A binomial identity from Mathematical Reflections Here is the problem:
Let $m,n$ be positive integers with $n>m$. Prove that
$\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$
This problem is O243 of Mathematical Reflections. A solution had been published using complex integration (https://www.awesomemath.org/assets/PDFs/MR5sol(1).pdf). However, I would like to see a solution using the difference operator, if any, since the form of the summand brings this to mind.

Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor (m+n)/2 \rfloor}
{n\choose k} (-1)^k {m+n-2k\choose n-1}.$$
As the link in the OP seems no longer to be working I present the complex variable proof.
Observe  that  in  the   second  binomial  coefficient  we  must  have
$m+n-2k\ge  n-1$ in  order  to avoid  hitting  the zero  value in  the
product in  the numerator  of the binomial  coefficient, so  the upper
limit for the sum is in fact $m+1\ge 2k$ with the sum being
$$\sum_{k=0}^{\lfloor (m+1)/2 \rfloor}
{n\choose k} (-1)^k {m+n-2k\choose n-1}.$$
Introduce
$${m+n-2k\choose n-1} = {m+n-2k\choose m+1-2k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+n-2k}}{z^{m+2-2k}} \; dz.$$
This integral correctly encodes the  range for $k$ being zero when $k$
is larger than $\lfloor (m+1)/2 \rfloor.$ Therefore we may let $k$ go
to infinity in the sum and obtain for $n\gt m$
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+n}}{z^{m+2}} 
\sum_{k\ge 0} {n\choose k} (-1)^k \frac{z^{2k}}{(1+z)^{2k}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+n}}{z^{m+2}} 
\left(1-\frac{z^2}{(1+z)^2}\right)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1+z)^{n-m} z^{m+2}} 
(1+2z)^n \; dz.$$
This produces the closed form
$$\sum_{q=0}^{m+1} {n\choose q} 2^q (-1)^{m+1-q}
{m+1-q+n-m-1\choose n-m-1}
\\ = (-1)^{m+1} \sum_{q=0}^{m+1} {n\choose q} (-1)^q 2^q
{n-q\choose n-m-1}.$$
This is
$$(-1)^{m+1} \sum_{q=0}^{m+1} {n\choose q} (-1)^q 2^q
{n-q\choose m+1-q}.$$
Introduce
$${n-q\choose m+1-q} 
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-q}}{z^{m+2-q}} \; dz$$
which once  more correctly  encodes the range  with the pole  at $z=0$
disappearing when $q\gt m+1.$ Therefore we may extend the range to $n$
to get
$$\frac{(-1)^{m+1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{m+2}} 
\sum_{q=0}^{n} {n\choose q} (-1)^q 2^q \frac{z^q}{(1+z)^q}
\; dz
\\ = \frac{(-1)^{m+1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{m+2}} 
\left(1-2\frac{z}{1+z}\right)^n \; dz
\\ = \frac{(-1)^{m+1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{m+2}} 
\frac{(1-z)^n}{(1+z)^n} \; dz
\\ = \frac{(-1)^{m+1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1-z)^{n}}{z^{m+2}} \; dz
\\ = (-1)^{m+1} {n\choose m+1} (-1)^{m+1}
= {n\choose m+1}.$$
