Prove that for any natural value of n the value of the expression (n+2)^2-(n-2)^2 is a multiple of 8.

We have the expression:

(n + 2)^2 - (n - 2)^2

Let´s break the parentheses:

(n + 2)^2 = n^2 + 4*n + 4

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

Then:

(n + 2)^2  - (n - 2)^2 = (n^2 + 4*n + 4) - (n^2 - 4n + 4) =

                                 = (n^2 - n^2) + (4 - 4) + (4n - (-4n)) = 4n - (-4n) = 8*n

Then for any natural value of n, 8*n will be a multiple of 8.

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