**Proof: n^2 - 1 is divisible by 8 for odd positive integers**

To prove that n^2 - 1 is divisible by 8 for odd positive integers, we will use the principle of mathematical induction.

**Basis Step:**
Let's start by testing the hypothesis for the smallest odd positive integer, n = 1.
n^2 - 1 = 1^2 - 1 = 1 - 1 = 0.
0 is divisible by 8, so the hypothesis holds true for n = 1.

**Inductive Step:**
Now, assume that the hypothesis holds true for an odd positive integer k, i.e., k^2 - 1 is divisible by 8.

We need to prove that the hypothesis also holds true for the next odd positive integer, k + 2.

Let's substitute k + 2 in place of n in the expression n^2 - 1:
(k + 2)^2 - 1 = k^2 + 4k + 4 - 1 = (k^2 - 1) + 4k + 3.

Since we have assumed that k^2 - 1 is divisible by 8, let's rewrite the expression:
(k^2 - 1) + 4k + 3 = 8m + 4k + 3, where m is an integer.

Next, we need to prove that 8m + 4k + 3 is divisible by 8.

**Divisibility of 8:**
To prove that 8m + 4k + 3 is divisible by 8, we can rearrange the expression:
8m + 4k + 3 = 8m + 4k + 2 + 1.

Now, we can rewrite the expression as follows:
8m + 4k + 2 + 1 = 8m + 4(k + 1) + 1.

Since k is an odd positive integer, k + 1 is an even integer. Let's represent it as 2n, where n is an integer.

Therefore, we have:
8m + 4(k + 1) + 1 = 8m + 4(2n) + 1 = 8m + 8n + 1 = 8(m + n) + 1.

As we can see, the expression 8(m + n) + 1 has the form 8p + 1, where p is an integer.

Hence, 8m + 4k + 3 is divisible by 8, and our hypothesis holds true for the odd positive integer k + 2.

By the principle of mathematical induction, we can conclude that n^2 - 1 is divisible by 8 for all odd positive integers n.