n is an integer greater than 1 Prove algebraically that $n^2 - 2 - (n - 2)^2$ is always an even number. - Edexcel - GCSE Maths - Question 16 - 2017 - Paper 1

To prove that the expression is always even, we start with:

$n^2 - 2 - (n - 2)^2$

Next, let’s expand the term $(n - 2)^2$:

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

Now substituting this back into the expression:

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

This simplifies to:

$n^2 - 2 - n^2 + 4n - 4$

Combining like terms gives:

$4n - 6$

We can factor this expression as follows:

$4n - 6 = 2(2n - 3)$

Since both $4n$ and $-6$ are even numbers, the entire expression $2(2n - 3)$ is also even, as it is a multiple of 2. Hence, it can be concluded that the original expression is always even for any integer $n > 1$.