Example 4 Given the equation $a^{2} x^{2}-\left(3 a^{2}-8 a\right) x+$ $2 a^{2}-13 a+15=0$ (where $a$ is a non-negative integer) has at least one integer root. Then, $a=$ $\qquad$
Solution: Clearly, $a \neq 0$. Therefore, the original equation is a quadratic equation in $x$.
$$
\begin{aligned}
\Delta & =\left[-\left(3 a^{2}-8 a\right)\right]^{2}-4 a^{2}\left(2 a^{2}-13 a+15\right) \\
& =[a(a+2)]^{2}
\end{aligned}
$$

is a perfect square.
Thus, $x=\frac{\left(3 a^{2}-8 a\right) \pm a(a+2)}{2 a^{2}}$,

which gives $x_{1}=\frac{2 a-3}{a}=2-\frac{3}{a}, x_{2}=\frac{a-5}{a}=1-\frac{5}{a}$.
Therefore, by the method of multiple and divisor analysis, $a=1,3$ or 5.
2. When $\Delta=b^{2}-4 a c \geqslant 0$ and is not a perfect square, there are generally the following three approaches to consider.