Find all $n \in \mathbb{N}$ such that ${{2n}\choose{n}} \equiv (-1)^n\pmod{2n+1}$ 
Find all $n \in \mathbb{N}$ such that $${{2n}\choose{n}} \equiv (-1)^n\pmod{2n+1}.$$

I know that if $2n+1$ were prime number, then 
$${{2n}\choose{n}} = \frac{(2n) (2n-1) \cdots (n+1)}{n!} \equiv \frac{(-1)(-2)\cdots(-n)}{n!} = (-1)^n \pmod{2n+1}.$$
However, I'm not sure whether $\{ n \,|\, 2n+1 \text{ are prime}\}$ are the only possible solutions. 

The solutions are those $n \in \mathbb N$ for which $2n+1$ is either prime or a Catalan pseudoprime.
We say that $2n+1$ is a Catalan pseudoprime if it is a composite number and
$$(-1)^n\, C_n \equiv 2 \pmod{2n+1}$$
where $C_n$ is the $n$-th Catalan number, that is,
$$C_n = \frac 1 {n+1} \binom {2n} n.$$
Rewriting the definition, we see that this means
$$(-1)^n \frac 1 {n+1} \binom {2n} n \equiv 2 \pmod {2n+1}$$
and multiplying by $(-1)^n (n+1)$,
$$\binom {2n} n \equiv (-1)^n (2n + 2) \equiv (-1)^n \pmod {2n+1}$$
which is the original congruence.
The only known Catalan pseudoprimes are $5907$, $1194649$ and $12327121$. So the only other solutions to the congruence that we know of are $2953$, $597324$ and $6163560$.
