Show that for all primes \(p\) there is a \(n \in \N\) such that
\[2^n + 3^n + 6^n - 1 \equiv 0 \quad (mod \quad p)\]
This being a modulo problem with primes, we should keep in mind Fermat's Little Theorem:
\[p \space \text{prime}, a \in \N \Rightarrow a^p \equiv a \space (mod \space p)\]
Let's first find \(n\) for the cases \(p = 2\) and \(p = 3\).
\[p = 2; \quad 2 + 3 + 6 - 1 \equiv_2 10 \equiv_2 0\]
\[p = 3; \quad 2^2 + 3^2 + 6^2 - 1 \equiv_3 48 \equiv_3 0\]
Now, we can use Fermat's Little Theorem to state that, if \(p > 3\),
\[2^{p-1} \equiv 3^{p-1} \equiv 6^{p-1} \equiv 1 \space (mod \space p)\]
so
\[2^{p-2} \equiv_p 2^{-1}, 3^{p-2} \equiv_p 3^{-1}, 6^{p-2} \equiv_p 6^{-1},\]
We need \(p > 3\) for these inverses to be well defined. We therefore have
\[2^{p-2} + 3^{p-2} + 6^{p-2} \equiv_p 2^{-1} + 3^{-1} + 6^{-1} \equiv_p 6^{-1}(3+2+1) \equiv_p 6^{-1}(6) \equiv_p 1\]
Which gives us the final result
\[2^{p-2} + 3^{p-2} + 6^{p-2} - 1 \equiv_p 0\]
And we are done.