Edexcel IAL October 2022 P4 (WMA14/01) Q8 Proof by Contradiction
A student was asked to prove by contradiction that
“there are no positive integers x and y such that \( 3x^2 + 2xy – y^2 = 25 \)”
The start of the student’s proof is shown in the box below
Assume that integers x and y exist such that \( 3x^2 + 2xy – y^2 = 25 \)
\( \Rightarrow (3x – y)(x + y) = 25 \)
\( \Rightarrow (3x – y)(x + y) = 25 \)
If
\( (3x – y) = 1 \quad \text{and} \quad (x + y) = 25 \)
\( 3x – y = 1 \)
\( x + y = 25 \)
\( \Rightarrow 4x = 26 \Rightarrow x = 6.5, y = 18.5 \quad \text{Not integers} \)
Show the calculations and statements that are needed to complete the proof.
(4)
Solution to Question: Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume positive integers x, y exist.
- Factorisation: Factorise the left-hand side into two integer factors.
- Integer Factors of 25: The factors must be \( 1 \times 25 \), \( 5 \times 5 \), or \( 25 \times 1 \).
- Simultaneous Equations: Solving the resulting systems of linear equations and checking if x and y are integers.
Step-by-Step Solution:
- Form the Assumption and Factorise
Assume that positive integers x and y exist such that \( 3x^2 + 2xy – y^2 = 25 \). Factorise the left-hand side:
\( (3x – y)(x + y) = 25 \) - Identify Possible Factor Pairs
Since x and y are positive integers, \( x + y \) must be a positive integer factor of 25. The factors of 25 are 1, 5, and 25. Also, \( x + y > 3x – y \). The possible pairs are:
\( 3x – y = 1 \quad \text{and} \quad x + y = 25 \)
\( 3x – y = 5 \quad \text{and} \quad x + y = 5 \)
\(3x – y = 25\) and \(x + y = 1\) (This pair is impossible since \(x + y\) must be less than \(3x – y\) if \(x,y > 0\)). - Solve Case 1:
\( 3x – y = 1 \text{ and } x + y = 25 \)
Add the two equations:
\( (3x – y) + (x + y) = 1 + 25 \Rightarrow 4x = 26 \)
\( x = \frac{26}{4} = 6.5 \)Since x is not an integer, this case fails.
- Solve Case 2:
\( 3x – y = 5 \text{ and } x + y = 5 \)
Add the two equations:
\( (3x – y) + (x + y) = 5 + 5 \Rightarrow 4x = 10 \)
\( x = \frac{10}{4} = 2.5 \)Since x is not an integer, this case also fails.
- Identify the Contradiction
All possible pairs of integer factors lead to non-integer solutions for x and y. This contradicts the initial assumption that positive integers x and y exist.
Final Answer
Since all scenarios fail to produce positive integers x and y, the assumption is incorrect. Therefore, there are no positive integers x and y such that \( 3x^2 + 2xy – y^2 = 25 \).