A Level Math RevisionProve algebraically that the difference between the squares of any two consecutive odd integers is always a multiple of $8$.
Mark scheme and worked answer
[M1] for defining two consecutive odd integers, e.g., $2n - 1$ and $2n + 1$, where $n \in \mathbb{Z}$.
[M1] for setting up the difference of their squares: $(2n + 1)^2 - (2n - 1)^2$.
[A1] for expanding and simplifying correctly: $(4n^2 + 4n + 1) - (4n^2 - 4n + 1) = 8n$.
[R1] for concluding that since $n$ is an integer, $8n$ is a multiple of $8$. [AG]