Question

prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8

Answer 1

Answer: See explanation

Step by step

explanation:

Suppose that N is  any  integer.

Let the numbers (2N +1) and (2N +3) be consecutive odd numbers.

So the difference of their squares is:

(2N + 3) ² - (2N + 1) ² = (2N) ² + 12N + 9 - [(2N) ² + 4N + 1]

                                   = 4N² + ​​12N + 9 - [4N² + ​​4N + 1]

                                   = 4N² + ​​12N + 9 - 4N² - 4N - 1

                                   = 12N - 4N + 9 - 1

                                   = 8N + 8

                                   = 8 (N + 1)

                                  And this result is a multiple of 8.