Prove that the difference between the squares of consecutive odd numbers is a multiple of 8

I understand what the question means, but I am not sure on how to 'prove' it. This is an exam question worth 6 marks and I am really confused on how to start it/what to write. Can anybody do this for me by showing your working, so that I can understand and learn from it? Thank you

Consecutive odd numbers are of the form (2n - 1) and (2n + 1), with n an integer.

Take the square of both:
4n^2 - 4n + 1
4n^2 +4n + 1
Take the difference:
(2n + 1)^2 - (2n - 1)^2 = 4n^2 + 4n + 1 - 4n^2 + 4n - 1 = 8n
As n is an integer, this difference must be a multiple of 8.

Ok so for any whole number n, the expression: 2n+1 is always an odd number. The next odd number would then be 2n + 3 because every odd number is 2 away from the next one. Therefore:

(2n + 3)^2 - (2n + 1)^2 =

(4n^2 + 12n + 9) - (4n^2 + 4n + 1) =

4n^2 + 12n + 9 - 4n^2 - 4n - 1=

8n - 8 =

8(n -1)

Which means that no matter what, the solution will be evenly divisible by 8 because we declared than was a whole number so:

8(n-1)/8 = n -1 which is still a whole number.

ALL even numbers are of the form 2X (cos you can ALWAYS divide it by 2)
where X= 1 or 2 or 3 etc

Therefore ALL odd numbers are of the form 2X+1 (cos you can NEVER divide it by 2 and get an integer)

so take any odd number =(2X+1) - square this (F2=the result)
take the next odd number = (2X+1) +2 (cos (2X+1)+1 is the next EVEN number)= 2X+3 - square this (F1 = the result)

Now subtract the results F1-F2

That how to do it- now you do it- you learn nothing by me doing it

Let n be an integer.
2n + 1 and 2n + 3 represent 2 consecutive odd numbers.
The difference between the squares of these consecutive odd numbers is
(2n + 3)² - (2n + 1)² = (2n + 3 + 2n + 1)(2n + 3 - 2n - 1)
. . . . . . . . . . . . . . .= (4n + 4)*2
. . . . . . . . . . . . . . .= 8(n + 1)
Since n is an integer, n + 1 is another integer and 8*(n + 1) is a multiple of 8.
QED

Let the two consecutive odd numbers be,
(2n + 1)
(2n - 1)

Then the difference in the squares
= (2n + 1)^2 - (2n - 1)^2
= 4n^2 + 4n + 1 - (4n^2 - 4n + 1 )
= 8n

So it is always a multiple of 8

2n + 1 = an odd number where n = 0,1,2,3...

[2(n+1)+1]^2 - (2n+1)^2

(2n + 3)^2 - (2n+1)^2

4n^2 + 12n + 9 - 4n^2 - 4n - 1

8n + 8

8(n+1)

Q.E.D