Question

# Show that ${n}^2 - 1$ is divisible by 8, if n is an odd positive integer.

Hint: So, we know that odd positive integers are 3,5,7,..

Read the question once again. So we have to square an odd positive integer and subtract 1 from it, right?

Let's look at example, We’ll take 3

So, $3^2$ is 9 and 9-1=8.

Similarly, We’ll take 5

So, $5^2$ is 25 and 25-1=24

If you look at these two examples, 8,24 are divisible by 8. Hence, we can easily conclude that ${n}^2 - 1$ is divisible by 8 for any odd positive odd integer.

Any odd positive number is in the form of (4p + 1) or (4p + 3) for some integers p.

Let, n = 4p + 1

${n^2} - 1$ = ${(4p + 1)^2}$ - 1

On simplifying RHS we’ll get,

$16{p^2} + 1 + 8p - 1$

$\Rightarrow$ 8p(2p + 1)

$\Rightarrow {n^2} - 1$ is divisible by 8

Now, let n = (4p + 3)

${n^2} - 1= {{(4p + 3)^2} - 1}$

$\Rightarrow 16{p^2} + 9 + 24p - 1$

$\Rightarrow 16{p^2} + 8 + 24p$

$\Rightarrow 8\left( {2{p^2} + 3p + 1} \right)$

$\Rightarrow {n^2} - 1{\text{ is divisible by 8}}$

Hence, ${n^2} - 1$ is divisible by 8 if n is an odd positive integer.

NOTE: - In this type of question, always write the general points of odd positive integers and then think practically and solve them. First, understand the problem correctly, then the solution becomes easy for you.