Question

Show that nsquare-1 is divisible by 8 , if n is an odd positive integer please solve the problem
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Answer 1

Solution:-

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 .

or........

given:- n^(2-1) is divisible by 8 , if n is an odd positive integer.

=> If n = 4q + 3,

when n^(2 - 1 ) = (4q + 3)^(2 - 1)

= 16q² + 24q + 9 - 1

= 8(2q² + 3q + 1)

which is divisible by 8.

So, it is clear that n^(2 - 1 ) is divisible by 8, if n is an odd positive integer.

note:- ( ^ ) this sign used for power.

I hope it helps you...

Answer 2

Answer:

pls mark as the brainliest answer

Step-by-step explanation:

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