(n² − 1) is divisible by 8, if n is

(a) any natural number
(b) any integer
(c) any odd positive integer
(d) any even positive integer

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Solution

(c) any odd positive integer

$(n^{2} - 1)$ is divisible by 8 if $n$ is an odd positive integer.
We know that an odd positive integer $n$ is of the form (4 $q$ + 1) or (4 $q$ + 3) for some integer $q$ .
Case 1: When $n = (4 q + 1)$ , we have:
$(n^{2} - 1) = {(4 q + 1)^{2} - 1}$ = {^{$16 q^{2} + 1 + 8 q - 1} = 16 q^{2} + 8 q$} = $8 q (2 q + 1)$ ,
which is clearly divisible by 8.
Case 2: When $n = 4 q + 3$ , we have:
$(n^{2} - 1) = {(4 q + 3)^{2} - 1} = {16 q^{2} + 9 + 24 q - 1} = 16 q^{2} + 24 q + 8$ = $8 (2 q^{2} + 3 q + 1)$ ,
which is clearly divisible by 8.

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