n is an integer greater than 1. Prove algebraically that n^2-2-(n-2)^2 is always an even number

We want to show that this gives an even number, so we need to get it into a form that it is 2 multiplied by some positive integer. We start by expanding the (n-2)² out, remember that (n-2)²=(n-2)(n-2). Then expand the brackets to get (n-2)(n-2)=n² -4n + 4. We can then sub this back in to our original equation to get n² - 2 - (n-2)² = n² - 2 - (n² -4n + 4), being careful to keep the part we have substituted in, inside the brackets. Then, we can take this out of the brackets to get n² - 2 - (n² -4n + 4) = n² - 2 - n² +4n - 4. Then we can collect like terms and cancel out the n² and -n² to get n² - 2 - n² +4n - 4 = 4n - 6. We then notice that we can factorise 4n-6, as both 4 and 6 are divisible by 2. So we have 4n - 6 = 2(2n-3), which is in the required form, so we have shown n² - 2 - (n-2)² is even for all integer n greater than 1. Note that for n greater than 1, (2n-3) is greater than 0, so this always works.