N57 (46-4, Poland) The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
a_{n}=2^{n}+3^{n}+6^{n}-1 \quad(n=1,2,3, \cdots) .
$$

Find all positive integers that are coprime with every term of this sequence.
We first prove the following conclusion: For any prime $p$ not less than 5, we have
$$
2^{p^{\cdots 2}}+3^{p^{\cdots 2}}+6^{p-2}-1 \equiv 0 \quad(\bmod p) .
$$

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,
$$
2^{p-1} \equiv 1 \quad(\bmod p), 3^{p-1} \equiv 1 \quad(\bmod p), 6^{p-1} \equiv 1 \quad(\bmod p),
$$

Thus,
$$
3 \cdot 2^{p-1}+2 \cdot 3^{p-1}+6^{p-1} \equiv 3+2+1=6 \quad(\bmod p),
$$

which means
$$
6 \cdot 2^{p-2}+6 \cdot 3^{p-2}+6 \cdot 6^{p-2} \equiv 6 \quad(\bmod p),
$$

Therefore,
$$
2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0 \quad(\bmod p) .
$$

So (1) holds, and thus,
$$
a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0 \quad(\bmod p) .
$$

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.