$ p | 2^{(p-1)/2}-1 $
p is prime. I can prove that if $(p-1)/2$ is composite, then $ 2^{(p-1)/2}-1 $ is also composite. I cannot manage to deduce anything about $p$. I'm trying to come up with some set of criteria that $p$ must fulfil such that $ p | 2^{(p-1)/2}-1 $
I've experimented with $2^{(p-1)/2} \equiv 1 \pmod p$ but didn't really get anywhere. I think this question is leading me into quadratic reciprocity, so I shouldn't need quadratic reciprocity to solve it (so should, presumably, be able to do it with things like the fundamental theorem of arithmetic, Fermat's Little Theorem, and things of that nature). I've also tried treating this as $ (2^{(p-1)/4})(2^{(p-1)/4}) \equiv 1\pmod p$ which looked like it was going somewhere for a while.
This is supposed to be the easy one mark intro to a much bigger question and should presumably take about sixty seconds. I've spent a few hours on it so far and got nowhere. I am terrible at number.
Edited to confirm that p is prime.