TITLE: Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold?
QUESTION [21 upvotes]: 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
  $$ \sum_{n>10^5} \frac1{2n(n-1)+1} < 5\cdot 10^{-6}. $$

REPLY [2 votes]: Confirmed up to $n = 1e6$. No example found.
But note that there is probably nothing special about the number $2$. The same holds (also up to $n = 1e6$) with $2$ replaced by $3, 4, 5, 8$.