Algebraic Proof · Study Notes · Edexcel IGCSE 4MA1 Higher

Use algebra to prove identities and properties of integers. A* topic — uncommon but worth marks when it appears.

At a glance

Proof = demonstration valid for ALL cases (use letters, not numbers).
Even integers: $2n$ . Odd: $2n + 1$ .
Consecutive: $n, n+1$ , or symmetric $n-1, n, n+1$ .
Show LHS = RHS for identities.
Show divisibility / oddness / evenness.
Conclude with 'QED', ' $\therefore$ ', or 'as required'.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.9 — Prove algebraic identities.
2.9 — Prove properties of integers (oddness, evenness, divisibility).
2.9 — Use general variables to argue for all cases.

Proving identities

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Expand both sides until they match. State QED.

An identity is an equation true for ALL values of the variable.

To prove an identity:

Take the LHS.
Expand and simplify.
Show it equals the RHS.

Example. Prove $(2n + 1)^{2} - (2n - 1)^{2} = 8n$ .

LHS: $(2n + 1)^{2} - (2n - 1)^{2}$ $= (4n^{2} + 4n + 1) - (4n^{2} - 4n + 1)$ $= 4n^{2} + 4n + 1 - 4n^{2} + 4n - 1$ $= 8n$ .

LHS = RHS. QED.

Difference of squares trick.

Often $(a + b)^{2} - (a - b)^{2}$ simplifies via the identity $(a + b)^{2} - (a - b)^{2} = 4ab$ . (Where $a = 2n, b = 1$ gives $4(2n)(1) = 8n$ .)

Worked qualitative. Why prove identities algebraically?

Numerical examples are not proofs.
$\sin^{2} + \cos^{2} = 1$ for $\theta = 30°, 45°, 60°$ — but this is just verification.
Algebraic proof works for ALL angles.

Edexcel tip. Always work from one side to the other. Don't 'meet in the middle' — that's circular reasoning.

Identity: true for all values.
Expand and simplify LHS.
Match to RHS.
QED at the end.

Properties of integers

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Set up with letters; manipulate to show desired structure.

Standard variable choices:

Set	Variable choice
Two consecutive integers	$n, n+1$
Three consecutive integers	$n-1, n, n+1$ (symmetric) or $n, n+1, n+2$
Two consecutive even integers	$2n, 2n+2$
Two consecutive odd integers	$2n+1, 2n+3$
Even integer	$2n$
Odd integer	$2n + 1$

Even is a block of 2n; odd adds one; symmetric n−1, n, n+1 makes sums tidy.

Example: Prove three consecutive integers sum to a multiple of $3$ .

Let them be $n-1, n, n+1$ .
Sum: $(n-1) + n + (n+1) = 3n$ .
$3n$ is divisible by $3$ . QED.

Example: Difference of squares of two consecutive integers is odd.

Two consecutive: $n, n+1$ .
Difference of squares: $(n+1)^{2} - n^{2} = 2n + 1$ .
$2n + 1$ is odd. QED.

Example: Sum of squares of consecutive odds is even.

Two consecutive odd: $2n+1, 2n+3$ .
Sum of squares: $(2n+1)^{2} + (2n+3)^{2}$ $= 4n^{2} + 4n + 1 + 4n^{2} + 12n + 9$ $= 8n^{2} + 16n + 10 = 2(4n^{2} + 8n + 5)$ .
This is $2 \times$ integer = even. QED.

Worked qualitative. Why use $n-1, n, n+1$ instead of $n, n+1, n+2$ ?

Symmetric choice often simplifies.
$(n-1) + n + (n+1) = 3n$ is cleaner than $n + (n+1) + (n+2) = 3n + 3$ .
Both work; choose the simpler.

Edexcel tip. Always START by defining variables: 'Let $n$ be an integer'. This earns the M1.

Define variables symbolically.
Even = $2n$ . Odd = $2n+1$ .
Consecutive: $n, n+1$ or symmetric.
Show structure (factor out, manipulate).

'Show that' problems

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Algebra to verify a stated equation.

'Show that' questions ask you to verify an algebraic equation by manipulation.

Method:

Start from one side (usually the messier one).
Expand, simplify, factor as appropriate.
Reach the form on the other side.

