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Algebraic Proof (Edexcel IGCSE Maths B): Revision Note

Exam code: 4MB1

Written by: Roger B

Reviewed by: Jamie Wood

Updated on 20 January 2026

Algebraic proof

What is algebraic proof?

Algebraic proof means proving a result using algebra
- This is different to proving a result by individually testing all possible values
The proofs may require algebraic skills such as
- expanding brackets
- factorising
- collecting like terms
  - The difference of two squares factorisation can also be helpful

Examiner Tips and Tricks

Algebraic proof questions will usually involve proving things involving integers.

In these questions letters like k, m, and n are used to stand for 'any (positive) integer' rather than for 'any number'.

How do I show that a result is a multiple of a number?

To prove an expression is a multiple of k, show that it can be written as $k \times (integer)$
- This may require rewriting the expression
  - and then factorising out a k
- For example, $7 n^{2} + 14 n = 7 (n^{2} + 2 n)$ is a multiple of 7

How do I show that a result is a square number?

To prove an expression is a square number, show that it can be written as ${(integer)}^{2}$
- This may require rewriting the expression
  - and then factorising
- For example, $(2 n^{2} + 7) + (6 n + 2 - n^{2}) = n^{2} + 6 n + 9 = {(n + 3)}^{2}$ is a square number

How do I show that a result is odd or even?

To prove an expression is even, show that it can be written as $2 \times (integer)$
- This may require rewriting the expression
  - and then factorising out a 2
- For example, $2 n$ is even
  - and so is $2 (n^{2} - 3 n)$
To prove something is odd, show that it can be written as $2 \times (integer) + 1$ or $2 \times (integer) - 1$
- For example, $2 n + 1$ is odd
  - and so is $2 n + 2 (n - \frac{1}{2}) = 2 n + 2 n - 1 = 2 (2 n) - 1$
- This is because one more or one less than an even number is an odd number

Examiner Tips and Tricks

For the examples above, make sure the part inside the brackets marked as $(integer)$ really is an integer.

For example, $2 (n + \frac{1}{2})$ is not even as $\frac{1}{2}$ is not an integer

How do I show that a result is (or isn't) an integer?

If n is an integer, then
- Any positive power of n is also an integer
  - I.e., n², n³, n⁴, etc. are all integers
- Any linear sum or difference of regular integers with multiples or positive powers of n is also an integer
  - E.g. n+3, n²-2n, and 5n⁴-3n²-7 are all integers
To show that an expression is an integer, show that it can be written in one of those forms
To show that an expression is not an integer,
- show that it can be written as an integer plus or minus something that isn't an integer
  - For example $\frac{6 n + 1}{2} = 3 n + \frac{1}{2}$ is not an integer
  - Neither is $7 (n - 2) + π$
- or show that it can be written as an odd number divided by (a multiple of) 2
  - For example $\frac{2 n + 1}{2}$ is an odd number
  - This is because half of an odd number is never an integer

Examiner Tips and Tricks

Be sure to explain your reasoning in an algebraic proof question.

For example " $n^{2}$ is an integer, so $n^{2} - \frac{1}{3}$ is not"

It is also important to state your conclusion.

A good trick is to copy word for word the phrases used in the question
For example, "this proves that $7 n^{2} + 14 n$ is a multiple of 7"

Worked Example

Prove that ${(6 n + 5)}^{2} - {(6 n - 5)}^{2}$ is a multiple of 20 for all positive integer values of $n$ .

