{m=0}^{2n+\ell}(-1)^m \binom{2n+\ell}{m} \binom{n-m+\ell-1}{\ell-1} (2n-2m+\ell)^{2n+1} \end{align\*}$$


where the sum may be extended to get the final line using $\binom{n-m+\ell-1}{\ell-1} = \binom{\ell-\beta-1}{\ell-1} = 0$ if $m = n + \beta$ with $\beta \in \{1,\ldots, \ell-1\}$ and the equality of the summands for $m=n+\ell+\alpha$ and $m = n-\alpha$ for $\alpha \in \{0,1,\ldots,n\}$; this follows from $\binom{-1-\alpha}{\ell-1} = \binom{\ell+\alpha-1}{\ell-1}(-1)^{\ell-1}$ and $\binom{2n+\ell}{n+\ell+\alpha} = \binom{2n+\ell}{n-\alpha}$.


Therefore, up to the multiple $(-1)^n/2^{2n+1}$, the right-hand side is the $(2n+\ell)$th (backwards) iterated difference of the polynomial $y \mapsto \binom{y+\ell-1}{\ell-1}(2y+\ell)^{2n+1}$ of degree $2n+\ell$, evaluated at $y=n$. In fact the evaluation point makes no difference: we always get $(2n+\ell)!$ times its leading coefficient, namely $2^{2n+1}/(\ell-1)!$. Hence the right-hand side is 


$$(-1)^n \frac{(2n+\ell)!}{(\ell-1)!},$$ 


as required.


