TITLE: Prove by induction: $2^n + 3^n -5^n$ is divisible by $3$
QUESTION [6 upvotes]: 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...

REPLY [2 votes]: Another approach:
$$2^n - 5^n = (2 - 5)\cdot (2^{n-1} + 2^{n-2} \cdot 5 + \cdots + 2^{n - j - 1}5^j + \cdots + 5^{n - 1})$$
This is the difference of nth powers formula, which you can prove by induction if you like.
Then it's clear that $2^n - 5^n$ is divisible by $2 - 5 = -3$, so divisible by $3$. Therefore $2^n + 3^n - 5^n$ is also divisible by $3$.