# If a is an odd integer prove that a^2 − 1 is always divisible by 8.

## Solution

### Proof.

If a is divided by 4 then the possible set of remainders are 0, 1, 2 and 3.

Since a is odd so the remainder when divided by 4 cannot be 0 or 2.

So if a is an odd integer then the remainder when divided by 4 is 1 or 3.

#### Now we do case analysis:

Case 1: Let the remainder be 1.

So a = 4k + 1 for some integer k.

Thus a^{2} = (4k + 1)^{2} = 16k^{2} + 8k + 1 = 8(2k^{2} + k) + 1 So a^{2} − 1 = 8(2k^{2} + k) and so a^{2} − 1 is divisible by 8

Case 2: Let the remainder be 3.

So a = 4k + 3 for some integer k.

Thus a^{2} = (4k + 3)^{2} = 16k^{2} + 24k + 9 = 8(2k^{2} + 3k + 1) + 1 So a^{2} − 1 = 8(2k^{2} + 3k + 1) and so a^{2} − 1 is divisible by 8