Let $\displaystyle p>3$ be a prime. Show that $\displaystyle p|\sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}$.
it's also true for $\displaystyle p=2.$ anyway, this is just a simple application of Lucas' theorem. by this theorem we have:
$\displaystyle \sum_{i=1}^p \binom{ip}{p} \binom{(p-i+1)p}{p} \equiv \sum_{i=1}^p \binom{i}{1} \binom{p-i+1}{1}=\sum_{i=1}^p i(p-i+1) \equiv \sum_{i=1}^p i(i-1)$
$\displaystyle =\frac{p(p^2 - 1)}{3} \equiv 0 \mod p,$ because every prime which is $\displaystyle \neq 3$ is in the form $\displaystyle 3k \pm 1$ and thus $\displaystyle 3 \mid p^2-1. \ \Box$
Remark: you might say that i didn't apply Lucas' theorem correctly for $\displaystyle i=1$ and $\displaystyle i=p.$ but actually what i did
is perfectly fine because:
if $\displaystyle i=1,$ then by Lucas' theorem: $\displaystyle \binom{(p-i+1)p}{p}=\binom{p^2}{p} \equiv \binom{1}{0} \binom{0}{1} = 0 \equiv \binom{p}{1}=\binom{p-i+1}{1} \mod p.$
similarly if $\displaystyle i=p,$ then by Lucas' theorem: $\displaystyle \binom{ip}{p}=\binom{p^2}{p} \equiv \binom{1}{0} \binom{0}{1} = 0 \equiv \binom{p}{1}=\binom{i}{1} \mod p.$