Example. Show that $(x + 2)(x - 1)(x - 3) = x^{3} - 2x^{2} - 5x + 6$ .

LHS: Expand step by step. $(x + 2)(x - 1) = x^{2} + x - 2$ . $(x^{2} + x - 2)(x - 3) = x^{3} - 3x^{2} + x^{2} - 3x - 2x + 6 = x^{3} - 2x^{2} - 5x + 6$ .

LHS = RHS. QED.

Worked qualitative. What's the difference between 'solve' and 'show that'?

'Solve' asks for value of $x$ .
'Show that' asks you to PROVE a given equation/identity.
Different tasks; different approaches.

Edexcel tip. ALL working must be shown for 'show that' questions. Mark schemes specifically deduct for missing steps.

Start from one side.
Manipulate to reach the other.
ALL steps shown.
Conclude clearly.

Quick recap

Proof: general, not specific.
Use letters: $n$ for integers.
Even $2n$ , odd $2n+1$ .
Consecutive: $n, n+1, \ldots$ .
Expand both sides; show equality.
Conclude with QED.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Even = $2n$ .
Odd = $2n + 1$ .
Consecutive = $n, n+1$ or symmetric.
Multiple of 3 = $3n$ .
Square = $n^{2}$ .
Conclude with 'QED' or ' $\therefore$ '.

How it’s examined

Algebraic proof appears Paper 2H (3-5 marks). Less common than other Topic 2 areas but reliable for A* grades. Examiner reports flag (1) using specific examples instead of general, (2) missing the conclusion, (3) poor variable choice complicating the proof.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on algebraic proof, written exactly the way a tutor would explain them at the board.

1Prove an identity
Higher• proof, identity
▼
Question
Prove that $(2n + 1)^{2} - (2n - 1)^{2} = 8n$ for all integers $n$ .
Step-by-step solution
1. Step 1
  Expand the LHS: $(2n + 1)^{2} = 4n^{2} + 4n + 1$ .
2. Step 2
  $(2n - 1)^{2} = 4n^{2} - 4n + 1$ .
3. Step 3
  Subtract: $(4n^{2} + 4n + 1) - (4n^{2} - 4n + 1) = 8n$ .
4. Step 4
  LHS = RHS. QED.
Answer
$8n$
Examiner tip
Always START with the LHS, work to match RHS. State QED or '∎' to indicate proof is complete.
2Prove a divisibility property
Higher• Adapted from 4MA1/2H May/Jun 2024 Q22• proof, divisibility
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Question
Prove that the difference between the squares of two consecutive integers is always odd.
Step-by-step solution
1. Step 1
  Let two consecutive integers be $n$ and $n + 1$ .
2. Step 2
  Difference of squares: $(n + 1)^{2} - n^{2}$ .
3. Step 3
  Expand: $n^{2} + 2n + 1 - n^{2} = 2n + 1$ .
4. Step 4
  $2n + 1$ is one more than an even number, so it's ODD. QED.
Answer
Difference is $2n + 1$ , always odd.
Examiner tip
Define variables CAREFULLY. 'Two consecutive integers $n$ and $n + 1$ ' is the standard setup. Mark schemes credit this clarity.
3Prove an expression is a multiple of $n$
Higher• proof, multiple
▼
Question
Prove that the sum of three consecutive integers is always a multiple of $3$ .
Step-by-step solution
1. Step 1
  Let three consecutive integers be $n - 1, n, n + 1$ .
2. Step 2
  Sum: $(n - 1) + n + (n + 1) = 3n$ .
3. Step 3
  $3n$ is divisible by $3$ . QED.
Answer
Sum is $3n$ , a multiple of $3$ .
Examiner tip
Choosing variables symmetrically ( $n - 1, n, n + 1$ ) often simplifies proofs. Try this pattern when you see 'three consecutive'.
4Show that — algebraic manipulation
Higher• proof, show that
▼
Question
Show that $(x + 4)(x + 2)(x - 3) = x^{3} + 3x^{2} - 10x - 24$ .
Step-by-step solution
1. Step 1
  Expand the first two: $(x + 4)(x + 2) = x^{2} + 6x + 8$ .
2. Step 2
  Multiply by $(x - 3)$ :
  $(x^{2} + 6x + 8)(x - 3) = x^{3} - 3x^{2} + 6x^{2} - 18x + 8x - 24$
3. Step 3
  Simplify: $x^{3} + 3x^{2} - 10x - 24$ . ✓
Answer
Verified.
5Explaining why an expression is always odd [1 mark]
Getting started• Paper 1H/2H style — 1 mark• proof, odd and even
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Question
$n$ is an integer.

