n is an integer - OCR - GCSE Maths - Question 21 - 2017 - Paper 1

Let the first odd number be represented as $x = 2n + 1$, where $n$ is an integer. The next consecutive odd number would then be $x' = 2(n + 1) + 1 = 2n + 3$.

Now we calculate the squares of these two odd numbers:

• The square of the first odd number:

  $x^2 = (2n + 1)^2 = 4n^2 + 4n + 1$

• The square of the second odd number:

  $x'^2 = (2n + 3)^2 = 4n^2 + 12n + 9$

Next, we find the difference between these squares:

$ext{Difference} = (2n + 3)^2 - (2n + 1)^2$

Calculating this gives:

$(4n^2 + 12n + 9) - (4n^2 + 4n + 1) = 12n + 9 - 4n - 1 = 8n + 8$

This can be factored as:

$= 8(n + 1)$

Since $n$ is an integer, $n + 1$ is also an integer. Therefore, the difference between the squares of two consecutive odd numbers is always a multiple of 8.