<s>[INST] 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
 [/INST] 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}
</s>