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.
