$n^{2} - 1$ is divisible by 8 , if n is
(a) an integer (b) a natural number
(c) an odd integer (d) an even integer

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Solution

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

When n = 4m + 1,
$n^{2} - 1 = {(4 m + 1)}^{2} - 1 = 16 m^{2} + 8 m + 1 - 1 = 16 m^{2} + 8 m = 8 m (2 m + 1) \Rightarrow n^{2} - 1 is divisible b y 8 . When n = 4 m + 3, n^{2} - 1 = {(4 m + 3)}^{2} - 1 = 16 m^{2} + 24 m + 9 - 1 = 16 m^{2} + 24 m + 8 = 8 (2 m^{2} + 3 m + 1) \Rightarrow n^{2} - 1 is divisible by 8 .$

Thus, n² – 1 is divisible by 8 if n is an odd positive integer.

Hence, the correct answer is option C.

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