Prove that every prime number divides some number in the sequence $ a_n = 2^n+3^n+6^n-1 $

Question

Let $ (a_n) $ be a sequence of numbers such that for all natural numbers $ n $:

$$ a_n = 2^n+3^n+6^n-1 $$

Show that every prime number divides some number in that sequence.

Answer 1

Let $p$ be prime number. Consider $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$. Then \begin{align*} 6a_{p-2} & \equiv 6(2^{p-2}+3^{p-2}+6^{p-2})-6 \pmod{p}\\ & \equiv 3\cdot 2^{p-1}+2\cdot 3^{p-1}+6^{p-1}-6\pmod{p}\\ & \equiv 3+2+1 -6\pmod{p}\\ & \equiv 0 \pmod{p}. \end{align*}

For $p>3$, we will have $\gcd(6,p)=1$. So we can multiply by $6^{-1}$ on both sides to get $$a_{p-2} \equiv 0 \pmod{p}$$

Now you can deal with $p=2,3$ as special case.