Mathematical Induction divisibility $8\mid 3^{2n}-1$

Question

So I'm trying to use mathematical induction to show that for all integers $n \ge 1$ ,

$$ 8|(3^{2n} - 1)$$

(is divisible by 8)

I have my base case: [P(1)], $3^2 - 1 = 9 - 1 = 8$, since $8|8$, the base case proves true

Assume [P(k)], $ 8 | (3^{2k} - 1)$.

I know that I need to show [P(k+1)], $ 8| (3^{2(k+1)}-1) $, but I'm not sure how to prove this. I've only been using induction for summation, so how could I prove divisibility?

Answer 1

If you don't mind I'm gonna do some magic $$3^{2n}=(3^2)^n=(8+1)^n=\binom{n}{o}8^n+\binom{n}{1}8^{n-1}+\cdots+\binom{n}{n-1}8^{1}+\binom{n}{n}8^0$$

$$3^{2n}=\binom{n}{0}8^n+\binom{n}{1}8^{n-1}+\cdots+\binom{n}{n-1}8^{1}+1$$

$$3^{2n}-1=\binom{n}{0}8^n+\binom{n}{1}8^{n-1}+\cdots+\binom{n}{n-1}8=8(k)$$

$$\implies8|(3^{2n} - 1)$$

Answer 2

Hint: Note that: $$3^{2(n+1)} -1=9\cdot3^{2n}-1=(3^{2n}-1)+8\cdot3^{2n}$$

Answer 3

Here is an alternative approach to that given by AsdrubalBeltran.

Write $3^{2n}-1$ is base three notations as :

$$3^{2n}-1 = \underbrace{22\dots2_{base 3}}_{2n \text{ times}}$$

Then since $ 8 = 22_3 $, we can see the result at once. Now use induction on n.