Question

$n^2-1$ is divisible by 8, if n is

• A. An integer
• B. A natural number
• C. An odd integer
• D. An even integer

Answer 1

The correct option is C An odd integer

Let $a=n^2-1$
Here n can be even or odd
Case I: n = Even i.e., n = 2k where k is an integer
$\Rightarrow a=(2k{)}^2-1$
$\Rightarrow a=4k^2-1$
$Atk=-1,=4(-1{)}^2-1=3$, which is not divisible by 8
At $k=0,a=4(0{)}^2-1=0-1=-1$, which is not divisible by 8.
Case II: n = odd i.e., n = 2k + 1, where k is an odd integer
$\Rightarrow a=2k+1$
$\Rightarrow a=4k2+4k+1-1$
$\Rightarrow a=4k^2+4k$
$\Rightarrow a=4k(5+1)$
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