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

**(A)** an integer

**(B)** a natural number

**(C)** an odd integer

**(D)** an even integer

Given:

$n^2 - 1$ is divisible by 8.

To do:

We have to find the correct option.

Solution:

Let $a = n^2 - 1$ where $n$ can be even or odd.

When $n$ is even,

$n = 2k$, where $k$ is an integer.

This implies,

$a = (2k)^2 - 1$

$a = 4k^2 - 1$

For $k = -1$,

$a = 4(-1)^2 - 1$

$= 4 - 1$

$= 3$ which is not divisible by 8.

For $k = 0$,

$a = 4(0)^2 - 1$

$= 0 - 1$

$= -1$, which is not divisible by 8.

When $n$ is odd, $n = 2k + 1$, where $k$ is an integer,

This implies,

$a = (2k+1)^2-1$

$=4k^2+4k+1-1$

$=4k^2+4k$

$=4k(k+1)$

For $k = 1$,

$a=4(1)(1 + 1)$

$= 8$ which is divisible by 8.

Hence, we can conclude from the above two observations that if $n$ is odd, then $n^2 - 1$ is divisible by 8.

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