Prove $3^{2 n}-1$ is divisible by 8 .

Step 1 ：We are given the problem to prove that \(3^{2 n}-1\) is divisible by 8 for any positive integer n. This is a problem of mathematical induction.

Step 2 ：First, we check the base case when n=1. Substituting n=1 into the equation, we get \(3^{2*1}-1 = 8\), which is divisible by 8. So, the base case is true.

Step 3 ：Now, for the induction step, we assume that the statement is true for n=k (i.e., \(3^{2k}-1\) is divisible by 8), and we need to prove that the statement is also true for n=k+1 (i.e., \(3^{2(k+1)}-1\) is divisible by 8).

Step 4 ：We can write \(3^{2(k+1)}-1\) as \(3^{2k}*3^2-1\). If we subtract and add 1 inside the expression, we get \(3^{2k}*3^2-1+1-1 = (3^{2k}*3^2-1) + (1-1)\).

Step 5 ：The first part of the expression, \(3^{2k}*3^2-1\), is divisible by 8 by the induction hypothesis. The second part of the expression, \(1-1\), is obviously divisible by 8.

Step 6 ：Therefore, \(3^{2(k+1)}-1\) is divisible by 8, which completes the induction step.

Step 7 ：\(\boxed{\text{The statement } 3^{2 n}-1 \text{ is divisible by 8 for any positive integer n is true.}}\)