How to show $\binom{2p}{p} \equiv 2\pmod p$?

Question

how to prove $\forall p$ prime :

$\binom{2p}{p} \equiv 2 \pmod p$ we have:

$\binom{2p}{p} = \frac{2p (2p-1)(2p-3)...1}{p!p!}$ but how to continue?

Answer 1

It's very particular case of Lucas' theorem. Compute in $\mathbf{Z}_{p}[x]$ $$ (1 + x)^{2p} = \sum_{i=0}^{2p} \binom{2p}{i} x^{i}, $$ but also $$ (1 + x)^{2p} = ((1 + x)^{p})^{2} = (1 + x^{p})^{2} = 1 + 2 x^{p} + x^{2p} $$ and now compare the coefficients of $x^{p}$ in the two expressions.

Answer 2

Back to basics: $$ \begin{align} &{2p\choose p} \equiv 2 \mod p\\ \Leftrightarrow&\frac{(p+1)(p+2)...(p+p-1)(2p)}{p!} \equiv 2 \mod p\\ \Leftrightarrow&\frac{(p+1)(p+2)...(p+p-1)2}{(p-1)!} \equiv 2 \mod p\\ \Leftrightarrow&p\ \text{ divides }\ \left[\frac{(p+1)(p+2)...(p+p-1)2}{(p-1)!}-2\right]\tag{1} \\ \Leftrightarrow&p\ \text{ divides }\ (p-1)!\left[\frac{(p+1)(p+2)...(p+p-1)2}{(p-1)!}-2\right]\tag{2} \\ \Leftrightarrow&(p+1)(p+2)...(p+p-1)2 \equiv 2(p-1)! \mod p\\ \Leftrightarrow&(1)(2)...(p-1)2 \equiv 2(p-1)! \mod p\\ \Leftrightarrow&2(p-1)! \equiv 2(p-1)! \mod p, \end{align} $$ where $(1)\Leftarrow(2)$ because $(p-1)!$ does not contain the prime factor $p$.