Explain why $2n + 7$ is always an odd number. [1]
Step-by-step solution
1. Step 1
  An odd number is one more than a multiple of $2$ . Split $7$ as $6 + 1$ so that the even part can be factorised.
2. Step 2
  B1 Write the expression as $2 \times$ an integer plus $1$ , and conclude.
  $2n + 7 = 2n + 6 + 1 = 2(n + 3) + 1$
Answer
$2n + 7 = 2(n + 3) + 1$ , which is one more than a multiple of $2$ , so it is odd.
Examiner tip
Examiner tip: An equally good explanation in words is " $2n$ is even, and even $+$ odd ( $7$ ) is odd". Either way, give a reason.

Common mistake: Substituting a value such as $n = 1$ to get $9$ and saying "it is odd". One example does not explain why it is always odd.
6Proving a multiple of 3 with a negative bracket [2 marks]
Getting started• Paper 1H/2H style — 2 marks• proof, multiples, expanding brackets
▼
Question
$n$ is an integer.

Prove that $5(n + 2) - 2(n - 4)$ is always a multiple of $3$ . [2]
Step-by-step solution
1. Step 1
  M1 Expand both brackets. The $-2$ multiplies both terms of $(n - 4)$ , and $-2 \times -4 = +8$ .
  $5(n + 2) - 2(n - 4) = 5n + 10 - 2n + 8 = 3n + 18$
2. Step 2
  A1 Take out the factor $3$ and conclude.
  $3n + 18 = 3(n + 6)$
Answer
$3(n + 6)$ , which is $3$ times an integer, so it is always a multiple of $3$ .
Examiner tip
Examiner tip: Treat the minus sign as part of the multiplier: expand $-2(n - 4)$ , not $2(n - 4)$ followed by a subtraction you might forget. Then factorise, because $3n + 18$ on its own does not show the multiple.

Common mistake: Writing $-2n - 8$ , giving $3n + 2$ , which is not a multiple of $3$ at all. Sign errors with negatives, such as $-8m + 5m$ , are among the errors examiners report most often.
7The middle of three consecutive integers [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• proof, consecutive numbers, expanding brackets
▼
Question
Prove that, for any three consecutive integers, the square of the middle integer is $1$ more than the product of the other two. [3]
Step-by-step solution
1. Step 1
  M1 Call the middle integer $n$ , so the three integers are $n - 1$ , $n$ and $n + 1$ . Form the difference between the square of the middle one and the product of the other two.
  $n^{2} - (n - 1)(n + 1)$
2. Step 2
  M1 Expand the product (a difference of two squares), keeping it in a bracket.
  $= n^{2} - (n^{2} - 1)$
3. Step 3
  A1 Simplify and conclude.
  $= n^{2} - n^{2} + 1 = 1$
Answer
$n^{2} - (n - 1)(n + 1) = 1$ , so the square of the middle integer is always $1$ more than the product of the other two.
Examiner tip
Examiner tip: Putting $n$ in the middle makes the outer pair $n - 1$ and $n + 1$ , whose product is simply $n^{2} - 1$ . Check with $4$ , $5$ , $6$ : $5^{2} = 25$ and $4 \times 6 = 24$ .

Common mistake: Writing $(n - 1)(n + 1) = n^{2} + 1$ , or dropping the bracket so that $n^{2} - n^{2} - 1 = -1$ . Either slip reverses the conclusion.
8Proving an identity with algebraic fractions [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• proof, algebraic fractions, identity
▼
Question
Show that $\dfrac{x + 1}{x - 1} - \dfrac{x - 1}{x + 1} \equiv \dfrac{4x}{x^{2} - 1}$ [4]
Step-by-step solution
1. Step 1
  M1 Write both fractions over the common denominator $(x - 1)(x + 1)$ .
  $\frac{(x + 1)^{2} - (x - 1)^{2}}{(x - 1)(x + 1)}$
2. Step 2
  M1 Expand both squares in the numerator, keeping the second in a bracket.
  $(x^{2} + 2x + 1) - (x^{2} - 2x + 1)$
3. Step 3
  A1 Simplify the numerator.
  $x^{2} + 2x + 1 - x^{2} + 2x - 1 = 4x$
4. Step 4
  A1 Expand the denominator as a difference of two squares, giving the right-hand side.
  $\frac{4x}{(x - 1)(x + 1)} = \frac{4x}{x^{2} - 1}$
Answer
$\dfrac{4x}{x^{2} - 1}$ , as required.
Examiner tip
Examiner tip: Do the expansion and the collecting in separate lines. The examiners note that doing both in one untidy step is where most errors creep in.

