Explanation:
Exp: C
An odd integer.
Let a= n\(^{2}\) -1
Here n can be ever or odd.
Case I:
n = Even i.e., n = 2k. where k is an integer.
⇒ a = (2k)\(^{2}\) - 1
⇒ a = 4k\(^{2}\) - 1
At k = -1, = 4 (-1)\(^{2}\) -1 = 4 - 1 = 3, which is not divisible by 8.
At k = 0, a = 4 (0)\(^{2}\) -1 = 0 - 1 = -1, which is not divisible by 8, which is not
Case II:
n = Odd i.e., n = 2k + 1, where k is an odd integer.
⇒ a = 2k + 1
⇒ a = (2k + 1)2 - 1
⇒ a = 4k\(^{2}\) + 4k + 1 - 1
⇒ a = 4k\(^{2}\) + 4k
⇒ a = 4k(k + 1)
At k = -1, a = 4 (-1)(-1 + 1) = 0 which is divisible by 8.
At k = 0, a = 4 (0)(0 + 1) = 4 which is divisible by 8.
At k = 1, a = 4 (1)(1 + 1) = 8 which is divisible by 8.
Hence, we can conclude from above two cases, if n is odd, then n\(^{2}\) - 1 is divisible by 8.
Hard **[Real Numbers]**
