When $p$ is prime, one easily checks that ${2p \choose p} \equiv 2 \mod p$. Moreover, if $p \le 31$, we have even ${2p \choose p} \equiv 2 \mod p^2$. Is this true for all prime numbers? Is it easy to prove?
View the computation of of the first central binomial coefficients and of the ratios ${2n \choose n}-2$ by $n^2$ and by $n^3$
![computation of first binomial coefficients and ratios [2n \choose n]-2 by n^2 and by n^3](https://i.stack.imgur.com/FmMwn.png)
If $p>3$, this even holds $\pmod {p^3}$.
Proof: Wolstenholmes' Theorem tells us that $$\binom {2p-1}{p-1}\equiv 1\pmod {p^3}$$
But we have $$\binom {2p}p=2\times \binom {2p-1}{p-1}$$
by direct computation, so we are done.
Note $1$: This strengthening of the desired claim is false for $p≤3$ but it is easy to check that the $p^2$ claim holds for those two primes. Numerical evidence suggests that nothing special holds for $p^4$)
Note $2$: As the OP has noted in a comment, a proof of Wolstenholmes can be found here