If $p$ is a prime number, prove that ${2p \choose p} - 2$ is divisible by $p^2$ without using Wolstenholme's theorem.

Question

If $p$ is a prime number, prove that ${2p \choose p} - 2$ is divisible by $p^2$ without using Wolstenholme's theorem.

My approach so far has been: If $0 < k < p$, then ${p \choose k}$ is divisible by $p$. How to proceed?

Answer 1

Use the fact you noted, along with Vandermonde's identity (with $n=m=r=p$).

Answer 2

Assume that we have a parliament with $2p$ politicians, $p$ politicians belonging to the left wing (numbered from $0$ to $p-1$) and $p$ politicians belonging to the right wing (numbered from $0$ to $p-1$).
We wonder in how many ways me may

• select $p$ politicians from the parliament to join a committee, in such a way that there is at least one politician from each wing in this committee.

It is clear that such number of ways is given by $\binom{2p}{p}-2$. On the other hand, we may put an equivalence relation on the sets of such committees, such that the committee formed by the politicians $\{l_1,l_2,\ldots\}$ from the left wing and the politicians $\{r_1,r_2,\ldots\}$ from the right wing is considered the same as the committee formed by the politicians $\{l_1+a,l_2+a,\ldots\}$ from the left wing and the politicians $\{r_1+b,r_2+b,\ldots\}$ from the right wing (addition is performed $\pmod{p}$) for any $(a,b)\in\{0,\ldots,p-1\}^2$. Since every equivalence class has $p^2$ elements, $$ p^2\mid \binom{2p}{p}-2.$$