Example 6 Find integers $0<a_{1}<a_{2}<\cdots<a_{n}$, such that
$$
x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)+1
$$

can be factored over the integers.
【Analysis】Let $a_{0}=0$. Assume
$$
x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)+1=P(x) Q(x) \text {, }
$$

where $P(x)$ and $Q(x)$ are polynomials with integer coefficients, and
$1 \leqslant \operatorname{deg} P(x), \operatorname{deg} Q(x) \leqslant n$.
Let $x=a_{i}(i=0,1, \cdots, n)$. Then
$$
P\left(a_{i}\right) Q\left(a_{i}\right)=1 \text {. }
$$

Since $P\left(a_{i}\right)$ and $Q\left(a_{i}\right)$ are integers, we have
$$
P\left(a_{i}\right)-Q\left(a_{i}\right)=0 \text {. }
$$

Let $F(x)=P(x)-Q(x)$. Then $F(x)=0$ has $n+1$ roots $a_{0}, a_{1}, \cdots, a_{n}$.
Thus $F(x) \equiv 0 \Rightarrow P(x) \equiv Q(x)$
$$
\Rightarrow x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)+1=P^{2}(x) \text {. }
$$

Since the right-hand side of the above equation is a $2 \operatorname{deg} P(x)$ degree polynomial and the left-hand side is an $n+1$ degree polynomial, we know that $n$ must be a positive odd number.
When $n=1$,
$$
x\left(x-a_{1}\right)+1=(x-1)^{2} \text {, }
$$

we get $a_{1}=2$.
When $n=3$,
$$
x\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)+1=P^{2}(x) .
$$

Thus the quadratic polynomial $P(x)$ satisfies $P\left(a_{i}\right)= \pm 1$.
Combining $0=a_{0}\frac{n+1}{2}=\operatorname{deg} P(x)$, we reach a contradiction.