What is A6: Identities and proof?

An equation is satisfied by particular values, whereas an identity states that two expressions are equivalent for every permitted value.

How many marks are A6 questions worth?

In verified past-paper appearances, A6 questions on Edexcel GCSE Maths 1MA1 have been worth 2–7 marks.

Which past exam papers has A6 appeared in?

A6 has appeared in 8 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Know the difference between an equation and an identity; argue mathematically to show algebraic expressions are equivalent, and use algebra to support and construct arguments and proofs

Identities and proof

Worked answers, methods and verified real exam appearances for A6 on Edexcel GCSE Maths 1MA1.

Explanation

An equation is satisfied by particular values, whereas an identity states that two expressions are equivalent for every permitted value.
To show expressions are equivalent, expand or factorise one side until it matches the other.
Foundation tier: use algebra to support and construct an argument by defining quantities with variables, translating the claim and linking the resulting algebra back to it.
Higher tier: extend this to a proof by using a general form, such as $2n$ for an even integer or $2n+1$ for an odd integer, and reasoning that covers every permitted case.
Checking examples alone is not a proof.

Worked example

Higher tier: prove algebraically that the sum of two consecutive integers is odd.

1.Let the first integer be $n$ , so the next is $n+1$ .
2.Their sum is $n+(n+1)=2n+1$ .
3. $2n$ is even for every integer $n$ , so $2n+1$ is odd.

Answer: The sum has form $2n+1$ , so it is odd.

Common mistakes

Don't treat expressions that agree for one value as equivalent without simplifying them generally.
Don't check several numerical cases and call the pattern a proof (Higher tier).
Don't finish with algebra and fail to state why its form proves the claim (Higher tier).

Exam tip

Foundation tier: show each algebraic step and link the result to the argument. Higher tier: a “prove” question needs a general variable-based argument, not examples.

Worked practice

Tier 1 · Easy

Show that

4(n+2)-3(n-1)

is equivalent to

n+11

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$4(n+2)-3(n-1)\equiv n+11$	2	Expand both brackets, remembering that the subtraction acts on both terms: $4n+8-3n+3=n+11$ . Therefore the two expressions are equivalent.

Tier 2 · Standard

Prove algebraically that the sum of two consecutive integers is odd.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$n+(n+1)=2n+1$ , so the sum is odd	3	Let the first integer be $n$ , so the next is $n+1$ . Their sum is $n+(n+1)=2n+1$ . Since $2n$ is even for every integer $n$ , $2n+1$ is odd, proving the claim.

Tier 3 · Hard

An odd number is written as

2n+1

. Demonstrate algebraically that squaring it leaves remainder

1

after division by

8

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$(2n+1)^2=8k+1$ for an integer $k$	4	Write an odd integer as $2n+1$ . Then $(2n+1)^2=4n^2+4n+1=4n(n+1)+1$ . One of the consecutive integers $n$ and $n+1$ is even, so $n(n+1)=2k$ for some integer $k$ . Therefore the square is $4(2k)+1=8k+1$ .

Tier 1 · Easy

Find the values of

a

and

b

that make

6(2n-1)+5\equiv an+b

an identity.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$a=12$ $b=-1$	2	Expand and collect the constant terms: $6(2n-1)+5=12n-6+5=12n-1$ . Therefore $a=12$ and $b=-1$ .

Tier 2 · Standard

Priya says that

(x+2)^2

is equivalent to

x^2+4

. Show that Priya is not correct.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$(x+2)^2=x^2+4x+4$ , so it is not equivalent to $x^2+4$ (or a counterexample: $x=1$ gives $9\ne5$ )	3	Write the square as two factors and expand: $(x+2)^2=(x+2)(x+2)=x^2+2x+2x+4=x^2+4x+4$ . The term $4x$ is missing from Priya's expression, so her statement is not correct.

Tier 3 · Hard

A row of

n

squares is made with matchsticks. Four matchsticks are counted for each square, then one shared matchstick is removed for each of the

n-1

joins. Show that the number of matchsticks is

3n+1

(2)

(Total for Question 6 is 2 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$4n-(n-1)=3n+1$ matchsticks	2	Counting four matchsticks for each of the $n$ squares gives $4n$ . There are $n-1$ shared joins, so subtracting these gives $4n-(n-1)=4n-n+1=3n+1$ .

Tier 2 · Standard

Show that

2(2n-3)-1

is equivalent to

4n-7

. Hence work out the value of

n

for which

2(2n-3)-1=45

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$2(2n-3)-1=4n-6-1=4n-7$ $n=13$	3	Expand the bracket to get $2(2n-3)-1=4n-6-1=4n-7$ , so the expressions are equivalent. Then $4n-7=45$ gives $4n=52$ , so $n=13$ .

Tier 3 · Hard

Find the value of

k

that makes

3(2x-5)+k(x+1)\equiv10x-11

an identity. Show that both the

x

-coefficient and the constant term then match.

(3)

(Total for Question 8 is 3 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$k=4$ Check: $x$ -coefficient $6+4=10$ and constant $4-15=-11$ , both matching the right side.	3	Expanding the left side gives $(6+k)x+(k-15)$ . Matching the coefficient of $x$ gives $6+k=10$ , so $k=4$ . The constant term is then $4-15=-11$ , which also matches the right side.

Tier 3 · Hard

Higher only: Prove algebraically that the product of any three consecutive integers is divisible by

6

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	If $n=3k$ , then $n$ is a multiple of $3$ ; if $n=3k+1$ , then $n+2=3(k+1)$ ; if $n=3k+2$ , then $n+1=3(k+1)$ . $n(n+1)(n+2)$ also contains an even factor, so it is divisible by $2\times3=6$ .	4	Write the integers as $n$ , $n+1$ and $n+2$ . Every integer has one of the forms $3k$ , $3k+1$ or $3k+2$ . In these three cases respectively, $n$ , $n+2$ or $n+1$ is a multiple of $3$ . Also, if $n$ is even then the product contains the even factor $n$ ; if $n$ is odd then $n+1$ is even. Therefore $n(n+1)(n+2)$ always contains factors $2$ and $3$ , so it is divisible by $6$ .

Q10

Tier 3 · Hard

Higher only: A two-digit positive integer has tens digit

a

and non-zero units digit

b

. A second number is formed by interchanging the digits. Prove that the sum of the two numbers is a multiple of

11

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$(10a+b)+(10b+a)=11(a+b)$ , so the sum is a multiple of $11$	4	The original number is $10a+b$ and the number with its digits interchanged is $10b+a$ . Their sum is $10a+b+10b+a=11a+11b=11(a+b)$ . Since $a+b$ is an integer, the sum is a multiple of $11$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	3H	Q14	3	Allowed	Higher	QP MS
2023-06	3H	Q13	3	Allowed	Higher	QP MS
2023-06	3H	Q15	3	Allowed	Higher	QP MS
2019-06	1H	Q21	7	Non-calculator	Higher	QP MS
2023-06	3H	Q12	3	Allowed	Higher	QP MS
2019-06	1H	Q13	2	Non-calculator	Higher	QP MS
2019-11	3H	Q15	3	Allowed	Higher	QP MS
2022-11	1H	Q16	4	Non-calculator	Higher	QP MS

Other points in A Algebra · equations and reasoning

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