Algebra

Identities and algebraic proof

Q: How do you write an odd number in algebra?

Use 2n + 1, where n is any integer. An even number is 2n.

Q: What is the difference between an equation and an identity?

An equation is only true for particular values, like 2x = 10. An identity is true for every value, like 2(x + 5) \equiv 2x + 10.

8 past-paper questions, easiest first. Try each one, then reveal the answer.

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Identities and algebraic proof: what you need to know

An algebraic proof shows something is true for every number, not just the ones you try. Foundation questions ask you to show two expressions are equal; Higher questions ask you to prove facts about odd, even and consecutive numbers.

Foundation students need identities and showing expressions are equivalent; proofs about odd, even and consecutive numbers are mainly Higher.

Key facts

An equation is true for some values; an identity ( $\equiv$ ) is true for all values.
Any integer: $n$ . Even: $2n$ . Odd: $2n + 1$ . Consecutive: $n$ , $n + 1$ , $n + 2$ .
Consecutive even numbers: $2n$ and $2n + 2$ . Consecutive odd numbers: $2n + 1$ and $2n + 3$ . Any two odd numbers: $2m + 1$ and $2n + 1$ (different letters).
To show something is even, write it as $2 \times$ (an integer); a multiple of 8 as $8 \times$ (an integer).

Showing an identity

Start with the more complicated side.
Expand and simplify until it looks exactly like the other side.

Writing a proof (Higher)

Write the numbers algebraically, e.g. $2n + 1$ for an odd number.
Form the expression the question describes, then expand and simplify.
Factorise to show the required form, e.g. $8(\ldots) + 2$ .
Finish with a sentence explaining what this proves.

Worked examples

Example 1 Foundation · grade 4

Show that $3(2x + 4) - 2(x - 1) \equiv 4x + 14$

Show the solution

Start with the left-hand side.
$3(2x + 4) = 6x + 12$
$-2(x - 1) = -2x + 2$
$6x + 12 - 2x + 2 = 4x + 14$ , which is the right-hand side, so the identity is true.

Example 2 Higher · grade 7

Prove that the sum of the squares of any two consecutive odd numbers is always 2 more than a multiple of 8.

Show the solution

Let the numbers be $2n + 1$ and $2n + 3$ .
$(2n + 1)^2 = 4n^2 + 4n + 1$
$(2n + 3)^2 = 4n^2 + 12n + 9$
Sum: $8n^2 + 16n + 10$
$= 8(n^2 + 2n + 1) + 2$
$8(n^2 + 2n + 1)$ is a multiple of 8, so the sum is always 2 more than a multiple of 8.

Common mistakes

Testing a few numbers. Examples never prove a statement is always true.
Using $n$ and $n + 2$ for two odd numbers; they might both be even. Use $2n + 1$ and $2n + 3$ .
Sign errors when subtracting a squared bracket: put it in brackets before expanding.
Stopping too early, e.g. at $4(2n^2 + \ldots)$ when a multiple of 8 is needed. Keep factorising, or explain why the bracket is even.

Exam tips

The final mark is for the conclusion, so write a sentence linking your algebra to the statement.
Show the factorised form, such as $8(\ldots)$ , rather than just saying the terms divide by 8.

Questions people ask

How do you write an odd number in algebra?

Use $2n + 1$ , where $n$ is any integer. An even number is $2n$ .

What is the difference between an equation and an identity?

An equation is only true for particular values, like $2x = 10$ . An identity is true for every value, like $2(x + 5) \equiv 2x + 10$ .

Can I prove something by trying examples?

No. Examples can show a statement is false, but a proof needs algebra that works for every number.

Exam questions: Identities and algebraic proof (8, easiest first)

Identities and algebraic proof2 marksGrade 6No calculatorHigher

June 2019 · Paper 1 Higher · Q13

Prove that subtracting an integer from its square always gives an even number.

Show answer

Answer	$n^2 - n = n(n-1)$ , a product of two consecutive integers, one of which is even, so it is always even	2 marks

Worked solution Checked against the mark scheme

$n^2 - n = n(n - 1)$

$n$ and $n - 1$ are consecutive integers, so one of them is even.

An even number times any integer is even, so $n^2 - n$ is never odd.

Watch out: Testing a few values of n instead of giving a general argument.

Identities and algebraic proof4 marksGrade 7No calculatorHigher

November 2022 · Paper 1 Higher · Q16

Prove an algebraic identity for the difference of two squared expressions, then judge a claim about odd numbers that follows from it.

Show answer

(a)	Proof that both sides are equal	3 marks
(b)	Yes: $2m + 1$ and $2n - 1$ are any odd numbers, and the right-hand side has a factor of 4	1 mark

Worked solution Checked against the mark scheme

(a) $(2m + 1)^2 - (2n - 1)^2 = 4m^2 + 4m + 1 - (4n^2 - 4n + 1)$

$= 4m^2 - 4n^2 + 4m + 4n = 4[(m + n)(m - n) + (m + n)]$

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

(b) Yes. $2m + 1$ and $2n - 1$ stand for any odd numbers, and the result has a factor of 4, so it is always a multiple of 4.

Watch out: Getting the sign of the constant wrong when subtracting $(2n - 1)^2$ .

