Show that: n²-1 is divisible by 8 if n is an odd positive integer. If n is an odd number, prove that n² – 1 is divisible by 8

Solution:

We know an odd number as (2Q + 1), where Q is a natural integer

so, n² -1 = (2Q + 1)² -1

n² -1 = 4Q² + 4Q + 1 -1

n² -1 = 4Q² + 4Q

When n is equal to 4q + 1,

Then n² – 1 = (4q + 1) is the result.

8q(2q + 1) -1= 16q² + 8q + 1 – 1 = 8q(2q + 1)-1

When n = 4q + 3, the result is n² – 1 = (4q + 3)² – 1 = 16q² + 24q + 9 – 1 = 8(2q² + 3q + 1), which is divisible by 8.

As a result of the aforementioned equations, n² – 1 is divisible by 8 if n is an odd positive integer.

When the number n equals 4q + 1,

The solution is n² – 1 = (4q + 1)

8q(2q + 1) = 16q² + 8q + 1 – 1 = 8q(2q + 1) = 16q² + 8q + 1 – 1 = 8q(2q + 1)

n² – 1 = (4q + 3) is the outcome for n = 4q + 3.

2 – 1 = 16q² + 24q + 9 – 1 = 8(2q2 + 3q + 1), which is divisible by 8. 2 – 1 = 16q² + 24q + 9 – 1 = 8(2q² + 3q + 1), which is divisible by 8.

If n is an odd positive integer, n² – 1 is divisible by 8 due to the above mentioned equations.