This identity came up in my research:
$$
\sum\_{m=1}^n m^2 \frac{(\frac{xy}n + m-1)\_{2m-1} (n+m-1)\_{2m-1}}{(x+m)\_{2m+1} (y+m)\_{2m+1}} = \frac{n^2}{(x^2-n^2) (y^2 - n^2)}.
$$
Here $n$ is a fixed positive integer and $x,y$ are variables. So for each $n$ this is an identity of rational functions in $x,y$. I denote by $(x)\_n$ the falling factorial (Pochhammer symbol)
$$
(x)\_n = x(x-1)\cdots (x-n+1).
$$
It may be worth pointing out that all the falling factorials that appear are of this form:
$$
(x+m-1)\_{2m-1} = x \prod\_{i=1}^{m-1} (x^2-i^2)
$$
This identity looked very foreign to me because of the quadratic argument $\frac{xy}n$ inside the falling factorial. I managed to prove it eventually, but I wonder if there is a simple/standard approach or if this is known and/or related to something interesting.


EDIT: If you prefer binomial identities, by taking the partial fraction expansions this reduces to
$$
\sum\_{m=\max(a,b)}^n \binom{2m}{m-a} \binom{2m}{m-b} \binom{\frac{ab}{n}+m-1}{2m-1} \binom{n+m-1}{2m-1} = 0
$$
for positive integers $a,b,n$ satisfying $a<n, b<n$.



Identity involving a quadratic term inside the Pochhammer symbol
Your identity is not just a curiosity. It is a special case of a result that has been used to obtain many quadratic and cubic identities for hypergeometric series. Perhaps the most general formulation is given by Warnaar (Summation and transformation formulas for elliptic hypergeometric series, Constr. Approx. 18 (2002), 479–502, Lemma 3.1). There it is formulated as a theta function identity. In the rational limit case, when one replaces a Jacobi theta function $\theta(x|\tau)$ by $x$, it becomes
\begin{multline\*}\sum\_{k=0}^n (a\_k^2-b\_k^2)(c\_k^2-d\_k^2)\prod\_{j=0}^{k-1}(a\_j^2-c\_j^2)(b\_j^2-d\_j^2)
\prod\_{j=k+1}^n(a\_j^2-d\_j^2)(b\_j^2-c\_j^2)\\=
\prod\_{j=0}^n(a\_j^2-c\_j^2)(b\_j^2-d\_j^2)-
\prod\_{j=0}^n(a\_j^2-d\_j^2)(b\_j^2-c\_j^2).
\end{multline\*}
Here, $a\_j$, $b\_j$, $c\_j$ and $d\_j$ are arbitrary sequences. Your identity seems to be the case $a\_j=x$, $b\_j=n+1$, $c\_j=(n+1)(j+1)/y$, $d\_j=j+1$. Since $d\_n=b\_j$, the first term on the right-hand side vanishes. Of course you also need to replace $n$ by $n-1$ and $k$ by $m-1$.


As Ira Gessel points out, this is an indefinite summation, so once you guess the identity the proof is trivial by induction on $n$.


(I removed my claim that there are typos in your formula. I missed that you use falling rather than rising factorials.)