Answer:

Method 1 (Expanding brackets)

Start by expanding the brackets

$\begin{array}{rcl} {(6 n + 5)}^{2} & = & (6 n + 5) (6 n + 5) \\ = & {(6 n)}^{2} + 6 n \times 5 + 5 \times 6 n + 5^{2} \\ = & 36 n^{2} + 60 n + 25 \end{array}$

$\begin{array}{rcl} {(6 n - 5)}^{2} & = & (6 n - 5) (6 n - 5) \\ = & {(6 n)}^{2} - 6 n \times 5 - 5 \times 6 n + {(- 5)}^{2} \\ = & 36 n^{2} - 60 n + 25 \end{array}$

Substitute that back into the expression

$\begin{array}{rcl} {(6 n + 5)}^{2} - {(6 n - 5)}^{2} & = & (36 n^{2} + 60 n + 25) - (36 n^{2} - 60 n + 25) \\ = & 36 n^{2} + 60 n + 25 - 36 n^{2} + 60 n - 25 \\ = & 120 n \end{array}$

Factorise out a 20

$= 20 (6 n)$

That algebraically proves your result

But to get full marks explain your reasoning and state your conclusion

${(6 n + 5)}^{2} - {(6 n - 5)}^{2} = 20 (6 n)$
This proves that ${(6 n + 5)}^{2} - {(6 n - 5)}^{2}$ is a multiple of 20.

Method 2 (Difference of two squares)

Use the difference of two squares, $a^{2} - b^{2} = (a + b) (a - b)$

Here $a = 6 n + 5$ and $b = 6 n - 5$

$\begin{array}{rcl} {(6 n + 5)}^{2} - {(6 n - 5)}^{2} & = & ((6 n + 5) + (6 n - 5)) ((6 n + 5) - (6 n - 5)) \\ = & (12 n) (10) \\ = & 120 n \end{array}$

Factorise out a 20

$= 20 (6 n)$

That algebraically proves your result

But to get full marks explain your reasoning and state your conclusion

${(6 n + 5)}^{2} - {(6 n - 5)}^{2} = 20 (6 n)$
This proves that ${(6 n + 5)}^{2} - {(6 n - 5)}^{2}$ is a multiple of 20.

Worked Example

The $n$ th term of a sequence is $a_{n}$ , where $a_{n} = \frac{2 n^{2} + 5 n + 2}{2 n + 4}$

By simplifying the expression for $a_{n}$ , or otherwise, explain why no term in the sequence is an integer.

Show algebraic working and clearly explain your answer.

Answer:

At first glance this may look like a question about sequences

But really it is an algebraic proof question
You need to prove that $\frac{2 n^{2} + 5 n + 2}{2 n + 4}$ is not an integer for any positive integer $n$

Start by factorising the top and bottom of the expression

$\begin{array}{rcl} a_{n} & = & \frac{2 n^{2} + 5 n + 2}{2 n + 4} \\ = & \frac{(2 n + 1) (n + 2)}{2 (n + 2)} \end{array}$

Cancel common factors

$\begin{array}{rcl} = & \frac{(2 n + 1) (n + 2)}{2 (n + 2)} \\ = & \frac{2 n + 1}{2} \end{array}$

From here there are two ways to write your conclusion

Method 1

$2 n + 1$ is odd because it is one more than an even number

and an odd number divided by 2 is never an integer

$a_{n} = \frac{2 n + 1}{2}$ . $2 n + 1$ is odd, and an odd number divided by 2 is not an integer. Therefore $a_{n}$ is not an integer for any value of $n$ .

Method 2

Split the expression into two fractions and simplify

$\begin{array}{rcl} a_{n} & = & \frac{2 n + 1}{2} \\ = & \frac{2 n}{2} + \frac{1}{2} \\ = & n + \frac{1}{2} \end{array}$

That is equal to an integer ( $n$ ) plus something that is not an integer $(\frac{1}{2})$

$a_{n} = n + \frac{1}{2}$ . $n$ is an integer, so $n + \frac{1}{2}$ is not an integer. Therefore $a_{n}$ is not an integer for any value of $n$ .

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Algebraic Proof (Edexcel IGCSE Maths B): Revision Note

Algebraic proof

What is algebraic proof?

Examiner Tips and Tricks

How do I show that a result is a multiple of a number?

How do I show that a result is a square number?

How do I show that a result is odd or even?

Examiner Tips and Tricks

How do I show that a result is (or isn't) an integer?

Examiner Tips and Tricks

Worked Example

Worked Example

Unlock more, it's free!

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Author: Roger B

Reviewer: Jamie Wood