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If 'n' is an odd positive integer, show that (n^2-1) 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.
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.
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