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

Explanation for the correct option:

Option (C). an odd integer

Let $x=n^2-1$.

Assume that $n$ is an odd integer.

Therefore let $n=2k+1$, where $k$ is an integer.

Hence,

$\begin{array}{rcl}x& =& n^2-1\\ & \Rightarrow & x={\left(2k+1\right)}^2-1\left[\because n=2k+1\right]\\ & \Rightarrow & x=4k^2+4k+1-1\\ & \Rightarrow & x=4k^2+4k\\ & \Rightarrow & x=4k\left(k+1\right)\end{array}$

At $k=-1$,

$\begin{array}{rcl}x& =& 4\left(-1\right)\left(-1+1\right)\Rightarrow x=0\end{array}$ which is divisible by 8.

At $k=0$,

$\begin{array}{rcl}x& =& 4\left(0\right)\left(0+1\right)\Rightarrow x=0\end{array}$ which is divisible by 8.

At $k=1$,

$\begin{array}{rcl}x& =& 4\left(1\right)\left(1+1\right)\Rightarrow x=8\end{array}$ which is divisible by 8.

At $k=2$,

$\begin{array}{rcl}x& =& 4\left(2\right)\left(2+1\right)\Rightarrow x=24\end{array}$ which is divisible by 8.

From the above observation, we can conclude that for every for every odd integer $n$, $n^2-1$ can be expressed as the multiple of $8$.

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

Hence, the correct option is (C) i.e. an odd integer.

Explanation for the incorrect options:

Option (A). an integer

Let $x=n^2-1$.

Assume that $n$ is an integer.

At $n=4$,

$x={\left(4\right)}^2-1\Rightarrow x=15$ which is not divisible by $8$.

Hence, Option (A) is incorrect.

Option (B). a natural number

Let $x=n^2-1$.

Assume that $n$ is a natural number.

At $n=4$,

$x={\left(4\right)}^2-1\Rightarrow x=15$ which is not divisible by $8$.

Hence, Option (B) is incorrect.

Option (D). an even integer

Let $x=n^2-1$.

Assume that $n$ is an even integer.

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

Hence,

$\begin{array}{rcl}x& =& n^2-1\\ & \Rightarrow & x={\left(2k\right)}^2-1\left[\because n=2k\right]\\ & \Rightarrow & x=4k^2-1\end{array}$

At $k=-1$,

$\begin{array}{rcl}x& =& 4{\left(-1\right)}^2-1\Rightarrow x=4-1=3\end{array}$ which is not divisible by 8.

At $k=0$,

$\begin{array}{rcl}x& =& 4{\left(0\right)}^2-1\Rightarrow x=-1\end{array}$ which is not divisible by 8.

At $k=1$,

$\begin{array}{rcl}x& =& 4{\left(1\right)}^2-1\Rightarrow x=3\end{array}$ which is not divisible by 8.

Hence, Option (D) is incorrect.

Hence, the correct option is (C) i.e. an odd integer.