Question

show that n²-1is divisible by 8, where n is any positive odd integer.​

Answer 1

Answer:

Q.E.D

Step-by-step explanation:

Let there be two cases, where odd positive integers can be 4m + 1 and 4m + 3.

Case 1: When n = 4m + 1

→ n² - 1 = (4m + 1)² - 1

→ 16m² + 8m + 1 - 1

→ 8(2m² + m)

→ 8q [where q = 2m² + m]

Case 2: When n = 4m + 3

→ n² - 1 = (4m + 3)² - 1

→ 16m² + 24m + 9 - 1

→ 8(2m² + 3m + 1)

→ 8q [where q = 2m² + 3m + 1]

Hence in both cases, divided n² - 1 is divisible by 8.

Answer 2

Solution :-

Any odd positive integer n can be written in form of 4q + 1 or 4q + 3.

If n = 4q + 1, when n² - 1 = (4q + 1)² - 1 = 16q² + 8q + 1 - 1 = 8q(2q + 1) which is divisible by 8.

If n = 4q + 3, when n² - 1 = (4q + 3)² - 1 = 16q² + 24q + 9 - 1 = 8(2q² + 3q + 1) which is divisible by 8.

So, it is clear that n2 - 1 is divisible by 8, if n is an odd positive integer.

Hope it Helps !!!