Let $P(n) = 2^n + 3^n - 5^n $.
I want to prove that $P(n)$ is divisible by $3$ for all integers $n\geq 1$.
The basis step for this proof is easy enough: $P(1)$ is divisible by $3$.
For the inductive step, I let $k$ be an arbitrary integer, then assume $P(k)$ is divisible by $3$, and set out to prove that $P(k+1)$ is divisible by $3$.
$ P(k) = 2^k + 3^k - 5^k $
$\begin{align*} P(k + 1) &= 2^{k+1} + 3^{k+1} - 5^{k+1}\\ &= 2*2^k + 3*3^k - 5*5^k \end{align*} $
I'm guessing that the best way to do this is to prove $P(k+1) - P(k)$ is divisible by $3$, but I'm not sure on that so this could be where I start to approach this wrong.. I'm not sure what else to try though.. $P(k+1) * P(k)$? But that wouldn't distribute very well would it?
So what I did is write out P(k+1) - P(k): $ P(k+1) - P(k) = 2^k + 2*3^k - 6*5^k $
At this point, I know that the second and third terms are divisible by 3, but I know that $2^k$ is not necessarily divisible by 3, so here I am stuck...