If n is an odd integer, then show that n^2 – 1 is divisible by 8.

Y are you taking some odd integers as 4m+1 or 4m+3.
Why no we take 2m+1 or 3m+1

Namasthe,
Y are you taking some odd integers as 4m+1 or 4m+3.
Why no we take 2m+1 or 3m+1. Is there any specific reason for taking as 4m+1 or 4m+3.

Because every odd integer will cover by 4m+1 or 4m+3.
You may also use 2m+1.

Solution:
Any odd integer n is of the form 4m + 1 or 4m + 3.

=> n² – 1 = (4m + 1)² – 1
= 16m² + 8m
= 8(2m² + m),
which is divisible by 8.
Also, n² – 1 = (4m + 3)² – 1
= 16m² + 24m + 8
= 8(2m² + 3m + 1),
which is divisible by 8.
Hence, n² – 1 is divisible by 8 for any odd integer n.

We know that any odd positive integer n can be written in form 4q + 1 or 4q + 3.

So, according to the question,

When n = 4q + 1,

Then n² – 1 = (4q + 1)² – 1 = 16q² + 8q + 1 – 1 = 8q(2q + 1), is divisible by 8.

When n = 4q + 3,

Then n² – 1 = (4q + 3)² – 1 = 16q² + 24q + 9 – 1 = 8(2q² + 3q + 1), is divisible by 8.

So, from the above equations, it is clear that, if n is an odd positive integer

n² – 1 is divisible by 8.

Hence Proved.