1. (20 points) Given that $p$ and $q$ are integers, and are the roots of the equation in $x$
$$
x^{2}-\frac{p^{2}+11}{9} x+\frac{15}{4}(p+q)+16=0
$$

Find the values of $p$ and $q$.
$$
\text { Three, 1. According to }\left\{\begin{array}{l}
p+q=\frac{p^{2}+11}{9}, \\
p q=\frac{15}{4}(p+q)+16,
\end{array}\right.
$$

it is known that $p+q>0, p q>0$. Therefore, $p, q$ are positive integers.
From equation (2), we have $16 p q=60(p+q)+16^{2}$, which means
$$
\begin{array}{l}
(4 p-15)(4 q-15)=16^{2}+15^{2}=481 . \\
\text { And } 481=1 \times 481=13 \times 37=-1 \times(-481) \\
=-13 \times(-37),
\end{array}
$$

For positive integers $p, q$, $4 p-15$ and $4 q-15$ cannot take the values -37 and -481.
Thus, from equation (1), there are only four possible scenarios:
$$
\left\{\begin{array}{l}
4 p-15=1,481,13,37, \\
4 q-15=481,1,37,13 .
\end{array}\right.
$$

Solving these, we get $\left\{\begin{array}{l}p=4,124,7,13, \\ q=124,4,13,7 .\end{array}\right.$
Substituting into equation (1), only $p=13, q=7$ satisfies the given conditions.
In this case, the original equation becomes $x^{2}-20 x+91=0$, which clearly has two roots 13 and 7.
Therefore, there is only one solution $p=13, q=7$.