Question

Question 3
$n^2-1$ 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 answer is C.
Let $a=n^2-1$
Here, n can be even or odd.
Case I
n is an even number i.e., n = 2k, where k is an integer.
$\Rightarrow a=(2k{)}^2-1$
$\Rightarrow a=4k^2-1$
$Whenk=-1,a=4(-1{)}^2-1=3$, which is not divisible by 8.
When $k=0,a=4(0{)}^2-1=0-1=-1$, which is not divisible by 8.
Case II
n is odd i.e., n = 2k + 1, where k is an integer.
$\Rightarrow a=(2k+1{)}^2-1$
$\Rightarrow a=4k^2+4k+1-1$
$\Rightarrow a=4k^2+4k$
$\Rightarrow a=4k(k+1)$
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.