Chinese Mathematical Olympiad

If $2 \mid pq$, we suppose that $p = 2$ without loss of generality, and then $q \mid 5^q + 25$. By Fermat's theorem we have $q \mid 5^q - 5$, so $q \mid 30$, where $(2, 3)$ and $(2, 5)$ are solutions [(2, 2) does not fit].
If $5 \mid pq$, we suppose that $p = 5$ without loss of generality and then $5q \mid 5^q + 5^5$. By Fermat's theorem we have $q \mid 5^{5^q - 1}$, so $q \mid 313$, where $(5, 5)$ and $(5, 313)$ are solutions.
Otherwise, we have $pq \mid 5^{p-1} + 5^{q-1}$, and so
$$5^{p-1} + 5^{q-1} \equiv 0 \pmod{p}.$$
By Fermat's theorem, $5^{p-1} \equiv 1 \pmod{p}$,
and because of the above, $5^{q-1} \equiv -1 \pmod{p}$.
Denote by $p-1=2^k(2r-1)$, $q-1=2^l(2s-1)$, where $k, l, r, s$ are positive integers.
If $k \le l$, because of the previous congruences we get
$$\begin{aligned} 1 &= 1^{2^{l-k}(2s-1)} \equiv (5^{p-1})^{2^{l-k}(2s-1)} \\ &= 5^{2^l(2r-1)(2s-1)} = (5^{q-1})^{2r-1} = (-1)^{2r-1} \\ &= -1 \pmod{p}, \end{aligned}$$
a contradiction of $p \ne 2$. So $k > l$.
But we have $k < l$ by a similar argument — a contradiction.
Therefore, all possible pairs of primes $(p, q)$ are $(2, 3)$, $(3, 2)$, $(2, 5)$, $(5, 2)$, $(5, 5)$, $(5, 313)$ and $(313, 5)$.