Question

Prove that $n^2-1$ divisible by 8 whenever n is an odd positive integer.

Short answer

8 divides $n^2-1$ whenever n is a positive integer

Step-by-step solution

Step: 1

If n=0,

0 - 0 = 0

it is true for n=0.

Step: 2

Let P(k) be true.

$k^2-1$

We need to prove that P(k+1) is true.

Step: 3

$(k+2{)}^2-1=\left(k^2-1\right)+(4k+4)$

It is divisible by 8.

It is true for P(k+1) is true.