Question

prove that n^2 -1 is divisible by 8 if n is an odd positive integer

Answer 1

Step-by-step explanation:

Since n is an odd positive integer.

(i)

Let us assume that n = 4m + 1.

Then:

n² - 1 = (4m + 1)² - 1

         = 16m² + 1 + 8m - 1

         = 16m² + 8m

         = 8m(2m + 1)

Thus, n² - 1 is divisible by 8.

(ii)

Let us assume n = 4m + 3.

Then:

n² - 1 = (4m + 3)² - 1

        = 16m² + 9 + 24m - 1

        = 16m² + 24m + 8

        = 8(2m² + 3m + 1)

        = 8(2m + 1)(m + 1)

Thus, n² - 1 is divisible by 8.

Hope it helps!

Answer 2

If n is odd and positive, we define

n=2k+1 where k is a non-negative integer.

from which we substitute, expand and factor:

n²-1

=(2k+1)²-1

=4k²+4k+1-1

=4k²+4k

=4k(k+1)

Since k is a non-negative integer, we have two possible cases:

1. k is odd, in which case (k+1) is even, and equal to 2q (q=non-negative integer)

=> n²-1=4k(2q)=8kq (where both k and q are non-negative integers)

therefore 8 divides n²-1

2. k is even, then k=2q (q=non-negative integer)

=> n²-1=4(2q)(k+1)=8q(k+1) (where both k and q are non-negative integers)

therefore 8 divides n²-1

Since in both cases, 8 divides n²-1, therefore it is proved that 8 divides n²-1 in for all positive values of n.