| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | September |
| Marks | 4 |
| Topic | Proof |
| Type | Divisibility proof for all integers |
| Difficulty | Moderate -0.5 This is a straightforward proof requiring students to represent consecutive integers as n and n+1, expand (n² + (n+1)²) to get 2n² + 2n + 1, factor as 2n(n+1) + 1, then observe n(n+1) is always even (product of consecutive integers) to conclude the form is 4k+1. While it requires algebraic manipulation and understanding of even/odd properties, it's a standard proof technique with clear structure, making it slightly easier than average. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps |
4 Prove that the sum of the squares of any two consecutive integers is of the form $4 k + 1$, where $k$ is an integer.
\hfill \mbox{\textit{OCR H240/02 2018 Q4 [4]}}