A6AlgebraHigherGrades 6–9

Algebraic Proof

"Show that the sum of any two consecutive odd numbers is a multiple of 4." Testing $3 + 5 = 8$ and $11 + 13 = 24$ suggests it is true, but examples can never prove a statement about every number. Algebra can: write the numbers with a letter, simplify, and the result holds for all of them at once.

40 min AQA GCSE Maths

The big picture

Proof questions reward a clear structure: define the numbers algebraically, carry out the operation, factorise to reveal the property, and finish with a sentence. AQA mark schemes usually give the last mark only for that concluding statement.

Two other ideas complete the topic. An identity, written with $\equiv$ , is true for all values and is proved by showing both sides are the same. A statement is disproved by a single counterexample.

What you'll learn

Write even, odd and consecutive numbers algebraically
Prove statements about integers
Show that an identity is true
Use counterexamples to disprove statements

Writing numbers with letters

For any integer $n$ : $2n$ is even, $2n + 1$ is odd, and $n$ , $n + 1$ , $n + 2$ are consecutive integers.

Consecutive odd numbers: $2n + 1$ and $2n + 3$ . Consecutive even numbers: $2n$ and $2n + 2$ .

A multiple of $k$ is $k \times$ (an integer).

Your turnSame method, new numbers — try it before moving on.

If $2n$ is an even number, write an expression for the next even number after it.

Your answer

Show worked solution

Solution

1Even numbers go up in 2s.
2The next even number is $2n + 2$ (or $2(n + 1)$ ).

Answer: $2n + 2$

Tip — Use a different letter for numbers that are not connected: two unrelated even numbers are $2m$ and $2n$ .

Proofs about integers

Form the expression, simplify it, and factorise to show the required property.

End with a sentence linking the algebra back to the claim.

Example 1Prove that the sum of two consecutive odd numbers is a multiple of 4.

1Let the numbers be

2n + 1

and

2n + 3

2Sum:

4n + 4

3Factorise:

4(n + 1)

AnswerSince

4(n + 1)

is 4 times an integer, the sum is always a multiple of 4.

Example 2Prove that

(2n + 1)^{2} - (2n - 1)^{2}

is always a multiple of 8.

1Expand:

(4n^{2} + 4n + 1) - (4n^{2} - 4n + 1)

2Simplify:

8n

Answer

8n

is 8 times an integer, so the expression is a multiple of 8.

Why this works

The second result says the difference between the squares of two consecutive odd numbers is always a multiple of 8. Checking: $7^{2} - 5^{2} = 24$ and $9^{2} - 7^{2} = 32$ . Algebra shows it will never fail.

Identities

To prove $A \equiv B$ , simplify one side until it is identical to the other — do not rearrange across the $\equiv$ sign as if solving.

Example 1Show that

(x + 5)^{2} - 10x \equiv x^{2} + 25

1Left side:

x^{2} + 10x + 25 - 10x

2Simplify:

x^{2} + 25

AnswerThe left side equals the right side, so the identity holds.

Your turnSame method, new numbers — try it before moving on.

Find the value of $a$ if $(x + 3)^2 - a \equiv x^2 + 6x + 4$ .

Your answer

Show worked solution

Solution

1 $(x + 3)^2 = x^2 + 6x + 9$ .
2 $9 - a = 4$ , so $a = 5$ .

Answer: $a = 5$

Counterexamples

To show a statement is false, find one case where it does not hold.

"The sum of two prime numbers is always even" is false: $2 + 3 = 5$ .

Your turnSame method, new numbers — try it before moving on.

Someone claims $n^2 + n + 41$ is prime for every positive integer $n$ . What is the smallest positive integer $n$ that disproves this?

Your answer

Show hint

Try a value of $n$ that makes every term share a factor of 41.

Show worked solution

Solution

1 $40^2 + 40 + 41 = 1600 + 81 = 1681 = 41^2$ .
2That is not prime, so $n = 40$ is a counterexample (every smaller $n$ does give a prime).

Answer: $n = 40$

Tip — Show the working for a counterexample and say why it breaks the claim. A bare number is not enough.

Think like an examiner

State what $n$ represents.
Show the factorised form that proves the property.
Finish with a concluding sentence.

Remember these

2n,\ 2n + 1

Even and odd numbers.

n,\ n + 1

Consecutive integers.

k(\dots)

A multiple of k.

Watch out for these

Using numerical examples as a proof.

Examples cannot cover every case; use algebra.

Writing two consecutive odd numbers as

2n + 1

and

2n + 2

2n + 2

is even; use

2n + 1

and

2n + 3

Stopping at

4n + 4

Factorise to

4(n + 1)

and explain.

Expanding

-(2n - 1)^{2}

without changing every sign.

Bracket the expansion and subtract all of it.

Stretch yourself

Prove that the product of two consecutive integers, added to the larger integer, is always a square number.

Hint — Let the integers be $n$ and $n + 1$ , then factorise.

Reveal solution

SolutionLet the integers be

n

and

n + 1

. Then

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

, taking out the common factor

(n + 1)

. Since

n + 1

is an integer, the result is always a square number. Example:

4 \times 5 + 5 = 25 = 5^{2}

✓.

\blacksquare

Questions students ask

Why can’t examples prove a statement?

There are infinitely many cases; examples only check the ones you tried.

What is the difference between an identity and an equation?

An identity is true for all values; an equation is true only for particular ones.

How many counterexamples do I need?

One is enough to disprove a statement.

Key takeaways

Represent integers with letters: $2n$ , $2n + 1$ , $n + 1$ .
Simplify and factorise to reveal the property.
Conclude with a clear statement.
Prove identities by transforming one side into the other.
A single counterexample disproves a claim.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 63 marks

Prove that the difference between the squares of two consecutive integers is always odd.[3]
Show mark scheme
- M1Uses $n$ and $n + 1$ : $(n + 1)^2 - n^2$
- A1Simplifies to $2n + 1$
- C1Concludes: $2n$ is even, so $2n + 1$ is odd

Question 2Grade 83 marks

Prove that $(3n + 1)^2 - (n - 1)^2$ is a multiple of 8 for all positive integers $n$ .[3]
Show mark scheme
- M1Expands both: $9n^2 + 6n + 1$ and $n^2 - 2n + 1$
- M1Subtracts correctly: $8n^2 + 8n$
- A1Writes $8(n^2 + n)$ and states that this is a multiple of 8

Question 3Grade 93 marks

Prove that the sum of the squares of any three consecutive integers is always 2 more than a multiple of 3.[3]
Show mark scheme
- M1Uses $(n - 1)^2 + n^2 + (n + 1)^2$ or equivalent
- M1Expands and simplifies to $3n^2 + 2$
- A1Concludes: $3n^2$ is a multiple of 3, so the sum is 2 more than a multiple of 3

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

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