Common mistake: Using $x^{2} - 1$ as the denominator but not multiplying each numerator by the other bracket, or cancelling the $x$ terms across the minus sign, e.g. $\dfrac{x + 1}{x - 1} = -1$ . You can only cancel common factors, never terms.
9Squares of numbers that are not multiples of 3 [5 marks]
Stretch• Paper 1H/2H style — 5 marks• proof, multiples, cases
▼
Question
Prove that the square of any integer that is not a multiple of $3$ is always $1$ more than a multiple of $3$ . [5]
Step-by-step solution
1. Step 1
  B1 An integer that is not a multiple of $3$ is either $1$ more or $2$ more than a multiple of $3$ , so it is $3k + 1$ or $3k + 2$ for some integer $k$ .
2. Step 2
  M1 Square the first case.
  $(3k + 1)^{2} = 9k^{2} + 6k + 1$
3. Step 3
  A1 Take out the factor $3$ from the first two terms.
  $= 3(3k^{2} + 2k) + 1$
4. Step 4
  M1 Square the second case.
  $(3k + 2)^{2} = 9k^{2} + 12k + 4$
5. Step 5
  A1 Split $4$ as $3 + 1$ , factorise and conclude for both cases.
  $= 9k^{2} + 12k + 3 + 1 = 3(3k^{2} + 4k + 1) + 1$
Answer
In both cases the square is $3 \times$ an integer $+ 1$ , so it is always $1$ more than a multiple of $3$ .
Examiner tip
Examiner tip: When a statement is about numbers that are "not a multiple of $3$ ", split into the cases that cover every such number. Using $3k - 1$ instead of $3k + 2$ is equally good and gives $3(3k^{2} - 2k) + 1$ .

Common mistake: Proving only the $3k + 1$ case. Numbers such as $2$ , $5$ and $8$ are not of that form, so half the numbers are left unproved and the final mark is lost.
10Disproving one claim and proving another [5 marks]
Stretch• Paper 1H/2H style — 5 marks• proof, counter-example, odd and even
▼
Question
Zara says,

"If the sum of two whole numbers is even, then both numbers must be even."

(a) Give an example to show that Zara is wrong. [1]

(b) Prove that the sum of the squares of any two odd numbers is even but is never a multiple of $4$ . [4]
Step-by-step solution
1. Step 1
  B1 (a) Try two odd numbers. $3 + 5 = 8$ , which is even, but neither $3$ nor $5$ is even.
2. Step 2
  M1 (b) Write two general odd numbers with different letters, because they need not be equal or consecutive, and add their squares.
  $(2m + 1)^{2} + (2n + 1)^{2}$
3. Step 3
  M1 Expand both squares.
  $= 4m^{2} + 4m + 1 + 4n^{2} + 4n + 1$
4. Step 4
  A1 Collect terms and take out the factor $4$ from every term except the constant.
  $= 4m^{2} + 4m + 4n^{2} + 4n + 2 = 4(m^{2} + m + n^{2} + n) + 2$
5. Step 5
  A1 Conclude both parts. $4(\ldots) + 2 = 2[2(m^{2} + m + n^{2} + n) + 1]$ is a multiple of $2$ , so it is even. It is also $2$ more than a multiple of $4$ , so it leaves remainder $2$ when divided by $4$ and is never a multiple of $4$ .
Answer
(a) e.g. $3 + 5 = 8$ is even, but $3$ and $5$ are odd. (b) $(2m + 1)^{2} + (2n + 1)^{2} = 4(m^{2} + m + n^{2} + n) + 2$ , which is even, and is always $2$ more than a multiple of $4$ , so it is never a multiple of $4$ .
Examiner tip
Examiner tip: "Never a multiple of $4$ " is proved by showing the remainder: write the result as $4 \times$ an integer $+ 2$ . Check with $3$ and $5$ : $9 + 25 = 34 = 4 \times 8 + 2$ .

