Mathematical induction proof that $8$ divides $3^{2n} - 1$

Question

I'm struggling with this question: prove the following using simple mathematical induction. $$ 8 \mid (3^{2k} - 1) $$

What I've got so far is:
$$ 3^{2k+2} - 1 = 3^{2k} \cdot 3^{2} - 1 $$

From here, I'm not entirely sure where to go, please advise.

Answer 1

$$ 3^{2k}\times 3^2-1 = 3^{2k}\times (8+1)-1 = 3^{2k}\times 8 + 3^{2k} -1 $$is then a multiple of $8$, because $3^{2k}-1$ is. This end the induction step.

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Another proof:

$$ 3^{2k} - 1 = (9-1) (1+3^2+\cdots + 3^{2k-2}) $$because of the geometric sum.

Answer 2

Not inductive (already covered by mookid):

$3^2 \equiv 1 \pmod 8 \implies 3^{2n} \equiv (3^2)^n \equiv 1^n \equiv 1 \pmod 8$.

Answer 3

For induction:

(i) if $k=1$, then ok, because $8|(3^2-1)$, i.e., $8|8$. Therefore, the induction base is true.

(ii) Suppose that the statement is true for $k=n$, i.e, is true that $$ 8|(3^{2n}-1).$$ Then $\exists m\in \mathbb{N}$ such that $8m=3^{2n}-1$.

Then, for $k=n+1$ we have $$3^{2(n+1)}-1=3^{2n}\cdot 3^2-1=9\cdot 3^{2n}-1-8+8=9(3^{2n}-1)+8= $$ $$=9\cdot (8m)+8=8(9m+1), $$ i.e., $$8|(3^{2(n+1)}-1). $$ Then, for (i) and (ii) the result follow for induction.