13. Given that $p$ and $q$ are both prime numbers, and that the quadratic equation in $x$, $x^{2}-(8 p-10 q) x+5 p q=0$, has at least one positive integer root. Find all prime number pairs $(p, q)$.
13. From the sum of the two roots of the equation being $8 p-10 q$, we know that if one root is an integer, the other root is also an integer. From the product of the two roots being $5 p q$, we know that the other root is also a positive integer.

Let the two positive integer roots of the equation be $x_{1}$ and $x_{2} \left(x_{1} \leqslant x_{2}\right)$. By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=8 p-10 q, \\
x_{1} x_{2}=5 p q .
\end{array}
$$

From equation (2), we get the following possible cases for $x_{1}$ and $x_{2}$:
$$
\left\{\begin{array}{l}
x_{1}=1,5, p, q, 5 p, 5 q, \\
x_{2}=5 p q, p q, 5 q, 5 p, q, p,
\end{array}\right.
$$

Substitute $x_{1}+x_{2}=1+5 p q, 5+p q, p+5 q, q+5 p$ into equation (1).

When $x_{1}+x_{2}=1+5 p q$, $1+5 p q=8 p-10 q$, but $1+5 p q>10 p>8 p-10 q$, so there is no solution in this case.
When $x_{1}+x_{2}=5+p q$, $5+p q=8 p-10 q$, then we have $(p+10)(q-8)=-85$.
Since $p$ and $q$ are both primes, it is only possible that
$$
\left\{\begin{array}{l}
q-8=-5,-1, \\
p+10=17,85 .
\end{array}\right.
$$

Therefore, $(p, q)=(7,3)$.
When $x_{1}+x_{2}=p+5 q$, $p+5 q=8 p-10 q$, so $7 p=15 q$, which is impossible.

When $x_{1}+x_{2}=q+5 p$, $q+5 p=8 p-10 q$, so $3 p=11 q$, thus, we have $(p, q)=(11,3)$.
In summary, the prime pairs $(p, q)$ that satisfy the conditions are $(7,3)$ or $(11,3)$.