Find all the primes $p,q$ such that $2^{p-q}+1\equiv0\pmod{pq}$
I'm not sure how to start this. I am guessing Fermat's little theorem has something to do with this as $2^p\equiv 2\pmod{p}$ and $2^q\equiv 2\pmod{q}$. but I can't make any progress... Any help is appreciated thanks!
Note first that you can rewrite your starting equation as two equations via the Chinese Remainder theorem. Now you need to examine $2^{p-q}\equiv -1 \pmod{p}, 2^{p-q} \equiv -1 \pmod{q}$. Now you can apply Fermat's little theorem and take reciprocals to get to the equations $2^{p-1} \equiv -1 \pmod{q}, 2^{q-1}\equiv -1 \pmod{p}$.
Since $p-1$ and $q-1$ are even, these equations say that $-1$ is expressible as the even power of something in both $(\Bbb Z/p\Bbb Z)^\times$ and $(\Bbb Z/q\Bbb Z)^\times$. You should pursue determining which primes $p,q$ for which this is true, as a start.