Well, the title does not tell the whole story; the complete question is:

Are there any primes of the form $p=2n(n-1)+1$, with integer $n\ge 1$, such that $$ \binom{2n}{n} \equiv 2\pmod p ? $$

Primes $p=2n(n-1)+1$ are not that rare: say, of all numbers of this form up to $2\cdot10^{10}$, some 12.7% are prime; however, for none of these primes the congruence above holds true.

A simple heuristic is as follows. Let's say that a prime is *bad* if it satisfies the congruence. If the binomial coefficients $\binom{2n}{n}$ were distributed uniformly modulo $p$, then the probability that a particular prime $p$ is bad would be about $1/p$. Also, the probability that bad primes exist does not exceed the sum of probabilities for all primes $p=2n(n-1)+1>2\cdot10^{10}$ to be "bad". Hence, this probability is at most
$$ \sum_{n>10^5} \frac1{2n(n-1)+1} < 5\cdot 10^{-6}. $$

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