College Algebra, Chapter 9, 9.5, Section 9.5, Problem 20

Prove that $3^{2n} - 1$ is divisible by 8 for all natural numbers $n$.

Let $P(n)$ denote the statement $3^{2n} - 1$ is divisible by 8

Step 1: $P(1)$ is true, since $3^{2(1)} - 1$ or $8$ is divisible by 8

Step 2: Suppose $P(k)$ is true. Now


$
\begin{equation}
\begin{aligned}

3^{2(k + 1)} - 1 =& 3^{2k + 2} -1
\\
\\
=& 3^{2k} \cdot 3^2 - 1
\\
\\
=& 3^{2k} \cdot 9 - 1
\\
\\
=& 3^{2k} \cdot (8 + 1) - 1
\\
\\
=& 8 \cdot 3^{2k} + 3 ^{2k} - 1

\end{aligned}
\end{equation}
$


By the induction hypothesis, we see that $3^{2k} - 1$ is divisible by $8$ and $8 \cdot 3^{2k}$ is clearly divisible by $8$ because the statement is in multiple of $8$.

So, $P(k + 1)$ follows from $P(k)$. Thus, by the principle of mathematical induction $P(n)$ holds for all $n$.