Question

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

Answer 1

Answer:

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Step-by-step explanation:

We know ,

odd number in the form of (2Q +1) where Q is a natural number ,

so, n² -1 = (2Q + 1)² -1

= 4Q² + 4Q + 1 -1

= 4Q² + 4Q

now , checking :

Q = 1 then,

4Q² + 4Q = 4(1)² + 4(1) = 4 + 4 = 8 , it is divisible by 8.

Q =2 then,

4Q² + 4Q = 4(2)² + 4(2) =16 + 8 = 24, it is also divisible by 8 .

Q =3 then,

4Q² + 4Q = 4(3)² + 4(3) = 36 + 12 = 48 , divisible by 8

It is concluded that 4Q² + 4Q is divisible by 8 for all natural numbers.

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

Answer 2

Answer:

Explanation:

To Prove :

• n² -1 is divisible by 8, if n is an odd positive integer.

Proof :

Let assume that, any odd positive number is in the form of (4R + 1) & (4R + 3) for some integer R.

Now,

Let, n = 4R + 1

=> n² - 1

=> (4R + 1)² - 1

=> 16R² + 8R + 1 - 1

=> 16R² + 8R

=> 8R(2R + 1)

.°. n² - 1 is divisible by 8.

Similarly,

Let, n = 4R + 3

=> n² - 1

=> (4R + 3)² - 1

=> 16R² + 24R + 9 - 1

=> 16R² + 24R + 8

=> 8(2R² + 3R + 1)

.°. n² - 1 is divisible by 8.

Hence :

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