TITLE: Prove that $\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}$
QUESTION [9 upvotes]: This problem derives from an expression of probability in random walk. I hope to 

prove that 
  \begin{equation*}
\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}
\end{equation*}

It seems quite right, I verified this equation from $n=1$ to $6$ without any idea to prove it.
This equality is found by @Sangchul Lee.

REPLY [2 votes]: We seek to verify that
$$\sum_{l=0}^{n} {n\choose l}^2 (x+y)^{2l} (x-y)^{2n-2l} 
= \sum_{l=0}^{n} {2l\choose l} {2n-2l\choose n-l} 
x^{2l}y^{2n-2l}.$$
Now we see on the LHS that the  powers of $x$ and $y$ always add up to
$2n$ and the exponent on $x$ determines the one on $y.$ Extracting the
coefficient on $[x^q][y^{2n-q}]$ we obtain
$$\sum_{l=0}^{n} {n\choose l}^2
\sum_{p=0}^q {2l\choose p} (-1)^{2n-2l-(q-p)} {2n-2l\choose q-p}
\\ = \sum_{l=0}^{n} {n\choose l}^2
\sum_{p=0}^q {2l\choose p} (-1)^{q-p} 
[z^{q-p}] (1+z)^{2n-2l}
\\ = [z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} 
\sum_{p=0}^q {2l\choose p} (-1)^p z^p.$$
We  may extend  $p$ to  infinity  because with  $p\gt q$  there is  no
contribution to $[z^q]$, getting
$$[z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} 
\sum_{p\ge 0} {2l\choose p} (-1)^p z^p
\\ = [z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} 
(1-z)^{2l}
\\ = [z^q] (-1)^q [w^n] (1+w(1-z)^2)^n (1+w(1+z)^2)^n
\\ = [z^q] [w^n] (1+w(1-z)^2)^n (1+w(1+z)^2)^n.$$
Re-write this as
$$[z^q] [w^n] ((w(1+z^2)+1)^2 - 4 w^2 z^2)^n
\\ = [z^q] [w^n]
\sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} w^{2p} z^{2p}
(w(1+z^2)+1)^{2n-2p}
\\ = [z^q] 
\sum_{p=0}^n {n\choose p} (-1)^p 2^{2p}  z^{2p}
[w^{n-2p}] (w(1+z^2)+1)^{2n-2p}
\\ = [z^q] 
\sum_{p=0}^n {n\choose p} (-1)^p 2^{2p}  z^{2p}
{2n-2p\choose n-2p} (1+z^2)^{n-2p}.$$
We observe at this point that we  get zero here when $q$ is odd, which
agrees  with the  target formula.   We are  thus justified  in putting
$q=2l$ to get
$$[z^l] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p}  z^{p}
{2n-2p\choose n-2p} (1+z)^{n-2p}
\\ = \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p}
{2n-2p\choose n-2p} {n-2p\choose l-p}.$$
Note that
$${n\choose p} {2n-2p\choose n-2p} {n-2p\choose l-p}
= \frac{(2n-2p)!}{p! \times (n-p)! \times 
(l-p)! \times (n-l-p)!}
\\ = {l\choose p} \frac{(2n-2p)!}{(n-p)! \times 
l! \times (n-l-p)!}
= {l\choose p} {2n-2p\choose n-p} {n-p\choose l}.$$
Re-indexing we get for the sum
$$(-1)^n 2^{2n} \sum_{p=0}^n {l\choose n-p} {2p\choose p}
{p\choose l} (-1)^p 2^{-2p}
\\ = (-1)^n 2^{2n} \sum_{p=0}^n  {2p\choose p}
(-1)^p 2^{-2p} [z^{n-p}] (1+z)^l 
[w^l] (1+w)^p
\\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l]
\sum_{p=0}^n  {2p\choose p}
(-1)^p 2^{-2p} z^p (1+w)^p.$$
We  may  once  more  extend  $p$  to  infinity  because  there  is  no
contribution from  the sum term  to the coefficient  extractor $[z^n]$
when $p\gt n,$ obtaining
$$(-1)^n 2^{2n} [z^n] (1+z)^l [w^l]
\sum_{p\ge 0}  {2p\choose p}
(-1)^p 2^{-2p} z^p (1+w)^p
\\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l]
\frac{1}{\sqrt{1+z(1+w)}}
\\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l]
\frac{1}{\sqrt{1+z+wz}}
\\ = (-1)^n 2^{2n} [z^n] (1+z)^{l-1/2} [w^l]
\frac{1}{\sqrt{1+wz/(1+z)}}
\\ = (-1)^n 2^{2n} [z^n] (1+z)^{l-1/2} 
{2l\choose l} (-1)^l 2^{-2l} z^l \frac{1}{(1+z)^l}
\\ = (-1)^{n-l} 2^{2n-2l} {2l\choose l}
[z^{n-l}] \frac{1}{\sqrt{1+z}}
\\ = (-1)^{n-l} 2^{2n-2l} {2l\choose l}
{2n-2l\choose n-l} (-1)^{n-l} 2^{-(2n-2l)}
\\ = {2l\choose l} {2n-2l\choose n-l}.$$
This  is the  claim. Credit  goes to  the Egorychev  method which  was
presented here in formal power series notation.