Think about Question 1 The sequence $a_{1}, a_{2}, \cdots$ is defined as follows: $a_{n}=2^{n}+3^{n}+6^{n}-1, n=1,2,3, \cdots$ Find all positive integers that are coprime to every term of this sequence.
Solution 1: We first prove the following conclusion: For any prime $p$ not less than 5, we have
$$
2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p)
$$

Since $p$ is a prime not less than 5, we have $(2, p)=1,(3, p)=1,(6, p)=1$. By Fermat's Little Theorem, $2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), 6^{p-1} \equiv 1(\bmod p)$.
Therefore, $3 \cdot 2^{p-1}+2 \cdot 3^{p-1}+6^{p-1} \equiv 3+2+1=6(\bmod p)$
That is, $6 \cdot 2^{p-2}+6 \cdot 3^{p-2}+6 \cdot 6^{p-2} \equiv 6(\bmod p)$
Hence, $2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p)$
So (1) holds, and thus
$$
a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p).
$$

Also, $a_{1}=10, a_{2}=48$.
For any positive integer $n$ greater than 1, it must have a prime factor $p$. If $p \in\{2,3\}$, then $\left(n, a_{2}\right)>1$; if $p \geqslant 5$, then $\left(n, a_{p-2}\right)>1$, so no positive integer greater than 1 meets the requirement.
However, 1 is coprime with all positive integers, so the only positive integer that meets the requirement is 1.

Solution 2: Let $b_{n}=a_{n}+1=2^{n}+3^{n}+6^{n}, n=-1,0,1,2, \cdots$
Then $b_{0}=3, b_{-1}=1$. By the characteristic root method, it is easy to know
$$
b_{n+3}=11 b_{n+2}-36 b_{n+1}+36 b_{n}, n=-1,0,1, \cdots
$$
(If $f(x)=(x-2)(x-3)(x-6)$, then $2^{n} f(2)+3^{n} f(3)+6^{n} f(6)=0$ leads to the above equation)
For any prime $p$ greater than 3, we have $(p, 6)=1$. Let $\bar{m}$ be the remainder of the integer $m$ divided by $p$. Since there are only $p^{3}$ different 3-tuples $\left(\overline{m_{1}}, \overline{m_{2}}, \overline{m_{3}}\right)$, by the pigeonhole principle, there must exist $i, j,-1 \leqslant i<j$, such that $\left(\overline{b_{i}}, \overline{b_{i+1}}, \overline{b_{i+2}}\right)=\left(\overline{b_{j}}\right.$, $\left.\overline{b_{j+1}}, \overline{b_{j+2}}\right)$, then
$$
\begin{array}{l}
36 b_{i-1} \equiv b_{i+2}-11 b_{i+1}+36 b_{i} \equiv b_{j+2}-11 b_{j+1}+36 b_{j} \equiv 36 b_{j-1}(\bmod p) \\
p\left|36\left(b_{i-1}-b_{j-1}\right) \Rightarrow p\right|\left(b_{i-1}-b_{j-1}\right) \Rightarrow\left(\overline{b_{i-1}}, \overline{b_{i}}, \overline{b_{i-1}}\right)=\left(\overline{b_{j-1}}, \overline{b_{j}}, \overline{b_{j-1}}\right)
\end{array}
$$

By induction, we can get

In particular,
$$
\left(\overline{b_{-1}}, \overline{b_{0}}, \overline{b_{l}}\right)=\left(\overline{b_{j-i-1}}, \overline{b_{j-1}}, \overline{b_{j-i+1}}\right)
$$

Thus,
$$
\begin{array}{c}
b_{j-i-1} \equiv b_{-1}=1(\bmod p) \\
p\left|b_{j-i-1}-1 \Rightarrow p\right| a_{j-i-1}
\end{array}
$$

If $j-i-1=0$, then $a_{0}=2, p \mid 2$, which is a contradiction, so $j-i-1 \geqslant 1$. Since $a_{2}=48$, the only positive number that is coprime with every term of this sequence is 1.