Prove by induction that for every positive integer $n$, $3^{2n} - 1$ is divisible by 8.

1. Problem Description

Prove by induction that for every positive integer $n$, $3^{2n} - 1$ is divisible by

8.

2. Solution Steps

We will prove this by induction.

Base Case: $n=1$

$3^{2(1)} - 1 = 3^2 - 1 = 9 - 1 = 8$.

Since 8 is divisible by 8, the base case holds.

Inductive Hypothesis:

Assume that $3^{2k} - 1$ is divisible by 8 for some positive integer $k$. This means $3^{2k} - 1 = 8m$ for some integer $m$.

Inductive Step:

We want to show that $3^{2(k+1)} - 1$ is divisible by

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

From the inductive hypothesis, $3^{2k} = 8m + 1$.

Substituting this into our expression, we get

$9 \cdot (8m + 1) - 1 = 72m + 9 - 1 = 72m + 8 = 8(9m + 1)$.

Since $9m+1$ is an integer, $8(9m+1)$ is divisible by

8.

Therefore, by induction, $3^{2n} - 1$ is divisible by 8 for all positive integers $n$.

3. Final Answer

$3^{2n} - 1$ is divisible by 8 for all positive integers $n$.