Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$... Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$.
So far I have the base case completed, and believe I am close to completing the proof itself.
Base case:$(n=1)$
$3^1 + 7^1 - 2 = 8/8 = 1 $
Inductive Hypothesis: Assume that $3^n +7^n −2$ is divisible by 8 for all positive integers n.
Induction step $(n+1)$ case:
$$ 3^{n+1} + 7^{n+1} - 2 $$
$$3(3^{n}) + 7(7^{n}) - 2$$
$$3^n + 7^n = 8x $$
-It  seems to me that this could be the end of the proof because whatever the answer is would be a multiple of 8: but I am unsure, any help is appreciated. 
 A: It holds for $n=1,2$.
If it holds for $1,2,\dots,n$, then
\begin{align}
&3^{n+1}+7^{n+1}-2\\
&=3^2\cdot3^{n-1}+7^2\cdot7^{n-1}-2\\
&=(8+1)\cdot3^{n-1}+(48+1)\cdot7^{n-1}-2\\
&=8\cdot(3^{n-1}+6\cdot7^{n-1})+3^{n-1}+7^{n-1}-2\\
\end{align}
Therefore it also holds for $n+1$.
So it holds for all $n\in \mathbb N$.
A: We have that $3^n+7^n-2=8k$ for some $k$. Now we substitute this expression into $3(3^n)+7(7^n)-2$ to get:

$3(8k-7^n+2)+7(7^n)-2=24k+4(7^n)+4=24k+4(7^n+1)$ where $7^n+1$ is even so we may rewrite $24k+4(7^n+1)=24k+4(2m)=24k+8m$ which is divisible by $8$.
A: $\begin{array}\\
3^{n+1}+7^{n+1}-2
&=3^{n+1}-1+7^{n+1}-1\\
&=(3-1)\sum_{k=0}^n 3^k+(7-1)\sum_{k=0}^n 7^k\\
&=\sum_{k=0}^n (2\ 3^k+6\ 7^k)\\
&=\sum_{k=0}^n (6\ 3^k+6\ 7^k-4\ 3^k)\\
&=\sum_{k=0}^n (6(3^k+7^k-2)+12-4\ 3^k)\\
&=\sum_{k=0}^n (6(3^k+7^k-2)+4(3-3^k))\\
\end{array}
$
Since
$3-3^k$
is even for $k \ge 0$
(being the difference of
two odd numbers),
$4(3-3^k)$
is divisible by $8$.
Therefore,
if
$3^k+7^k-2$
is divisible by $8$
for
$0 \le k \le n$,
then
$3^{n+1}+7^{n+1}-2$
is also divisible by $8$.
