Question

please answer with explanation​

Answer 1

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p. Let n = 4p+ 1, (n2 – 1) = (4p + 1)2 – 1 = 16p2 + 8p + 1 = 16p2 + 8p = 8p (2p + 1) ⇒ (n2 – 1) is divisible by 8. (n2 – 1) = (4p + 3)2 – 1 = 16p2 + 24p + 9 – 1 = 16p2 + 24p + 8 = 8(2p2 + 3p + 1) ⇒ n2– 1 is divisible by 8. Therefore, n2– 1 is divisible by 8 if n is an odd positive integer.