2. Find all prime pairs $(p, q)$ such that $p q \mid\left(5^{p}+5^{q}\right)$.
(Fu Yunhao)
2. If $21 p q$, let's assume $p=2$. Then
$$
2 q\left|\left(5^{2}+5^{q}\right) \Rightarrow q\right|\left(5^{q}+25\right) \text {. }
$$

By Fermat's Little Theorem, we know $q \mid\left(5^{q}-5\right)$, thus $q \mid 30$, i.e., $q=2,3,5$.
It is easy to verify that the prime pair $(2,2)$ does not meet the requirements, while $(2,3)$ and $(2,5)$ do.
If $p q$ is odd and $51 p q$, let's assume $p=5$. Then
$$
5 q\left|\left(5^{5}+5^{q}\right) \Rightarrow q\right|\left(5^{q-1}+625\right) \text {. }
$$

When $q=5$, the prime pair $(5,5)$ meets the requirements.
When $q \neq 5$, by Fermat's Little Theorem, we have $q \mid\left(5^{q-1}-1\right)$, hence $q \mid 626$. Since $q$ is an odd prime, and the only odd prime factor of 626 is 313, so $q=313$.
Upon verification, the prime pair $(5,313)$ meets the requirements.
If $p$ and $q$ are neither 2 nor 5, then
$$
p q \mid\left(5^{p-1}+5^{q-1}\right) \text {. }
$$

Thus, $5^{p-1}+5^{q-1} \equiv 0(\bmod p)$.
By Fermat's Little Theorem, we get
$$
5^{p-1} \equiv 1(\bmod p) \text {. }
$$

From equations (1) and (2), we have
$$
5^{q-1} \equiv -1(\bmod p) \text {. }
$$

Let $p-1=2^{k}(2 r-1), q-1=2^{l}(2 s-1)$ $(k, l, r, s$ are positive integers $)$.
If $k \leqslant l$, then from equations (2) and (3), it is easy to see
$$
\begin{array}{l}
1=1^{2^{l-k}(2 s-1)} \equiv\left(5^{p-1}\right)^{2^{l-k}(2 s-1)} \\
\pm 5^{2^{l}(2 r-1)(2 t-1)}=\left(5^{q-1}\right)^{2 r-1} \\
\equiv(-1)^{2 r-1} \equiv-1(\bmod p),
\end{array}
$$

which contradicts $p \neq 2$. Therefore, $k>l$.
Similarly, $k<l$ leads to a contradiction.
Thus, there are no $(p, q)$ that meet the requirements in this case.
In summary, the prime pairs $(p, q)$ that satisfy the problem's requirements are $(2,3),(3,2),(2,5),(5,2),(5,5)$, $(5,313),(313,5)$.