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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.