| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof about integers |
| Difficulty | Standard +0.8 This is a proof by contradiction involving Diophantine equations and parity arguments. While the technique is standard A-level fare, students must recognize that 4p² is always even, making the left side have the same parity as -q², and systematically work through cases (q odd/even) to reach contradiction. This requires more mathematical maturity than routine algebraic manipulation, placing it moderately above average difficulty. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assume there are positive integers \(p\) and \(q\) such that \((2p+q)(2p-q) = 25\) | M1 | 2.1 – For the key step in setting up the contradiction and factorising |
| If true then either \(2p+q=25\), \(2p-q=1\) or \(2p+q=5\), \(2p-q=5\) | M1 | 2.2a – Award for deducing either of the above statements. Note: ignore any reference to \(2p+q=1\), \(2p-q=25\) as this could not occur for positive \(p\) and \(q\) |
| Solutions are \(p=6.5, q=12\) or \(p=2.5, q=0\). Award for one of these. | A1 | 1.1b – For correctly solving one of the given statements. For first case, only need to find \(p=6.5\) to show contradiction. For second case, only need to find either \(p\) or \(q\). |
| This is a contradiction as there are no integer solutions, hence there are no positive integers \(p\) and \(q\) such that \(4p^2 - q^2 = 25\) | A1 | 2.1 – For a complete and rigorous argument with both possibilities and a correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4p^2 - q^2 = 25 \Rightarrow q^2 = 4p^2 - 25\) with \(4p^2\) even, or \(4p^2 = q^2 + 25\) with \(q^2\) (or \(q\)) odd | M1 | 2.1 |
| Sets \(q = 2n \pm 1\) and expands \((2n \pm 1)^2 = 4p^2 - 25\) | M1 | 2.2a |
| Proceeds to \(4p^2 = 4n^2 + 4n + 26 = 4(n^2+n+6)+2\) or \(p^2 = n^2 + n + \frac{13}{2}\) | A1 | 1.1b |
| This is a contradiction as \(4p^2\) must be a multiple of 4, or \(p^2\) must be an integer. Concludes there are no positive integers \(p\) and \(q\) such that \(4p^2 - q^2 = 25\) | A1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p\) even, \(q\) odd: \(4p^2 - q^2 = 4(2m)^2-(2n+1)^2 = 16m^2-4n^2-4n-1\) | — | One less than a multiple of 4, so cannot equal 25 |
| \(p\) odd, \(q\) odd: \(4p^2 - q^2 = 4(2m+1)^2-(2n+1)^2 = 16m^2+16m-4n^2-4n+3\) | — | Three more than a multiple of 4, so cannot equal 25 |
| M1: Set up contradiction and consider BOTH cases where \(q\) is odd | M1 | No requirement for evens case |
| A1: Correct work and deduction for one of the two scenarios | A1 | — |
| A1: Correct work and deductions for both scenarios with final conclusion | A1 | — |
## Question 16:
**Main Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume there are positive integers $p$ and $q$ such that $(2p+q)(2p-q) = 25$ | M1 | 2.1 – For the key step in setting up the contradiction and factorising |
| If true then either $2p+q=25$, $2p-q=1$ or $2p+q=5$, $2p-q=5$ | M1 | 2.2a – Award for deducing either of the above statements. Note: ignore any reference to $2p+q=1$, $2p-q=25$ as this could not occur for positive $p$ and $q$ |
| Solutions are $p=6.5, q=12$ or $p=2.5, q=0$. Award for one of these. | A1 | 1.1b – For correctly solving one of the given statements. For first case, only need to find $p=6.5$ to show contradiction. For second case, only need to find either $p$ or $q$. |
| This is a contradiction as there are no integer solutions, hence there are no positive integers $p$ and $q$ such that $4p^2 - q^2 = 25$ | A1 | 2.1 – For a complete and rigorous argument with both possibilities and a correct conclusion |
**Alt 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4p^2 - q^2 = 25 \Rightarrow q^2 = 4p^2 - 25$ with $4p^2$ even, or $4p^2 = q^2 + 25$ with $q^2$ (or $q$) odd | M1 | 2.1 |
| Sets $q = 2n \pm 1$ and expands $(2n \pm 1)^2 = 4p^2 - 25$ | M1 | 2.2a |
| Proceeds to $4p^2 = 4n^2 + 4n + 26 = 4(n^2+n+6)+2$ or $p^2 = n^2 + n + \frac{13}{2}$ | A1 | 1.1b |
| This is a contradiction as $4p^2$ must be a multiple of 4, or $p^2$ must be an integer. Concludes there are no positive integers $p$ and $q$ such that $4p^2 - q^2 = 25$ | A1 | 2.1 |
**Alt 2 (odd/even approach):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p$ even, $q$ odd: $4p^2 - q^2 = 4(2m)^2-(2n+1)^2 = 16m^2-4n^2-4n-1$ | — | One less than a multiple of 4, so cannot equal 25 |
| $p$ odd, $q$ odd: $4p^2 - q^2 = 4(2m+1)^2-(2n+1)^2 = 16m^2+16m-4n^2-4n+3$ | — | Three more than a multiple of 4, so cannot equal 25 |
| M1: Set up contradiction and consider BOTH cases where $q$ is odd | M1 | No requirement for evens case |
| A1: Correct work and deduction for one of the two scenarios | A1 | — |
| A1: Correct work and deductions for both scenarios with final conclusion | A1 | — |\begin{enumerate}
\item Prove by contradiction that there are no positive integers $p$ and $q$ such that
\end{enumerate}
$$4 p ^ { 2 } - q ^ { 2 } = 25$$
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q16 [4]}}