(8) Let $p$ and $q$ be two different prime numbers, then the remainder when $p^{q-1}+q^{p-1}$ is divided by $p \cdot q$ is $\qquad$
81 Prompt: Since $p$ and $q$ are different prime numbers, by Fermat's theorem, we know that $p^{q-1} \equiv 1(\bmod q)$.
Also, $q^{p-1} \equiv 0(\bmod q)$, so $p^{q-1}+q^{p-1} \equiv 1(\bmod q)$.
Similarly, $p^{q-1}+q^{p-1} \equiv 1(\bmod p)$.
Since $[p, q]=p q$, it follows that $p^{q-1}+q^{p-1} \equiv 1(\bmod p \cdot q)$.