6. Find all prime pairs $(p, q)$ such that $p q \mid\left(5^{p}-2^{p}\right)\left(5^{q}-2^{q}\right)$.
6. Solution: (1) When one of $p, q$ satisfies $a \mid 5^{a}-2^{a} (a=p$ or $q)$, let's assume $p \mid 5^{p}-2^{p}, p$ is obviously not 2,5, so $(p, 2)=(p, 5)=1$. By Fermat's Little Theorem: $5^{p-1} \equiv 2^{p-1} \equiv 1(\bmod p)$, so $p \mid 3 \Rightarrow p=3$.
Since $p \cdot q \mid\left(5^{p}-2^{p}\right)\left(5^{q}-2^{q}\right)$, we have $3 q \mid 9 \times 13\left(5^{q}-2^{q}\right)$, so $q=3$ or 13.
Therefore, the pairs $(p, q)$ that satisfy the condition are $(3,13)$, $(3,3)$, or $(13,3)$.
(2) When $p \times 5^{p}-2^{p}, q \times 5^{q}-2^{q}$, it is easy to see that $p\left|5^{q}-2^{q}, q\right| 5^{p}-2^{p}$.
Assume $p \geqslant q$, then $q \mid 5^{p}-2^{p}$,
and since $p \geqslant q$, we have $(p, q-1)=1$. By Bézout's Theorem, there exists $v \in \mathbf{N}_{+}$,
such that $p u-(q-1) v=1$. From (1), we get:
$q \mid 5^{\rho \alpha}-2^{\rho \alpha}$, i.e., $q \mid 5^{(q-1) v+1}-2^{(q-1) v+1}$.
It is also easy to see that $5^{q-1} \equiv 2^{q-1} \equiv 1(\bmod q)$, so $q \mid 5-2 \Rightarrow q=3$, which contradicts $q \times 5^{q}-2^{q}$!
In conclusion, the pairs $(p, q)$ that satisfy the condition are $(3,3)$, $(3,13)$, or $(13,3)$.