Common mistake: In (a), giving two even numbers, such as $2 + 4 = 6$ . That agrees with Zara, so it does not show she is wrong. In (b), using $2n + 1$ for both numbers proves the result only when the two odd numbers are equal.

Question 1
Paper 1H/2H style — 1 mark1 mark
Mo says,

"If $n$ is an odd number, then $n^{2} + 2$ is always a prime number."

Give an example to show that Mo is wrong. [1]
Model answer
When $n = 5$ : $5^{2} + 2 = 27 = 3 \times 9$

$27$ is not prime, so Mo is wrong.
Why this scores
Mark scheme:
- B1 for an odd value of $n$ with $n^{2} + 2$ shown not to be prime, e.g. $n = 5$ giving $27$ , or $n = 7$ giving $51 = 3 \times 17$
Why this scores full marks: The values $n = 1$ and $n = 3$ give $3$ and $11$ , which are prime, so the script keeps going until the statement fails, then shows the factors of $27$ to prove it is not prime.
Question 2
Paper 1H/2H style — 2 marks2 marks
Prove that the product of any two even numbers is a multiple of $4$ . [2]
Model answer
Let the even numbers be $2m$ and $2n$ , where $m$ and $n$ are integers.

$2m \times 2n = 4mn$

$mn$ is an integer, so $4mn$ is a multiple of $4$ .
Why this scores
Mark scheme:
- M1 for $2m \times 2n$ with different letters
- A1 for $4mn$ with a conclusion
Why this scores full marks: Two different letters make the numbers general. Writing $2n \times 2n = 4n^{2}$ proves the result only for a number multiplied by itself.
Question 3
Paper 1H/2H style — 2 marks2 marks
Show that $(x - 4)^{2} - (x^{2} + 16) \equiv -8x$ [2]
Model answer
$(x - 4)^{2} - (x^{2} + 16) = x^{2} - 8x + 16 - x^{2} - 16$

$= -8x$
Why this scores
Mark scheme:
- M1 for $x^{2} - 8x + 16$
- A1 for a complete, correct argument leading to $-8x$
Why this scores full marks: The square is expanded in full, with the middle term $-8x$ , and the second bracket is subtracted as a whole so both $x^{2}$ and $16$ cancel. Writing $(x - 4)^{2} = x^{2} + 16$ or $x^{2} - 16$ scores 0.
Question 4
Paper 1H/2H style — 3 marks3 marks
Prove that the sum of any two consecutive odd numbers is a multiple of $4$ . [3]
Model answer
Let the numbers be $2n + 1$ and $2n + 3$ .

$(2n + 1) + (2n + 3) = 4n + 4$

$= 4(n + 1)$ , which is a multiple of $4$ .
Why this scores
Mark scheme:
- M1 for two consecutive odd numbers in terms of one letter, e.g. $2n + 1$ and $2n + 3$ , or $2n - 1$ and $2n + 1$
- A1 for $4n + 4$ (or $4n$ )
- A1 for $4(n + 1)$ oe with a conclusion
Why this scores full marks: Consecutive odd numbers differ by $2$ , so one letter is enough and the second number is $2n + 3$ . The factorised form $4(n + 1)$ is what shows the multiple of $4$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
$n$ is an integer.

Prove that $(n + 3)^{2} - n^{2}$ is always an odd multiple of $3$ . [3]
Model answer
$(n + 3)^{2} - n^{2} = n^{2} + 6n + 9 - n^{2}$

$= 6n + 9$

$= 3(2n + 3)$

$2n + 3 = 2(n + 1) + 1$ is odd, so $3(2n + 3)$ is $3 \times$ an odd number: an odd multiple of $3$ .
Why this scores
Mark scheme:
- M1 for $n^{2} + 6n + 9$
- A1 for $6n + 9$ or $3(2n + 3)$
- A1 for $3(2n + 3)$ with a reason that $2n + 3$ is odd
Why this scores full marks: The question asks for two things, a multiple of $3$ and odd. The factor $3$ answers the first, and explaining that $2n + 3$ is odd answers the second.
Question 6
Paper 1H/2H style — 4 marks4 marks
The $n$ th term of a sequence is $n^{2} + n$ .

