Example 10 (Recommended by the Soviet Ministry of Education in 1990) Find such values of $a$ that the roots of the equation $x^{2}-$ $a x+9 a=0$ are integers.

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Let $x_{1}, x_{2}$ be the integer roots of the equation $x^{2} - a x + 9 a = 0$. Then, by Vieta's formulas, we have
$$
\begin{array}{l}
x_{1} + x_{2} = a, \\
x_{1} x_{2} = 9 a .
\end{array}
$$

Thus, $x_{1} x_{2} = 9(x_{1} + x_{2})$,
$$
\begin{array}{l}
x_{1} x_{2} - 9 x_{1} - 9 x_{2} + 81 = 81, \\
(x_{1} - 9)(x_{2} - 9) = 81 .
\end{array}
$$

Note that 81 has the following 6 factorizations:
$$
81 = 1 \cdot 81 = (-1) \cdot (-81) = 3 \cdot 27 = (-3) \cdot (-27) = 9 \cdot 9 = (-9) \cdot (-9) \text{.}
$$

Therefore, $a = x_{1} + x_{2} = (x_{1} - 9) + (x_{2} - 9) + 18$, which gives us the following 6 numbers:
$\pm 82 + 18, \pm 30 + 18, \pm 18 + 18$, i.e.,
$$
a = 100, -64, 48, -12, 36, 0 \text{.}
$$

Thus, when $a$ takes any of these 6 integer values, the given equation has integer roots.