Prove that $\binom{2n}{n}\equiv (-1)^n \pmod{(2n+1)}$ if and only if $2n+1$ is a prime number.

Question

I have conjectured that $$\binom{2n}{n}\equiv (-1)^n \pmod{(2n+1)}$$ if and only if $2n+1$ is a prime number, based on a short program that I wrote verifying this up to $n=100$.

I know that by Wilson's Theorem, $(2n)!\equiv -1 \pmod{(2n+1)}$ if and only if $(2n+1)$ is a prime number, which is as close as I can get. Any hints on how to proceed with either direction of the proof would be appreciated as I am rather stuck.

Edit: Never mind. The "only if" part is actually false. $n=2953$ is a counterexample

Answer 1

If $(2n + 1)$ is indeed prime, we are working in a field and can consider $(n!)^2$ directly, since it is invertible. Note that, mod $(2n + 1)$, we have $$ n! = n \times \cdots \times 1 = (-1)^n \times -n \times \cdots \times -1 = (-1)^n \times (n+1) \times \cdots \times 2n, $$ and so $$ (n!)^2 = (-1)^n\times (2n!), $$ which is precisely what you wanted to prove.

Answer 2

To do the "if" part, use Wilson's theorem together with the fact that $2n\times(2n-1)\times\cdots(n+1)\equiv (-1)\times(-2)\times\cdots(-n)\equiv (-1)^nn!$ mod $2n+1$. I'll carry on thinking about the "only if" :)