Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$...

Question

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.

Answer 1

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 $n\in \mathbb N$.

Answer 2

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 $8$.

Answer 3

$\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 $8$.