Mathematics · May / June 2022 · Paper 41 · Question 9

For question 9 many variations on the proof are possible. Below is a general outline

with some examples, which cover many cases. If you see an approach you do not know

how to score, consult your team leader.

M1: Will be scored for setting up an algebraic statement in terms of a variable (integer)

k or any other variable aside n that engages with divisibility by 4 in some way and can

lead to a contradiction and is scored at the point you can see each of these elements. A

formal statement of the assumption is not required at this stage.

A1: Scored for a correct statement from which it is possible to draw a contradiction.

dM1; For making a complete argument that leads to a (full) contradiction of the initial

statement, though may be allowed if there are minor gaps or omissions.

A1: Correct and complete work with contradiction drawn and conclusion made. There

must have been a statement of assumption at the start for which to draw the

contradiction, though it may not be technicality a correct assumption as long as a

2

relevant assumption has been made. E.g. Accept “Assume

n −

2

is divisible be 4 for

all n”

9

n

2

−=

2

4

k

(Assume that there is an n with

n − is divisible by 4 so)

2

2

then

n

2

=

4

k

+

2

=

2(2

k

+

1)

(so is even)

2n is a multiple of 4

Hence

n is even so n (=2m) is even hence

2

n is a multiple of 4 then

2

n

2

−

2

=

4

m

2

−

2

=

2(2

m

2

−

1)

As

cannot be a multiple of 4

(as 2m － 1 is odd) so there is a contradiction.

So the original assumption has been shown false.

n −

2

2

Hence “

is never divisible by 4” is true for all n *

Notes

M1: Sets up an algebraic statement in terms of a variable (integer) k or any other variable aside n that engages

with divisibility by 4 in some way and can lead to a contradiction. No need for explicit statement of

assumption accept if just a suitable equation is set up. In this case supposing divisibility by 4 by stating

n

2

−

2

=

4

k

A1: Reaches

n

2

=

2(2

k

+

1)

dM1: For a complete argument that leads to a contradiction. See scheme. Allow if minor details are omitted as

long as the overall argument is clear.

2n is a multiple of 4 so

Accept explanations such as “as

n is even then n is even hence

2

n − cannot be a

2

2

multiple of 4 (as 4 does not divide 2)”

A1*: Draws the contradiction to their initial assumption and concludes the statement is true for all n. There must

have been a clear assumption at the start that is contradicted, and all working must have been correct. For

the assumption be generous with the technicality as long as a relevant assumption has been made. E.g.

2

Accept “Assume

n −

2

is divisible be 4 for all n”

9

n −

2

2

2

Alt 1

n −

2

(Assume that

is divisible by 4 for some n,) so

is an integer. Then if n

4

2

2

n

−

2

(2 )

m

−

2

=

is even n = 2m (m integer) so

(oe with odd)

4

4

1

2

=

m

−

(which is not an integer)

2

1

2

Since m2 is an integer,

m −

is not, hence n cannot be even, but if n is odd then

2

(

)

2

2

m

1

2

2

2

+

−

1

n

−

=

=

m

2

+

m

−

, which is again not an integer (since m2 + m

4

4

4

is)

Hence there is a contradiction (as n cannot be an integer)

n −

2

2

Hence “

is never divisible by 4” is true for all n *

Notes

M1: Sets up an algebraic statement in terms of a variable (integer) m or any other variable aside n that engages

with divisibility by 4 in some way and can lead to a contradiction. No need for explicit statement of

assumption accept if just a suitable equation is set up. In this Alt, consider case use of n = 2m or n = 2m +

2

n −

2

1 in

is sufficient

4

1

1

2

2

A1: Reaches

m −

for n even or

m

+

m

−

for n odd.

2

4

dM1: For a complete argument that leads to a contradiction in both cases. See scheme. Allow if minor details are

omitted as long as the overall argument is clear.

A1*: Draws the contradiction to their initial assumption and concludes the statement is true for all n. There must

have been a clear assumption at the start that is contradicted, and all working must have been correct. For

the assumption be generous with the technicality as long as a relevant assumption has been made. E.g.

n −

2

2

Accept “Assume

is divisible be 4 for all n”

9

n −

2

2

n

2

−=

2

4

k

(Assume that

is divisible by 4)

Alt 2

1 or

n

2

=

4

k

+

2

n

=

2 k

+

n

=

2 2

k

+

1

2

2

1

m

m

k

2

k

1

So for some integer m

+

=

+ =

but m2 is odd if m is odd so

2

2

2

m not an integer, or m2 is a multiple of 4 if m even, so odd=even

2

2

or 2k + 1 is odd, so does not have a factor 2 to combine with the

2 outside, hence

n must be irrational

Hence we have a contradiction.

n −

2

2

So “

is never divisible by 4” is true for all n *

Notes

M1: Sets up an algebraic statement in terms of a variable (integer) k or any other variable aside n that engages

with divisibility by 4 in some way and can lead to a contradiction. No need for explicit statement of

assumption accept if just a suitable equation is set up. In this case supposing divisibility by 4 by stating

n

2

−

2

=

4

k

1

n

=

2

k

+

A1: Reaches

or

n

=

2 2

k

+

1

2

dM1: For a complete argument that leads to a contradiction. See scheme. Allow if minor details are omitted as

1

long as the overall argument is clear. Must be a valid attempt to show that

2

k +

/

2 2

k + is not an

2

integer, and this method is a hard route.

A1*: Draws the contradiction to their initial assumption and concludes the statement is true for all n. There must

have been a clear assumption at the start that is contradicted, and all working must have been correct. For

the assumption be generous with the technicality as long as a relevant assumption has been made. E.g.

n −

2

2

Accept “Assume

is divisible be 4 for all n”

9

n −

2

2

(Assume that

is divisible by 4) then for n even we have (for some integer m)

Alt 3

(

)

n

2

−

2

=

4

m

2

−

2

or for n odd

n

2

−

2

=

4

m

2

+

m

−

1

or (

)

4

m −

2

2

4

m

2

+

m

−

1

2

2

Since 4 divides

n −

2

and

4m thus for n even, 4 must divide 2, a contradiction,

so n cannot be even, and also 4 divides (

)

4 m

2

so for n odd, 4 divides 1, also a

+

m

contradiction.

Hence we have a contradiction for both cases (and as n must be either even or odd).

n −

2

2

so “

is never divisible by 4” is true for all n *

Notes

M1: Sets up an algebraic statement in terms of a variable (integer) m or any other variable aside n that engages

with divisibility by 4 in some way and can lead to a contradiction. No need for explicit statement of

assumption accept if just a suitable equation is set up. In this case supposing using n odd or n even to form

n −

2

2

4×

integer

±

nonmulitple of 4

an expression for

of the form

or (

)

4

2

2

A1: Reaches

m −

4

m

2

+

m

−

1

dM1: For a complete argument that leads to a contradiction. See scheme. Allow if minor details are omitted as

long as the overall argument is clear. Both cases must be considered with a reason for the contradiction

given (not just stated not divisible by 4).

A1*: Draws the contradiction to their initial assumption and concludes the statement is true for all n. There must

have been a clear assumption at the start that is contradicted, and all working must have been correct. For

the assumption be generous with the technicality as long as a relevant assumption has been made. E.g.

Accept “Assume

n −

2

2

is divisible be 4 for all n”

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