You are given that n is a positive integer. By expressing (x^2n)-1 as a product of factors, prove that (2^2n)-1 is divisible by 3.

X²ⁿ-1 = (xⁿ+1)(xⁿ-1) Therefore we can say 2²ⁿ-1 = (2ⁿ+1)(2ⁿ-1) . As 2ⁿ is always even, a multiple of 3 is always either going to be 1 above or 1 below it, e.g. 3 is one below 4 and 9 is 1 above 8, therefore either (2ⁿ+1) or (2ⁿ-1) is going to be a multiple of 3, making the entire equation 2²ⁿ-1 divisible by 3 as (2ⁿ+1) and (2ⁿ-1) are multiplied together, and they keep their factors.