Example 8 Given that $P(x)$ is a polynomial with integer coefficients, satisfying $P(17)=10, P(24)=17$. If the equation $P(n)=n+3$ has two distinct integer solutions $n_{1}, n_{2}$, find the value of $n_{1} n_{2}$. ${ }^{[7]}$
(2005, American Invitational Mathematics Examination)
【Analysis】From the conditions of the problem, we cannot determine the zeros of the integer-coefficient polynomial $P(x)$. Let's construct another polynomial function $T(x)$ such that 17 and 24 are zeros of $T(x)$, and solve the problem using the zero-product property.
Solution Let $S(x)=P(x)-x-3$. Then
$S(17)=-10, S(24)=-10$.
Let $T(x)=S(x)+10=P(x)-x+7$. Hence, 17 and 24 are zeros of the integer-coefficient polynomial $T(x)$.

Assume $T(x)=(x-17)(x-24) Q(x)$, where $Q(x)$ is an integer-coefficient polynomial.

Let $n$ be an integer root that satisfies the equation $P(n)=n+3$. Then
$$
\begin{array}{l}
S(n)=0, \\
T(n)=(n-17)(n-24) Q(n)=10 .
\end{array}
$$

Therefore, $n-17$ and $n-24$ are factors of 10.
Thus, $\left\{\begin{array}{l}n-17=2, \\ n-24=-5\end{array}\right.$ or $\left\{\begin{array}{l}n-17=5, \\ n-24=-2 .\end{array}\right.$
Solving these, we get $n_{1}=19, n_{2}=22$.
Therefore, $n_{1} n_{2}=19 \times 22=418$.