Proof by induction that $3^{2n}-1$ is divisible by 8.

Question

Proof by induction

Let the property $P(n)$ be the sentence $$ 3^{2n}-1 \text{ is divisible by 8} $$

Base case:

Let $n = 0$ and we have $$ 3^{2 \cdot 0} - 1 = 3^0 - 1 = 1 - 1 = 0 $$ and $0$ is divisible by $8$ because $0 = 8 \cdot 0$. Hence $P(0)$ is true.

Induction step:

Let $k$ be any integer where $k \geq 0$ and suppose $P(k)$. That is, suppose $3^{2k}-1$ is divisible by $8$. We need to show that $P(k+1)$ is true. That is, we need to show that $3^{2(k + 1)}-1$ is divisible by $8$.

Now, \begin{align} 3^{2(k + 1)}-1 &= 9 \cdot 3^{2k}-1\\ &= (8 + 1) \cdot 3^{2k}-1\\ &= 8 \cdot 3^{2k} + 3^{2k}-1\\ &= 8 \cdot 3^{2k} + 8r \tag{ *** }\\ &= 8(3^{2k} + r) \end{align}

But $3^{2k} + r$ is an integer since $3^{2k}$ and $r$ are integers. Hence by the definition of divisibility $3^{2(k + 1)}-1$ is divisible by $8$.

*** Substituting for $8r$ because $3^{2k}-1$ is divisible by $8$ by the inductive hypothesis.