Proving that $8^n - 3^n$ is divisible by $5$ I really should be able to do this but I don't know why I can't figure it out.
My problem is that I have to prove $8^n - 3^n$ is divisible by $5$.
So what I did is I tried it for $n=1, n=2, n=3$ and so on and all the numbers I tried, the expression was divisible by 5.
So then what I did is I said "assume it's true for $n=g$" 
So the expression turns into $8^g - 3^g$ is divisible by 5
Then I wanted to try it for $g+1$ so:
$8^{g+1} - 3^{g+1} $
$(8^g * 8) - (3^g * 3)$
Where should I go from here?
 A: Since $8\equiv 3\pmod 5$, the expression reduces to $3^n-3^n$ modulo 5.
If you don't have modular arithmetic available, write $8^n$ as $(3+5)^n$ and expand using the binomial theorem. The $\binom n03^n5^0$ term is just what you subtract away, and every other term has at least one factor of $5$.
A: $8^{n+1}-3^{n+1}=8(8^{n})-3(3^{n})=8(8^{n}-3^{n})+3^{n} \times 8 - 3(3^{n})=8(8^{n}-3^{n})+3^{n} \times 5$. Could you continue from here?
A: The induction hypothesis can be written
$$
8^g-3^g=5k
$$
for some integer $k$. Therefore $8^g=3^g+5k$. Then
$$
8^{g+1}-3^{g+1}=8\cdot 8^g-3^{g+1}=
8\cdot 3^g+40k-3\cdot3^g=(8-3)\cdot 3^g+40k
$$
Can you finish?
A: Recall the difference of $n$th powers factorization.
\begin{equation}
a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n - 1})
\end{equation}
Therefore
\begin{equation}
8^n - 3^n = (8 - 3)(8^{n - 1} + 3 \cdot 8^{n - 2} + \cdots + 8 \cdot 3^{n - 2} + 3^{n - 1}) = 5M
\end{equation}
where $M = 8^{n - 1} + 3 \cdot 8^{n - 2} + \cdots + 8 \cdot 3^{n - 2} + 3^{n - 1}$. Since $M$ is an integer it is clear that $8^n - 3^n$ is divisible by 5.
A: $$8^n - 3^n = 5\sum_{k=0}^{n-1} 8^k \cdot 3^{n-k-1}$$
