Question

prove that n^2 - 1 is divisible by 8 where n is any odd positive integer.

Answer 1

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Answer 2

SOLUTION

Let a= n^2-1

Here n can be even or odd.

Case (1)

n= Even i.e , n= 2k (where k is an integer)

=) a= (2k)^2 -1

=) a= 4k^2 -1

At k = -1,

=) 4(-1)^2 = 4-1 = 3 (which not divisible by 8)

At k=0,

a= 4(0)^2-1 =0-1 = -1 (which not divisible by 8)

Case (2)

=) Odd I.e, n= 2k+1, where is an odd int.

=) a= 2k+1

=) a= (2k+1)^2 -1

=) a= 4k^2 +4k+1 -1

=) a= 4k^2 +4k

=) a= 4k(k+1)

At k= -1, a= 4(-1) (-1+1)=0 which is divisible by 8.

At k= 0 a= 4(0)(0+1)= 4 which is divisible by 8.

At k= 1, a= 4(1)(1+1) =8 which is divisible by 8.

Hence, we conclude from above two cases, if n is odd, then n^2 -1.

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