Prove that the sum of any two consecutive terms of the sequence is always twice a square number. [4]
Model answer
The $n$ th term is $n^{2} + n$ , and the $(n + 1)$ th term is $(n + 1)^{2} + (n + 1)$ .

$(n + 1)^{2} + (n + 1) = n^{2} + 2n + 1 + n + 1 = n^{2} + 3n + 2$

Sum $= n^{2} + n + n^{2} + 3n + 2 = 2n^{2} + 4n + 2$

$= 2(n^{2} + 2n + 1) = 2(n + 1)^{2}$

$(n + 1)^{2}$ is a square number, so the sum is twice a square number.
Why this scores
Mark scheme:
- M1 for the next term $(n + 1)^{2} + (n + 1)$ oe
- M1 for expanding and adding the two terms
- A1 for $2n^{2} + 4n + 2$
- A1 for $2(n + 1)^{2}$ with a conclusion
Why this scores full marks: The next term is found by replacing $n$ with $n + 1$ everywhere, in brackets. After taking out $2$ , the script recognises $n^{2} + 2n + 1$ as $(n + 1)^{2}$ . Check with the terms $6$ and $12$ : $6 + 12 = 18 = 2 \times 3^{2}$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
Prove that the square of any odd number is $1$ more than a multiple of $8$ . [4]
Model answer
Let the odd number be $2n + 1$ .

$(2n + 1)^{2} = 4n^{2} + 4n + 1$

$= 4n(n + 1) + 1$

$n(n + 1)$ is the product of two consecutive integers, so it is even: $n(n + 1) = 2k$ .

$4 \times 2k + 1 = 8k + 1$ , which is $1$ more than a multiple of $8$ .
Why this scores
Mark scheme:
- M1 for $(2n + 1)^{2} = 4n^{2} + 4n + 1$
- M1 for $4n(n + 1) + 1$ oe
- A1 for stating $n(n + 1)$ is even (product of consecutive integers)
- A1 for a complete argument concluding $8k + 1$ oe
Why this scores full marks: The algebra alone only gives $1$ more than a multiple of $4$ . The extra factor of $2$ comes from the fact that $n(n + 1)$ is even, and the script states that reason explicitly.
Question 8
Paper 1H/2H style — 5 marks5 marks
(a) Prove that $\dfrac{n}{n + 1} - \dfrac{n - 1}{n} \equiv \dfrac{1}{n(n + 1)}$ [4]

(b) Hence show that $\dfrac{99}{100} - \dfrac{98}{99} = \dfrac{1}{9900}$ [1]
Model answer
(a) $\dfrac{n}{n + 1} - \dfrac{n - 1}{n} = \dfrac{n \times n - (n - 1)(n + 1)}{n(n + 1)}$

$= \dfrac{n^{2} - (n^{2} - 1)}{n(n + 1)}$

$= \dfrac{n^{2} - n^{2} + 1}{n(n + 1)} = \dfrac{1}{n(n + 1)}$

(b) Put $n = 99$ : $\dfrac{99}{100} - \dfrac{98}{99} = \dfrac{1}{99 \times 100} = \dfrac{1}{9900}$
Why this scores
Mark scheme:
- M1 for a common denominator of $n(n + 1)$ with both numerators adjusted
- M1 for $(n - 1)(n + 1) = n^{2} - 1$
- A1 for a numerator of $1$ from $n^{2} - (n^{2} - 1)$
- A1 for a complete, correct argument
- B1 for substituting $n = 99$ into the identity, leading to $\dfrac{1}{9900}$
Why this scores full marks: The second numerator is subtracted as a bracket, so $-(n^{2} - 1)$ becomes $+1$ . Losing that bracket gives $-1$ , a sign error the examiners see often. In (b), "Hence" means using the identity with $n = 99$ , not working out the fractions on a calculator.
Question 9
Paper 1H/2H style — 5 marks5 marks
Ali says,

"The product of two consecutive even numbers is always a multiple of $16$ ."

