Conjecture:
$n^{2p-1}\equiv n\pmod {2p}$ for all $n\in \mathbb N$ and all odd primes $p$.
I started investigate the least $x_n$ such that $p_n^{p_{n+1}}\equiv x_n\pmod{p_{n+2}}$ and ended up with this for me unknown conjecture. I've tested it for a lot of random values of $n$ and $p$.
Surely $n^{2p-1}-n$ is always even then it remains to show that $p$ divides $n^{2p-1}-n$. If $p\mid n$ it's obvious and if $p\nmid n$ it is true according to Fermat's little theorem https://en.wikipedia.org/wiki/Fermat%27s_little_theorem