Q. let $p$ and $q$ be distinct odd prime numbers. Prove that for any $x \in \mathbb{Z} /pq$ we get $$x^{(pq-p-q+3)/2} \equiv x \mod pq$$
we have just learnt the chinese remainder theorem so I have tried to use it but I'm getting on where. Please help!
Hints: Little Fermat tells you that $$ x^p\equiv x\pmod p $$ and the corresponding congruence with $q$. As a consequence of this you can surely prove that $$ x^{1+k(p-1)}\equiv x\pmod p $$ for any natural number $k$. A key observation is that $$ \frac{pq-p-q+3}2=1+\frac{(p-1)(q-1)}2. $$ Stare at this for a while. It will give you bits that you can combine with the Chinese Remainder Theorem.