if n is an odd positive integer,

then prove that
n^2 -1 is divisible by 8

if n is odd, n=2m+1 for some m

n^2-1
= (2m+1)^2-1
= 4m^2 + 4m + 1 - 1
= 4m(m+1)

Now, either m or m+1 is even, so m(m+1) is a multiple of 2

So, 4m(m+1) is a multiple of 8

To prove that n^2 - 1 is divisible by 8 when n is an odd positive integer, we can use the concept of mathematical induction which involves two steps: the base case and the inductive step.

Base Case:
Let's start by verifying that the statement holds true for the smallest odd positive integer. When n = 1, we have:
n^2 - 1 = 1^2 - 1 = 0.
Since 0 divided by 8 gives 0 with no remainder, the statement is true for the base case.

Inductive Step:
Now, we assume that the statement holds true for an arbitrary odd positive integer k, where k^2 - 1 is divisible by 8. We need to prove that it also holds true for k + 2, meaning (k + 2)^2 - 1 is divisible by 8.

To do this, we expand (k + 2)^2 - 1:
(k + 2)^2 - 1 = (k^2 + 4k + 4) - 1 = k^2 + 4k + 3.

To prove that (k + 2)^2 - 1 is divisible by 8, we need to show that k^2 + 4k + 3 is also divisible by 8.

By the assumption, k^2 - 1 is divisible by 8, so we can rewrite it as k^2 - 1 = 8m, where m is an integer.

Now, let's consider k^2 + 4k + 3:
k^2 + 4k + 3 = k^2 - 1 + 4k + 4 = 8m + 4k + 4.

We can factor out 4 from 8m + 4k + 4:
8m + 4k + 4 = 4(2m + k + 1).

Since 2m + k + 1 is an integer, we can rewrite it as 2m' (where m' is an integer):
4(2m + k + 1) = 4(2m') = 8m'.

Therefore, k^2 + 4k + 3 is divisible by 8, which completes the inductive step.

By the principle of mathematical induction, we have proven that if n is an odd positive integer, n^2 - 1 is divisible by 8.