$a_n=2^n+3^n+6^n-1$. Find all positive integers that are primes to all terms of the sequence. Let the sequence $a_n=2^n+3^n+6^n-1, n\in\mathbb N_{> 0}$. Find all positive integers that are prime to all terms of this sequence. I have no idea how to approach this, but I know that I CAN'T use Euler's theorem or any generalization/extension of fermat's little theorem as it's on a later chapter in the textbook.
If possible, fully explain all steps.

Thought it could be useful to cite the solution from The IMO Compendium A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009:

We will show that $1$ is the only such number. It is sufficient to prove that for every prime number $p$ there exists some $a_m$ such that $p \mid a_m$. For $p=2,3$ we have $p\mid a_2=48$. Assume now that $p>3$. Applying Fermat's theorem, we have 
  $$
6a_{p-2}=3\cdot 2^{p-1}+2\cdot 3^{p-1}+6^{p-1}-6\equiv 3+2+1-6=0 \pmod{p}.
$$
  Hence $p \mid a_{p-2}$, i.e. $\gcd(p,a_{p-2})=p>1$. This completes the proof.

Notice this is very similar to @Exodd's answer, although it avoids fractions.
