AQA Paper 3 2019 June — Question 6 4 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof
TypeContradiction proof about integers
DifficultyStandard +0.8 Part (a) is trivial (1 mark for stating 6-8-10), but part (b) requires a modular arithmetic proof that odd squares are ≡1 (mod 4), leading to a contradiction in a²+b²=c². This is a non-routine proof requiring number theory insight beyond standard A-level content, though the technique is accessible once recognized.
Spec1.01a Proof: structure of mathematical proof and logical steps1.01d Proof by contradiction

The three sides of a right-angled triangle have lengths \(a\), \(b\) and \(c\), where \(a, b, c \in \mathbb{Z}\) \includegraphics{figure_6}
  1. State an example where \(a\), \(b\) and \(c\) are all even. [1 mark]
  2. Prove that it is not possible for all of \(a\), \(b\) and \(c\) to be odd. [3 marks]

Question 6:

AnswerMarks Guidance
6(a)States an appropriate even
Pythagorean triple2.2a B1
b=8
c=10

AnswerMarks
6(b)Begins an appropriate method of
proof assuming at least two sides
are odd eg states ‘assume a, b
odd’ or defines a, b (or c)
algebraically with different
AnswerMarks Guidance
unknowns3.1a B1
so a = 2m + 1 and b = 2n + 1
+
= + 24m + 1 + 2 + 4n + 1
(2𝑚𝑚+1) (2𝑛𝑛+1)
= 2 2
4𝑚𝑚 4 𝑛𝑛
2 2
wh2i(c2h𝑚𝑚 is e+ve2n𝑚𝑚, +so 2𝑛𝑛 is+ e2v𝑛𝑛en+, 1s)o c
is even. Therefore i2t is not possible
for all three to be ocdd.
Uses Pythagoras’ theorem with
at least two odd sides either in
AnswerMarks Guidance
words or algebraically1.1a M1
Completes rigorous argument to
prove the required result
AnswerMarks Guidance
CSO2.1 R1
Total4
QMarking instructions AO
Question 6:
--- 6(a) ---
6(a) | States an appropriate even
Pythagorean triple | 2.2a | B1 | a=6
b=8
c=10
--- 6(b) ---
6(b) | Begins an appropriate method of
proof assuming at least two sides
are odd eg states ‘assume a, b
odd’ or defines a, b (or c)
algebraically with different
unknowns | 3.1a | B1 | Assume a and b are odd
so a = 2m + 1 and b = 2n + 1
+
= + 24m + 1 + 2 + 4n + 1
(2𝑚𝑚+1) (2𝑛𝑛+1)
= 2 2
4𝑚𝑚 4 𝑛𝑛
2 2
wh2i(c2h𝑚𝑚 is e+ve2n𝑚𝑚, +so 2𝑛𝑛 is+ e2v𝑛𝑛en+, 1s)o c
is even. Therefore i2t is not possible
for all three to be ocdd.
Uses Pythagoras’ theorem with
at least two odd sides either in
words or algebraically | 1.1a | M1
Completes rigorous argument to
prove the required result
CSO | 2.1 | R1
Total | 4
Q | Marking instructions | AO | Mark | Typical solution
The three sides of a right-angled triangle have lengths $a$, $b$ and $c$, where $a, b, c \in \mathbb{Z}$

\includegraphics{figure_6}

\begin{enumerate}[label=(\alph*)]
\item State an example where $a$, $b$ and $c$ are all even. [1 mark]

\item Prove that it is not possible for all of $a$, $b$ and $c$ to be odd. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2019 Q6 [4]}}