Question

We know that any odd positive integer is of the form $4q+1$ or $4q+3$ for some integer $q.$
Thus, we have the following two cases.

• A. $n^2-1$ is divisible by 8
• B. $n^2+1$ is divisible by 8
• C. $n-1$ is divisible by 8
• D. $n+1$ is divisible by 8

Answer 1

The correct option is A $n^2-1$ is divisible by 8

When $n=4q+1$
In this case, we have
$n^2-1=(4q+1{)}^2-1=16q^2+8q+1-1$
$=8q(2q+1)=8r$ where $r=q(2q+1)$ is an integer
$\Rightarrow n^2-1$ is divisible by 8.
Case-II: When $n=4q+3$
In this case, we have
$n^2-1=(4q+3{)}^2-1=16q^2+24q+9-1=16q^2+24q+8$
$=8(2q^2+3q+1)=8(2q+1)(q+1)$
$=8r$ where $r=(2q+1)(q+1)$ is an integer.
$\Rightarrow n^2-1$ is divisible by 8
Hence $n^2-1$ is divisible by 8.