Prove that:
2p choose p is congruent to 2 modulus p^2
where p is any prime number
in general we have $\displaystyle \binom{2n}{n} = \sum_{k=0}^n \binom{n}{k}^2 = 2 + \sum_{k=1}^{n-1} \binom{n}{k}^2.$ now if $\displaystyle n = p$ is prime, then since $\displaystyle p \mid \binom{p}{k}, \ 1 \leq k \leq p-1,$ we'll have: $\displaystyle \binom{2p}{p} = 2 + \sum_{k=1}^{p-1} \binom{p}{k}^2 \equiv 2 \mod p^2. \ \ \ \square$