Showing $k^2-1$ is divisible by 8 when $k$ is an odd natural number Prove that $k^2-1$ is divisible by $8$ when $k$ is an odd natural number.
I am trying to prove this using induction.
Initial case: Let $k\in N$ such that $k=1$
              Then $k^2-1=1^1-1=0$.
$0\equiv 0\pmod 8$ so when $k=1$, $k^2-1$ is divisible by 8.
I am wondering how I would set up the inductive step since $k+1$ would be an even natural number. Would the inductive step in this case be showing that $((k+2)^2-1)$ is divisible by 8 instead of showing $((k+1)^2-1)$ is divisible by 8?

Since $k$ is odd, therefore $k \equiv 1,3,5,7 \pmod{8}$. In which case $k^2 \equiv 1 \pmod{8}$. Thus $8$ divide $k^2-1$.
