Answer
Let $P(n)$ denote the proposition: "$n^2 -1$ is divisible by $8$ whenever $n$ is an odd, positve integer."
$P(1)$ is true, since $(1)^2 -(1) = 0$.
Assume that $P(k)$ is true for an arbitrary integer $k$. Then to show $P(k+2)$ is also true,
\begin{align*}
(k+2)^2 - 1 &= k^2 + 4k +4 -1\\
&= k^2 -1 + 4k +4 \\
&= [k^2 -1] +4[k+1] \\
\end{align*}
Each of the bracketed terms is divisible by $8$ when $k$ is odd, so $P(k+2)$ is also true. Therefore by method of mathematical induction $P(n)$ is true - $n^2 - 1$ is divisible by $8$ whenever $n$ is an odd, positive integer.
Work Step by Step
First, we must give the proposition. We let $P(n)$ denote the proposition: "$n^2-1$ is divisible by $8$ whenever $n$ is an odd positive integer."
Since the proposition specifies "odd", we will start the base case with the first odd positive integer, $1$. We must prove our base case $P(1)$ is true by showing that the proposition is true when $n=1$. When $n=1$, $n^2 -1 = (1)^2 -(1) =0$. $0$ is divisible by $8$, so therefore $P(1)$ is true.
Finally, we must carry out the inductive step. Assume that $P(k)$ is true; that is, we assume that $k^2 -k$ is divisible by $8$ for an arbitrary, odd positive integer $k$. However, instead of showing $P(k+1)$ is true as usual, we must instead show $P(k+2)$ is true. This is because showing $P(k+1)$ is true would prove the statement true for ALL positive integers; showing $P(k+2)$ to be true would prove the statement true for EVERY OTHER integer, starting with $1$ (a set of numbers identical to the set of odd positive integers).
\begin{align*}
(k+2)^2 - 1 &= k^2 + 4k +4 -1\\
&= k^2 -1 + 4k +4 \\
&= [k^2 -1] +4[k+1] \\
\end{align*}
The first bracketed term is divisible by $8$, via our inductive hypothesis; the second term has a factor of 4, and whenever $k$ is odd it also has a factor of $2$ - this means that whenever $k$ is odd the second term is also divisible by $8$. This implies that the entire expression must also be divisible by $8$, and that $P(k+2)$ is true. Therefore by method of mathematical induction $P(n)$ is true - $n^2 -1$ is divisible by $8$ whenever $n$ is an odd, positive integer.
