Question

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

Answer 1

Hint:Given it is an odd integer. Find the general form of odd integer. Then find the way to substitute that into a given expression. Then solve the expression to conclude it is divisible by 8. For that you need to prove that the given expression is a multiple of 8. So, we need to convert the given expression into the form of 8k.

Complete step-by-step answer:
Odd number: A number which is not divisible by 2 is called an odd number.
Even number: A number which is divisible by 2 is called an even number.
When a number is divided by 2 there are 2 possibilities for remainder either 0 or 1. So, we can if the remainder is 0 then it is an even number. Given in the question n is an odd number. So, the remaining possibility is the remainder being 1. So, an odd number can be written as: $\text{n}\ =\ 2\text{Q+}1$.
By above condition we can write our ‘n’ in question as $2\text{Q+}1$.
Given expression is the question for which we need to check divisibility is given by (in terms of n): ${{n}^{2}}-1$
By substituting the value of n into this, we get ${{\left( 2\text{Q+}1 \right)}^{2}}-1$
By using general algebraic identity: ${{\left( a+b \right)}^{2}}\ =\ {{a}^{2}}+2ab+{{b}^{2}}$ we get $\ 4{{\text{Q}}^{2}}+4\text{Q}+1-1$
By simplifying the above expression, we get the $4\text{Q}\left( \text{Q+}1 \right)$.
So, there are 2 consecutive numbers $\text{Q,}\ \text{Q+}1$. So, one of them must be even. So, $\left( \text{Q} \right)\left( \text{Q+}1 \right)$is divisible by 2. So, we can write $\left( \text{Q} \right)\left( \text{Q+}1 \right)\ =\ 2K$.
So, the expressions become: $\ 4\left( 2K \right)\ =\ 8K$
So, it is multiple of 8.
Hence, we proved ${{n}^{2}}-1$ is divisible by 8 if n is odd.

Note: Whenever you have 2 consecutive numbers one of them is divisible by 2. Similarly, if we have 3 consecutive numbers then one of them is divisible by 3.