Prove by induction that a2 − 1 is divisible by 8 for all odd integers a

Proof by induction involves two steps: the base case and the inductive step.

Base Case: Let's start with the smallest odd integer, which is 1. 1^2 - 1 = 0, which is divisible by 8.

Inductive Step: Assume the statement is true for some odd integer a = 2k + 1, where k is an integer. That is, assume that (2k + 1)^2 - 1 is divisible by 8.

We need to prove that the statement is true for the next odd integer a = 2k + 3.

(2k + 3)^2 - 1 = 4k^2 + 12k + 9 - 1 = 4k^2 + 12k + 8 = 4k(k + 3) + 8.

We can see that 4k(k + 3) is divisible by 8 because it's a multiple of 4, and 8 is obviously divisible by 8.

Therefore, (2k + 3)^2 - 1 is divisible by 8.

Since the base case is true and the inductive step has been proven, we can conclude by induction that a^2 - 1 is divisible by 8 for all odd integers a.