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

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!

Answer 1

Observe that for $p>3$ we have $$ 6(2^{p-2}+3^{p-2}+6^{p-2}-1)\equiv 3+2+1-6\equiv 0\pmod p $$

Answer 2

The key hint is that rationally $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$. You need to do something analogous mod $p$. Since you need $n>0$, Fermat's little theorem will indeed be useful.

Answer 3

From FLT, $a^{p-2} \equiv a^{-1}$. Choosing $n = p-2$,

$$\begin{align} 2^n + 3^n + 6^n - 1& \equiv 2^{-1} + 3^{-1} + 6^{-1} - 1\\ & \equiv 6^{-1} \cdot 6 \cdot (2^{-1} + 3^{-1} + 6^{-1}) - 1\\ & \equiv 6^{-1} \cdot (3 + 2 + 1) - 1\\ & \equiv 6^{-1} \cdot 6 - 1\\ & \equiv 1 - 1 \equiv 0\\ \end{align}$$