(a) Give an example to show that Ali is wrong. [1]

(b) Prove that the product of two consecutive even numbers is always a multiple of $8$ . [4]
Model answer
(a) $2 \times 4 = 8$ , which is not a multiple of $16$ .

(b) Let the numbers be $2n$ and $2n + 2$ .

$2n(2n + 2) = 4n^{2} + 4n = 4n(n + 1)$

$n(n + 1)$ is the product of two consecutive integers, so it is even: $n(n + 1) = 2k$ .

$4 \times 2k = 8k$ , which is a multiple of $8$ .
Why this scores
Mark scheme:
- B1 for a correct counter-example, e.g. $2 \times 4 = 8$ or $4 \times 6 = 24$ (not $6 \times 8 = 48$ , which is $16 \times 3$ )
- M1 for $2n(2n + 2)$ oe
- M1 for $4n(n + 1)$ or $4n^{2} + 4n$
- A1 for stating that $n(n + 1)$ is even, with a reason
- A1 for a complete argument concluding $8k$ oe
Why this scores full marks: Part (a) needs only one pair that fails, with the check shown. In (b), the factor of $4$ comes from the two factors of $2$ , and the script gives the reason for the extra $2$ : one of $n$ and $n + 1$ is even.
Question 10
Paper 1H/2H style — 6 marks6 marks
Prove algebraically that the product of any four consecutive integers, plus $1$ , is always a square number. [6]
Model answer
Let the integers be $n$ , $n + 1$ , $n + 2$ and $n + 3$ .

Pair the outer and inner integers:

$n(n + 3) = n^{2} + 3n$

$(n + 1)(n + 2) = n^{2} + 3n + 2$

Let $u = n^{2} + 3n$ . Then

$n(n + 1)(n + 2)(n + 3) + 1 = u(u + 2) + 1$

$= u^{2} + 2u + 1 = (u + 1)^{2}$

$= (n^{2} + 3n + 1)^{2}$

$n^{2} + 3n + 1$ is an integer, so the product plus $1$ is always a square number.
Why this scores
Mark scheme:
- M1 for four consecutive integers in terms of one letter and a product of two of the brackets expanded
- M1 for $n^{2} + 3n$ and $n^{2} + 3n + 2$ (or for a full expansion attempt)
- M1 for $(n^{2} + 3n)(n^{2} + 3n + 2) + 1$ expanded or simplified, e.g. $n^{4} + 6n^{3} + 11n^{2} + 6n + 1$
- A1 for $u^{2} + 2u + 1$ oe
- A1 for $(n^{2} + 3n + 1)^{2}$
- A1 for a complete argument with a conclusion that it is the square of an integer
Why this scores full marks: Pairing $n$ with $n + 3$ and $n + 1$ with $n + 2$ gives two brackets that differ by $2$ , which turns a quartic into a perfect square. Expanding all four brackets also works: $n^{4} + 6n^{3} + 11n^{2} + 6n + 1 = (n^{2} + 3n + 1)^{2}$ , confirmed by expanding the square. Check: $1 \times 2 \times 3 \times 4 + 1 = 25 = 5^{2}$ .

4MA1 Algebraic Proof

At a glance

What you’ll learn

Proving identities

Properties of integers

'Show that' problems

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Prove an identity

2Prove a divisibility property

3Prove an expression is a multiple of $n$

4Show that — algebraic manipulation

5Explaining why an expression is always odd [1 mark]

6Proving a multiple of 3 with a negative bracket [2 marks]

7The middle of three consecutive integers [3 marks]

8Proving an identity with algebraic fractions [4 marks]

9Squares of numbers that are not multiples of 3 [5 marks]

10Disproving one claim and proving another [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Algebraic Proof

At a glance

What you’ll learn

Proving identities

Properties of integers

'Show that' problems

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Prove an identity

2Prove a divisibility property

3Prove an expression is a multiple of nnn

4Show that — algebraic manipulation

5Explaining why an expression is always odd [1 mark]

6Proving a multiple of 3 with a negative bracket [2 marks]

7The middle of three consecutive integers [3 marks]

8Proving an identity with algebraic fractions [4 marks]

9Squares of numbers that are not multiples of 3 [5 marks]

10Disproving one claim and proving another [5 marks]

Model AnswersFull mark-scheme-style answers

3Prove an expression is a multiple of $n$