# Prove $\binom{2p+1}{p}\equiv2$ mod $p$ when $p$ is any prime.

Prove $\binom{2p+1}{p}\equiv2$ mod $p$ when $p$ is any prime.

I started by attempting to construct the congruence as follows:

Since $(2p+1)!=(2p+1)(2p)(2p-1)!$ and $p|2p$, then $(2p+1)!\equiv0$ mod $p$. Since $p|p!$, then $2p!(p+1)!\equiv0$ mod $p$. Therefore, $(2p+1)!\equiv2p!(p+1)!$ mod $p$.

However, I run into a problem here, because since $p!$ and $(p+1)!$ are not relatively prime to $p$, I cannot divide them to the left side to get the congruence I want.

I also attempted to write out $\binom{2p+1}{p}$ and see if I can get cancellation, but I run into a problem with figuring out how to get p! to divide into the remaining parts of $(2p+1)!$ after I divide out $(p+1)!$.

Any suggestions on what I should try next?

• You should have a look at the Wikipedia page about Wolstenholme's theorem. – Jack D'Aurizio Mar 19 '17 at 15:08
• Lucas' theorem gives this more or less immediately. – Jyrki Lahtonen Mar 19 '17 at 15:11

## 3 Answers

$$\binom{2p+1}{p} = \frac{(p+2)(p+3)\cdots(2p+1)}{1\cdot2\cdots p}$$

Separating out the multiples of $p$ in the numerator and the denominator, this equals $$\frac{2p}{p}\frac{(p+2)(p+3)\cdots(2p-1)(2p+1)}{1\cdot2\cdots(p-1)}$$

And mod $p$ this is just $$2 \frac{2\cdot3\cdots(p-1)\cdot1}{1\cdot2\cdots(p-1)} =2$$

Hint: $$(u+v)^p\equiv u^p+v^p\pmod p$$ for any $u,v$. So examine $(1+x)^{2p+1}$ by examining $((1+x)^{2})^p(1+x)$.

How about a combinatorial / group theoretic proof?

We can define an action of $\mathbb{Z}/p\mathbb{Z}$ on a $2p+1$ element set by cyclically permuting the first $p$ elements, cyclically permuting the second $p$ elements and fixing the last point. Explicitly if we take out $2p+1$ element set to be $\{1,2,\dots, 2p+1\}$ the generator $1 \in \mathbb{Z}/p\mathbb{Z}$ sends $p$ to $1$, $2p$ to $p+1$, fixes $2p+1$, and sends every other $a$ to $a+1$.

This action induces an action on $p$ element subsets of our $2p+1$ element set. This splits up our set of $p$ element subsets into orbits, all of which must be of size $p$ or of size $1$.

But now by how we constructed the action it's clear that the only fixed subsets by this action are the first $p$ elements and the next $p$ after that.

Hence the total number of $p$ element subsets is $p\cdot \{\text{number of orbits of size p}\} +2$