Question

prove that n square minus 1 is divisible by 8 for any natural number N​

Answer 1

Here is your answer. Hope it help you ☺️

Answer 2

Answer:

It is not valid for all Natural numbers N, though It is true of odd numbers.

Step-by-step explanation:

Odd number is always of the form 4p + 1 or 4p + 3 where p belongs to Natural numbers.

Suppose,

         n = 4p + 1

         (n^2 – 1) = (4p + 1)^2 – 1 = 16p^2 + 8p + 1 = 16p^2 + 8p = 8p (2p + 1)

         Which is clearly divisible by 8

Suppose,

          n = 4p + 3

         (n^2 – 1) = (4p + 3)^2 – 1 = 16p^2 + 24p + 9 – 1 = 16p^2 + 24p + 8

                        = 8(2p^2 + 3p + 1)

          Which is also divisible by eight.

There for n^2 - 1 is divisible by eight when n is any odd number.