Example 3 Given the sequence $\left\{a_{n}\right\}$ with the general term formula
$$
a_{n}=2^{n}+3^{n}+6^{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Find the positive integers that are coprime to each term of this sequence. ${ }^{[3]}$
(46th IMO Shortlist Problem)
First, consider a prime number $p$ that is coprime with any $a_{n}$. Since $a_{2}=48$, any such $p \neq 2, 3$. Next, we prove that for any prime $p \geq 5$, $p \mid a_{p-2}$.

In fact, extend the definition to the set of negative integers.
When $n=-1$, $a_{-1}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1=0$.
When the prime $p \geq 5$,
$$
(2, p)=(3, p)=(6, p)=1 \text{. }
$$

By Fermat's Little Theorem,
$$
2^{p-1} \equiv 3^{p-1} \equiv 6^{p-1} \equiv 1(\bmod p) \text{. }
$$

Consider the periodicity of the sequence modulo $p$, we have $a_{k+p-1} \equiv a_{k}(\bmod p)$.
In particular, when $k=-1$,
$$
a_{p-2} \equiv a_{-1}(\bmod p) \Rightarrow a_{p-2} \equiv 0(\bmod p) \text{. }
$$

Here, $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$ are considered as the modular inverses of $2, 3, 6$ modulo $p$, respectively. At this point, $a_{k} \equiv a_{k+p-1}(\bmod p)$ for $k \geq 1$;
When $p \geq 5$, $\left(p, a_{p-2}\right)=p>1$.
Thus, any positive integer that has a prime factor does not satisfy the condition.
Therefore, the only positive integer that satisfies the condition is 1.
[Note] 1. To avoid the use of modular inverses and the extension of the sequence, the proof can be as follows:
$$
\begin{array}{l}
\text{When the prime } p \geq 5, \\
(6, p)=1 \\
\Rightarrow 3 \times 2^{p-1}+2 \times 3^{p-1}+6^{p-1} \\
\quad \equiv 3+2+1=6(\bmod p) \\
\Rightarrow 6 \times 2^{p-2}+6 \times 3^{p-2}+6 \times 6^{p-2} \equiv 6(\bmod p) \\
\Rightarrow 2^{p-2}+3^{p-2}+6^{p-2} \equiv 1(\bmod p) \\
\Rightarrow 2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p) \\
\Rightarrow p \mid a_{p-2} \text{. }
\end{array}
$$
(Turn to page 19)
(Continue from page 11)
2. The solution to this problem is due to a "coincidence":
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1=0 \text{. }
$$

By introducing the extended definition $a_{-1}$ and combining it with Fermat's Little Theorem, the handling of $a_{p-2}$ becomes natural.
3. Since $f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{1}{3}\right)^{x}+\left(\frac{1}{6}\right)^{x}-1$ is strictly monotonically decreasing for $x \in (0, +\infty)$, $f(x)=0$ if and only if $x=1$.