4. Find all prime numbers $p, q$ such that $\frac{\left(5^{p}-2^{p}\right)\left(5^{q}-2^{q}\right)}{p q}$ is a positive integer.
4. Lemma: Let $n_{0}$ be the smallest value of $n$ such that $s \mid 5^{n}-2^{n}, n \in \mathbf{N}^{*}, s$ is a prime. Then for all positive integers $n$ satisfying $s \mid 5^{n}-2^{n}$, we have $n_{0} \mid n$.

Proof of the lemma: If $n_{0} \nmid n$, let $n=q m_{0}+r,\left(1 \leqslant r \leqslant n_{0}-1\right)$. Since $s \mid 5^{n}-2^{n}$, we have $s \mid 5 m_{0}+r-2 m_{0}+r$. And since $s \mid 5^{n_{0}}-2^{n_{0}}$, we have $s \mid 5^{m_{0}}-2^{m_{0}}$, thus $s \mid 5^{m_{0}+r}-2^{m_{0}} \cdot 5^{r}$. Therefore, $s \mid 2^{m_{0}} \cdot 5^{r}-2^{m_{0}+r}=2^{m_{0}}\left(5^{r}-2^{r}\right)$. Since $(s, 2)=1$, we have $s \mid 5^{r}-2^{r}$, which contradicts the minimality of $n_{0}$. Hence, the lemma holds. Returning to the original problem:
(1) When one of $p, q$ is 3, without loss of generality, let $p=3$, then $\frac{\left(5^{p}-2^{p}\right)\left(5^{q}-2^{q}\right)}{p q}=\frac{117\left(5^{q}-2^{q}\right)}{3 q}=\frac{39\left(5^{q}-2^{q}\right)}{q}$ is an integer. By Fermat's Little Theorem, $q\left|5^{q}-5, q\right| 2^{q}-2$, so $5^{q}-2^{q} \equiv 5-2 \equiv 3(\bmod q)$. Since $q \mid 39\left(5^{q}-2^{q}\right)$, we have $q \mid 39 \times 3$, so $q=3$ or 13.
(2) If $p, q$ are not 3, by $5^{q}-2^{q} \equiv 3(\bmod q), 5^{p}-2^{p} \equiv 3(\bmod p)$, we know $p\left|5^{q}-2^{q}, q\right| 5^{p}-2^{p}$. Without loss of generality, assume $p \leqslant q$, let $b$ be the smallest value of $n$ such that $p \mid 5^{n}-2^{n}$. It is easy to see that $p \neq 2,5$, by Fermat's Little Theorem, $5^{p-1} \equiv 1(\bmod p), 2^{p-1} \equiv$ $1(\bmod p)$, so $p \mid 5^{p-1}-2^{p-1}$. By the lemma, $b|p-1, b| q$. Since $p \neq 3$, we have $b>1$. Since $q$ is a prime, we have $b=$ $q$, so $q \mid p-1$, thus $q \leqslant p-1$, which is a contradiction. Therefore, there are no such $p, q$ in this case.
In summary, the values of $(p, q)$ are $(3,3),(3,13),(13,3)$.