8. Let $p$ and $q$ be two different prime numbers. Then the remainder when $p^{q-1}+q^{p-1}$ is divided by $p q$ is $\qquad$
8. 1 .

Since $p$ and $q$ are different prime numbers, by Fermat's Little Theorem, we have
$$
p^{q-1} \equiv 1(\bmod q) \text {. }
$$

Also, $q^{p-1} \equiv 0(\bmod q)$, thus
$$
p^{q-1}+q^{p-1} \equiv(\bmod q) \text {. }
$$

Similarly, $p^{q-1}+q^{p-1} \equiv 1(\bmod p)$.
Therefore, $p^{q-1}+q^{p-1} \equiv 1(\bmod p q)$.