Identities and algebraic proof3 marksGrade 7CalculatorHigher

November 2019 · Paper 3 Higher · Q15

Prove with algebra that two consecutive even numbers squared and added always give a multiple of 4.

Show answer

Answer	$(2n)^2 + (2n+2)^2 = 4(2n^2 + 2n + 1)$ , a multiple of 4	3 marks

Worked solution Checked against the mark scheme

Let the numbers be $2n$ and $2n + 2$ .

$(2n)^2 + (2n + 2)^2 = 4n^2 + 4n^2 + 8n + 4 = 8n^2 + 8n + 4$

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

$2n^2 + 2n + 1$ is an integer, so the sum is a multiple of 4.

Watch out: Using $n$ and $n + 2$ , which need not be even.

Identities and algebraic proof4 marksGrade 7No calculatorHigher

June 2018 · Paper 1 Higher · Q12

Prove algebraically that squaring any odd number gives one more than a multiple of 4.

Show answer

Answer	$(2n+1)^2 = 4n^2 + 4n + 1 = 4(n^2 + n) + 1$	4 marks

Worked solution Checked against the mark scheme

Let an odd number be $2n + 1$ , where $n$ is an integer.

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

$4(n^2 + n)$ is a multiple of 4, so the square is 1 more than a multiple of 4.

Watch out: Checking a few numbers instead of using algebra, or writing the odd number as $n + 1$ .

Identities and algebraic proof3 marksGrade 7CalculatorHigher

November 2018 · Paper 3 Higher · Q15

Prove that the difference of the squares of two consecutive odd numbers is always a multiple of 8.

Show answer

Answer	$(2n+1)^2 - (2n-1)^2 = 8n$ , a multiple of 8	3 marks

Worked solution Checked against the mark scheme

Let the odd numbers be $2n - 1$ and $2n + 1$ , where $n$ is an integer.

$(2n+1)^2 - (2n-1)^2 = (4n^2 + 4n + 1) - (4n^2 - 4n + 1)$

$= 8n$

$8n$ is 8 times an integer, so it is always a multiple of 8.

Watch out: Using $n$ and $n + 2$ , which are not always odd.

Identities and algebraic proof4 marksGrade 7No calculatorHigher

Specimen · Paper 1 Higher · Q20

Prove algebraically that the difference of the squares of two consecutive integers equals their sum.

Show answer

Answer	$(n + 1)^2 - n^2 = 2n + 1 = n + (n + 1)$	4 marks

Worked solution Checked against the mark scheme

Let the integers be $n$ and $n + 1$ .

Difference of squares: $(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$

Sum: $n + (n + 1) = 2n + 1$

Both equal $2n + 1$ , so the statement is true for any two consecutive integers.

Watch out: Checking a few numerical examples instead of giving an algebraic proof.

Identities and algebraic proof3 marksGrade 8CalculatorHigher

June 2023 · Paper 3 Higher · Q13

Prove algebraically that adding two consecutive triangular numbers always gives a square.

Show answer

Answer	$\frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2} = (n+1)^2$	3 marks

Worked solution Checked against the mark scheme

The next triangular number is $\dfrac{(n+1)(n+2)}{2}$ .

Sum: $\dfrac{n(n+1) + (n+1)(n+2)}{2} = \dfrac{(n+1)(n + n + 2)}{2}$

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

This is a square number for every $n$ .

Watch out: Checking a few numerical examples instead of giving a general algebraic proof.

Identities and algebraic proof6 marksGrade 9CalculatorHigher

Specimen · Paper 3 Higher · Q22

Prove that the difference between the squares of any two terms of a given arithmetic sequence is always a multiple of 24.

Show answer

Answer	Terms are $6n + 1$ ; $(6a+1)^2 - (6b+1)^2 = 12(a - b)(3(a + b) + 1)$ , and $(a - b)(3(a + b) + 1)$ is always even, so the difference is a multiple of 24	6 marks

Worked solution Checked against the mark scheme

The $n$ th term is $6n + 1$ . Take two terms $6a + 1$ and $6b + 1$ .

$(6a + 1)^2 - (6b + 1)^2 = 36a^2 + 12a - 36b^2 - 12b$

$= 12(3a^2 - 3b^2 + a - b) = 12(a - b)(3a + 3b + 1)$

If $a$ and $b$ are both odd or both even, $a - b$ is even.

If one is odd and one even, $a + b$ is odd, so $3(a + b)$ is odd and $3(a + b) + 1$ is even.

Either way $(a - b)(3a + 3b + 1)$ is even, so the difference is $12 \times$ even $= 24k$ : a multiple of 24.

Watch out: Stopping at a multiple of 12 without showing the other factor is even.

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

June 2024 · Paper 3 Higher · Q10: Use a ratio and algebra to show the total number of counters in two bags is 4x.
June 2023 · Paper 3 Higher · Q15: Factorise a difference of two squares, then use it to show that a power of 2 minus 1 is the product of two consecutive odd numbers.
June 2022 · Paper 1 Higher · Q13: Prove that the semicircle on the hypotenuse of a right-angled triangle has the same area as the other two semicircles combined.
Specimen · Paper 2 Higher · Q13: Expand three linear brackets to prove a cubic identity.