Prove the following

Solution:

Let $a=n^{2}-1$

Here $n$ can be ever or odd.

Case I $n=$ Even $i . e ., n=2 k$, where $k$ is an integer.

$\Rightarrow \quad a=(2 k)^{2}-1$

$\Rightarrow \quad a=4 k^{2}-1$

At $k=-1,=4(-1)^{2}-1=4-1=3$, which is not divisible by 8 .

At $k=0, a=4(0)^{2}-1=0-1=-1$, which is not divisible by 8 , which is not

Case II $n=$ Odd i.e., $n=2 k+1$, where $k$ is an odd integer.

$\Rightarrow \quad a=2 k+1$

$\Rightarrow \quad a=(2 k+1)^{2}-1$

$\Rightarrow \quad a=4 K^{2}+4 k+1-1$

$\Rightarrow \quad a=4 k^{2}+4 k$

$\Rightarrow \quad a=4 k(k+1)$

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 .