Question

Show that n²- 1 is divisible by 8, if n is an odd positive integer.​

Answer 1

Answer:

We know ,

Odd number in the form of (2P +1) where P is a natural number ,

So, n² -1 = (2P + 1)² -1

= 4P² + 4P + 1 -1

= 4P² + 4P

Now , checking :

P = 1 then,

4P² + 4P = 4(1)² + 4(1) = 4 + 4 = 8 , it is divisible by 8.

P =2 then,

4P² + 4P = 4(2)² + 4(2) =16 + 8 = 24, it is also divisible by 8 .

P =3 then,

4P² + 4P = 4(3)² + 4(3) = 36 + 12 = 48 , divisible by 8

Hence, we conclude that 4P² + 4P is divisible by 8 for all natural number .

hence, n² -1 is divisible by 8 for all odd value of n .

⚡⚡⚡Hope it will help you.⚡⚡⚡

Answer 2

We know that an odd positive integer is of the form 4q+1 or 4q+3

So, n^2 - 1

=>(4q+1)^2 - 1

=>16q^2 + 1 + 8q - 1

=>8(2q^2 + q)

[it is divisible by 8 as no remainder is remaining]

Now, with 4q+3

(4q+3)^2 - 1

=>16q^2 + 9 + 24q - 1

=>16q^2 + 8 + 24q

=>8(2q^2 + 1 + 3q)

[it is too divisible by 8 as no remainder is remaining]

Hence n^2-1 is divisible by 8.

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