TITLE: Show that for every prime $p$, there is an integer $n$ such that $2^{n}+3^{n}+6^{n}-1$ is divisible by $p$.
QUESTION [14 upvotes]: So the problem states: Show that for every prime $p$, there is an integer $n$ such that $2^{n}+3^{n}+6^{n}-1$ is divisible by $p$. 
I was thinking about trying to prove this using the corollary to Fermat's Little Theorem, that for every prime $p$,  $a^{n-1}\equiv 1 \pmod {p}$, but I can't think about how to go about doing that.   
Any help would be greatly appreciated!

REPLY [5 votes]: From FLT, $a^{p-2} \equiv a^{-1}$.  Choosing $n = p-2$,
$$\begin{align}
2^n + 3^n + 6^n - 1& \equiv 2^{-1} + 3^{-1} + 6^{-1} - 1\\
& \equiv 6^{-1} \cdot 6 \cdot (2^{-1} + 3^{-1} + 6^{-1}) - 1\\
& \equiv 6^{-1} \cdot (3 + 2 + 1) - 1\\
& \equiv 6^{-1} \cdot 6 - 1\\
& \equiv 1 - 1 \equiv 0\\
\end{align}$$