How prove this $\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $ I think the following equality is true ($p\in \mathbb{N},p\ge 2$):
$$\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $$
when $p=2$, then
$$\sum_{k=0}^{n}\binom{n}{k}^2\binom{2n+k}{k}=\binom{2n}{n}^2$$
But I can't prove this
and I can't prove this  $p\ge 3$ ?  
Thank you for your help
 A: On p. 33 of this document the following appears:
"The following identity solves a problem on Page 122, Vol. 29, 1947 of Norsk Matematisk Tidsskrift."
$$\sum_{k=0}^n \binom{n}{k}\frac{\binom{x}{k}\binom{y}{k}}{\binom{x+y+n}{k}} = \frac{\binom{x+n}{n}\binom{y+n}{n}}{\binom{x+y+n}{n}}$$
followed by
$$\sum_{k=0}^n \binom{n}{k}\binom{x}{k}\binom{x+n+k+\alpha}{k+\alpha} = \binom{x+\alpha+n}{n}\binom{x+\alpha+n}{n+\alpha}.$$
Presumably the second equation can be derived from the first; in any case, setting $\alpha=0$ and $x=(p-1)n$ gives the identity you asked about.
To derive the latter identity, start with the left-hand side:
\begin{align}
  \binom{n}{k}\binom{x}{k}&\binom{x+n+k+\alpha}{k+\alpha}
     = \frac{n!x!(x+n+\alpha+k)!}{k!(n-k)!k!(x-k)!(\alpha+k)!(x+n)!} \\
 &= \frac{n!x!}{(n+x)!}\cdot\frac{(x+n+\alpha)!(x+n+\alpha+1)_k(-1)^k(-n)_k(-1)^k(-x)_k}
                          {n!x!\alpha!(\alpha+1)_k(1)_k k!} \\
 &= \frac{(x+n+\alpha)!}{\alpha!(x+n)!}\cdot\frac{(x+n+\alpha+1)_k(-n)_k(-x)_k}
                                               {(\alpha+1)_k(1)_kk!} \\
 &= \frac{(x+n+\alpha)!}{\alpha!(x+n)!}{{_3F_2}\left(\left.
 \begin{array}{ccc}
  -n, & -x, & x+n+\alpha+1 \ \\
      & 1, & \alpha+1
 \end{array}
\right\rvert\ {1}\right)}.\end{align}
Saalsch\"utz' formula applies, giving
\begin{align}\frac{(x+n+\alpha)!}{\alpha!(x+n)!}\cdot&\frac{(x+1)_n(-x-n-\alpha)_n}{(1)_n(-n-\alpha)_n}
      = \frac{(x+n+\alpha)!}{\alpha!(x+n)!}\cdot\frac{(x+n)!(x+n+\alpha)!\alpha!}
                                                    {x!n!(x+\alpha)!(n+\alpha)!}\\
 &= \frac{(x+n+\alpha)!}{n!(x+\alpha)!}\cdot\frac{(x+n+\alpha)!}{(n+\alpha)!x!} \\
 &= \binom{x+n+\alpha}{n}\binom{x+n+\alpha}{n+\alpha}.\end{align}
A: Let's (as @rogerl suggests) prove that
$$
\sum\binom nk\binom mk\binom{n+m+k}{n+m}=\binom{n+m}n^2.
$$
LHS can also be written as
$$
\sum(-1)^k\binom nk\binom mk\binom{-(n+m+1)}k
$$
so it's equal (oh, up to a sign) to the constant term of
$$
(1-Z^{-1})^n(1-W^{-1})^m(1-ZW)^{-(n+m+1)}
$$
i.e. to the residue
$$
\operatorname*{res}_{Z,W}\left\{(1-Z^{-1})^n(1-W^{-1})^m(1-ZW)^{-(n+m+1)}\frac{dZ}Z\frac{dW}W\right\}
$$
Now, like in a proof of Dixon's identity let's rewrite this in the form
$$
\operatorname*{res}_{Z,W}\left\{
\left(\frac{1-Z^{-1}}{1-ZW}\right)^n
\left(\frac{1-Z^{-1}}{1-ZW}\right)^m
\frac{dZ\,dW}{ZW(1-ZW)}\right\}
$$
and use the substitution $Z=\frac z{1-w}$, $W=\frac w{1-z}$. Since $\frac{dZ\,dW}{ZW(1-ZW)}=\frac{dz\,dw}{zw}$ and $\frac{1-Z^{-1}}{1-ZW}=-\frac z{(1-z)(1-w)}$ we see that our residue takes the form
$$
\pm\operatorname*{res}_{z,w}\left\{\frac{(1-z)^{n+m}(1-w)^{n+m}}{z^nw^m}\frac{dz}z\frac{dw}w\right\}=\pm\binom{n+m}n\binom{n+m}m.
$$
Bingo.
A: Suppose we seek to verify that
$$\sum_{k=0}^n {n\choose k}
{pn-n\choose k} {pn+k\choose k} = {pn\choose n}^2.$$
We use the integrals
$${pn-n\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z^{k+1}} \; dz$$
and
$${pn+k\choose k} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn+k}}{w^{k+1}} \; dw.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w} 
\sum_{k=0}^n {n\choose k} \frac{(1+w)^k}{z^k w^k}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w} 
\left(1+\frac{1+w}{zw}\right)^n
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w^{n+1}} 
(1+w+zw)^n
\; dw \; dz.$$
Expanding the binomial in the inner sum we get
$$\sum_{q=0}^n {n\choose q} w^q (1+z)^q$$
which yields
$$\sum_{q=0}^n {n\choose q}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n+q}}{z^{n+1}}
{pn\choose n-q}  \; dz
\\ = \sum_{q=0}^n {n\choose q}
{pn-n+q\choose n}
{pn\choose n-q}.$$
The inner term is
$${n\choose q}
{pn-n+q\choose n}
{pn\choose pn-n+q}
\\ = 
\frac{(pn)!}{q!\times (n-q)! \times
(pn-2n+q)! \times
(n-q)!}
\\ = {pn\choose n}
\frac{n! \times (pn-n)!}{q!\times (n-q)! \times
(pn-2n+q)! \times
(n-q)!}
\\ = {pn\choose n}
{n\choose q} {pn-n\choose n-q}.$$
Thus it remains to show that
$$\sum_{q=0}^n {n\choose q} {pn-n\choose n-q} = {pn\choose n}.$$
This can be done combinatorially or using the integral
$$\frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn-n}}{v^{n+1}}
\sum_{q=0}^n {n\choose q} v^q \; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn-n}}{v^{n+1}}
(v+1)^n \; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn}}{v^{n+1}}
= {pn\choose n}.$$
