G3.2 Suppose the equation $x^{2} y-x^{2}-3 y-14=0$ has only one positive integral solution $\left(x_{0}, y_{0}\right)$. If $x_{0}+y_{0}=b$, find the value of $b$.
$\begin{array}{l}(y-1) x^{2}=3 y+14 \\ x^{2}=\frac{3 y+14}{y-1}=\frac{3 y-3+17}{y-1}=3+\frac{17}{y-1}=3+1 \\ y=18, x=2 \\ b=20\end{array}$