Question

Show that n^2-1 is divisible by 8 if not is an odd +ve integer

Answer 1

Answer. 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,  (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.

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Answer 2

Answer:

Step-by-step explanation:

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