Let $p$ be a positive integer; assumeif $2p+1$ is prime then it is easily checked that $$(2p+1)\mid\left(\binom{2p}{p}+(-1)^{p-1}\right)$$$$(2p+1)\mid\left(\binom{2p}{p}+(-1)^{p-1}\right);$$
conversely I conjecture that if the above divisibility assumption holds, then $2p+1$ is a prime number. Is this true?