Prove that when the sum of the squares of any two consecutive odd numbers is divided by 8, the remainder is always 2Show clear algebraic working

Sure, let's start by defining our two consecutive odd numbers.

An odd number can be represented as 2n+1, where n is any integer. The next consecutive odd number would then be 2n+3.

The sum of the squares of these two numbers is:

(2n+1)^2 + (2n+3)^2

Expanding this gives:

4n^2 + 4n + 1 + 4n^2 + 12n + 9

Simplifying this gives:

8n^2 + 16n + 10

We want to find the remainder when this is divided by 8.

The remainder when 8n^2 is divided by 8 is 0, and the remainder when 16n is divided by 8 is also 0.

So, we only need to find the remainder when 10 is divided by 8, which is 2.

Therefore, the sum of the squares of any two consecutive odd numbers, when divided by 8, always leaves a remainder of 2.