show that (n^{2}-1) Is divisible by 8?

If ‘n’ is an odd positive integer, show that (n2-1) Is divisible by 8??

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.

Let n = 4p+ 1,

(n^{2} – 1) = (4p + 1)2 – 1 = 16p^{2} + 8p + 1 = 16p^{2} + 8p = 8p (2p + 1)

⇒ (n^{2} – 1) is divisible by 8.

(n^{2} – 1) = (4p + 3)2 – 1 = 16p^{2} + 24p + 9 – 1 = 16p^{2} + 24p + 8 = 8(2p^{2} + 3p + 1)

⇒ n^{2}– 1 is divisible by 8.

Therefore, n^{2}– 1 is divisible by 8 if n is an odd positive integer.