Mathematics · Oct / Nov 2022 · Paper 41 · Question 8
YMA01
As printed in the paper
The question as text
8. A student was asked to prove by contradiction that “there are no positive integers x and y such that 3x2 + 2xy – y2 = 25” The start of the student’s proof is shown in the box below. Assume that integers x and y exist such that 3x2 + 2xy – y2 = 25 ( ) ( ) = ⇒ 3 x − y x + y 25 ( ) = ( ) = If 3 x − y x + y 1 and 25 3 x − y = 1 . Not integers ⇒ 4 x = 26 ⇒ x = 6 5 . , y = 18 5 x + y = 25 Show the calculations and statements that are needed to complete the proof. (4)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
Solves one of
(
)
(
)
x
−
y
=
x
+
y
=
or 3
5 and
5
Correct solution of one.
✓
3
x
−
y
=
25
(
)
Either
4
x
=
26
x
=
6.5,
y
=−
5.5
x
+
y
=
1
✓
3
x
−
y
=
5
(
)
Or
4
x
=
10
x
=
2.5,
y
=
2.5
x
+
y
=
5
Solves both equations
Both solved correctly with a minimal reason given for the contradiction e.g
''not integers'' with conclusion ''hence there are no integers x and y such that
3
x
2
+
2
xy
−
y
2
=
25
''
Notes
M1: Attempts to solve one of the two possible cases.
Take as a minimum, one correct pair of equations followed by a value for x or a value for y
A1: Correctly solves one of the two possible cases.
To solve you need only find a value for x or y. Once a correct value is found you can ISW
dM1: Attempts to solve both possible cases
A1: Correctly solves the two possible cases and makes a concluding argument.
To score this mark (i) all calculations must be correct.
(ii) reason(s) for the contradiction must be written down. Allow ''not
integers'', ''''
and (iii) gives a concluding statement must be given. Allow for example '' hence
proven''
Ignore any possible cases which would give rise to negative numbers but satisfy
(
)(
)
3
x
−
y
x
+
y
=
25
E.g (
)
(
)
3
x
−
y
= −
5,
x
+
y
= −
5
Withhold the final mark only if they include cases which do not satisfy (
)(
3
x
−
y
x
+
E.g (
)
(
)
3
x
−
y
=
20,
x
+
y
=
5
2
−
λ
−
4
+
5
μ
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