The question is: Prove algebraically that the difference between the squares of any two

Question

consecutive even numbers is always a multiple of 4.

My working was: Let the numbers be n and n + 2
(n+2)^2 - (n)^2
=n^2 + 4n + 4 - n^2
=4n + 4
=4(n + 1)

But the answers say the numbers should be 2n + 2 and 2n, why is this or is mine right too?

Thanks

Answer 1

any even number is a multiple of 2.
When you pick any old n, it might not be even.

You have proven that any two numbers which differ by 2 have their squares differing by a multiple of 4.

So, technically you have not done what was asked, but have proved an even stronger result.

Using 2n, the proof is even easier:

(2n+2)^2 - (2n)^2
= 4(n+1)^2 - 4n^2
= 4((n+1)^2 - n^2)
QED

Answer 2

I understand thanks!