Example 25([46.4]) Let the sequence $a_{1}, a_{2}, \cdots$ be defined as follows: $a_{n}=2^{n}+3^{n}+6^{n}-1$ $(n=1,2, \cdots)$. Find all integers that are coprime with every term of this sequence.
It is evident that this problem can be reduced to finding all prime numbers that are coprime with each term of the sequence, i.e., primes that do not divide any term of the sequence. From \(a_{1}=10, a_{2}=48\), we know that we can consider primes \(p \geqslant 5\). For the integer sequence \(a^{n} (n=1,2, \cdots)\) and the relevant number theory results concerning division by a prime \(p\) are Fermat's Little Theorem (Chapter 1, § 4, Example 1):
$$a^{p} \equiv a(\bmod p)$$

and when \((a, p)=1\),
$$a^{p-1} \equiv 1(\bmod p)$$

Therefore, when the prime \(p \geqslant 5\), by equation (1), to check whether each term of the sequence is not divisible by \(p\), we only need to examine \(a_{1}, a_{2}, \cdots, a_{p-1}\). By equation (2), we have
$$a_{p-1} \equiv 1+1+1-1 \equiv 2(\bmod p)$$

Using the inverse modulo \(p\) (Chapter 3, § 1, Property VII), by equation (2), we have
$$a_{p-2} \equiv 2^{-1}+3^{-1}+6^{-1}-1(\bmod p)$$

This is equivalent to
$$6 a_{p-2} \equiv 6\left(2^{-1}+3^{-1}+6^{-1}-1\right) \equiv 3+2+1-6 \equiv 0(\bmod p)$$

which means \(p\) divides \(a_{p-2}\). Therefore, the integers that are coprime with every term of this sequence are only \(1, -1\).