Algebraic proof – GCSE Maths exam questions with mark scheme

1Grade 4–5Non-calculator2 marks

Prove that the sum of any three consecutive integers is a multiple of 3. [2]

for writing three consecutive integers algebraically and adding, e.g. $n + (n + 1) + (n + 2)$ (or $(n - 1) + n + (n + 1)$)M1
for $3n + 3 = 3(n + 1)$ (or $3n$) with a conclusion, e.g. 'which is a multiple of 3'C1
Answer: $n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)$, which is a multiple of 3
Examiner tip: Numerical examples (e.g. $4 + 5 + 6 = 15$) are not a proof – you must use algebra and write a concluding statement.

2Grade 4–5Non-calculator4 marks

(a) Show that $3(2x + 5) - 2(x - 1) \equiv 4(x + 4) + 1$ [2]

(b) Ravi says, '$n^2 + 3n + 1$ is a prime number for every positive integer $n$.'

Show that Ravi is wrong. [2]

(a) for expanding the left-hand side correctly: $6x + 15 - 2x + 2$ (or $4x + 17$)M1
(a) for showing both sides equal $4x + 17$, e.g. RHS $= 4x + 16 + 1 = 4x + 17$A1
(b) for substituting a value of $n$ that gives a non-prime, e.g. $n = 6$: $36 + 18 + 1 = 55$M1
(b) for showing the value is not prime, e.g. $55 = 5 \times 11$, so Ravi is wrongC1
Answer: (a) both sides $= 4x + 17$ (b) e.g. $n = 6$ gives $55 = 5 \times 11$, which is not prime
Examiner tip: In (b) one counterexample is enough, but you must SHOW it is not prime by giving a factor pair.

3Grade 6–7HigherNon-calculator3 marks

Prove that $(n + 5)^2 - (n + 1)^2$ is a multiple of 8 for all positive integer values of $n$. [3]

for expanding at least one bracket correctly, e.g. $n^2 + 10n + 25$ or $n^2 + 2n + 1$ (or for using the difference of two squares: $[(n + 5) - (n + 1)][(n + 5) + (n + 1)]$)M1
for $8n + 24$ (or $4(2n + 6)$)A1
for $8(n + 3)$ with a conclusion, e.g. '$8(n + 3)$ is a multiple of 8' (accept $\frac{8n+24}{8} = n + 3$, which is an integer, so a multiple of 8)C1
Answer: $8n + 24 = 8(n + 3)$, a multiple of 8
Examiner tip: Bracket the second square before subtracting: $-(n^2 + 2n + 1)$, so the $+1$ becomes $-1$.

4Grade 6–7HigherNon-calculator3 marks

Prove algebraically that the sum of any two consecutive odd numbers is a multiple of 4. [3]

for writing two consecutive odd numbers algebraically, e.g. $2n + 1$ and $2n + 3$ (or $2n - 1$ and $2n + 1$)M1
for adding to get $4n + 4$ (or $4n$)A1
for $4(n + 1)$ with a conclusion, e.g. 'which is a multiple of 4'C1
Answer: $(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1)$
Examiner tip: $n$ and $n + 2$ are not necessarily odd – use $2n + 1$ to guarantee an odd number.

5Grade 6–7HigherNon-calculator3 marks

Prove that $x^2 - 6x + 11 > 0$ for all values of $x$. [3]

for starting to complete the square: $(x - 3)^2 \ldots$M1
for $(x - 3)^2 + 2$A1
for a full explanation, e.g. '$(x - 3)^2 \ge 0$ for all $x$, so $(x - 3)^2 + 2 \ge 2 > 0$'C1
Answer: $x^2 - 6x + 11 = (x - 3)^2 + 2 \ge 2 > 0$
Examiner tip: Completing the square is not enough on its own – you must state that a square is never negative.

6Grade 8–9HigherNon-calculator4 marks

Prove that the product of two consecutive even numbers, plus 1, is always an odd square number. [4]

for writing two consecutive even numbers algebraically, e.g. $2n$ and $2n + 2$M1
for $2n(2n + 2) + 1$ expanded correctly: $4n^2 + 4n + 1$M1
for factorising to $(2n + 1)^2$A1
for a conclusion that explains both facts, e.g. '$2n + 1$ is odd, so $(2n + 1)^2$ is the square of an odd number, which is an odd square'C1
Answer: $2n(2n + 2) + 1 = 4n^2 + 4n + 1 = (2n + 1)^2$, the square of the odd number $2n + 1$
Examiner tip: Showing the result is a square earns most marks, but the final mark needs you to say WHY it is odd ($2n + 1$ is odd).

7Grade 8–9HigherNon-calculator4 marks

Prove that the sum of the squares of any three consecutive integers is never a multiple of 3. [4]

for writing three consecutive integers, e.g. $n - 1,\ n,\ n + 1$ (or $n,\ n + 1,\ n + 2$)M1
for the sum of their squares with brackets expanded correctly, e.g. $n^2 - 2n + 1 + n^2 + n^2 + 2n + 1$M1
for $3n^2 + 2$ (or $3n^2 + 6n + 5$)A1
for a conclusion, e.g. '$3n^2$ is a multiple of 3, so $3n^2 + 2$ is 2 more than a multiple of 3 and so cannot be a multiple of 3' (or $3(n^2 + 2n + 1) + 2$ with the same argument)C1
Answer: $(n - 1)^2 + n^2 + (n + 1)^2 = 3n^2 + 2$, which leaves remainder 2 when divided by 3
Examiner tip: Choosing $n - 1,\ n,\ n + 1$ makes the $n$ terms cancel; with $n,\ n + 1,\ n + 2$ you get $3n^2 + 6n + 5$ and must split it as $3(n^2 + 2n + 1) + 2$.

8Grade 1–3FoundationNon-calculator3 marks

Here are four pieces of algebra.

A $4(x + 3) \equiv 4x + 12$

B $5x - 2 = 18$

C $A = \pi r^2$

D $3a^2 + 2b$

(a) For each one, write down whether it is an expression, an equation, a formula or an identity. [2]

(b) Explain why $4(x + 3) \equiv 4x + 12$ is an identity but $4x + 12 = 20$ is not. [1]

(a) for at least two of the four correct: A identity, B equation, C formula, D expressionB1
(a) for all four correct (A identity, B equation, C formula, D expression)B1
(b) for a correct explanation, e.g. '$4(x + 3) \equiv 4x + 12$ is true for every value of $x$, but $4x + 12 = 20$ is only true for one value ($x = 2$)'. Accept 'the identity is always true / both sides are the same expression; the equation can be solved'. Do not accept 'because it has a $\equiv$ sign' on its ownC1
Answer: (a) A identity, B equation, C formula, D expression. (b) The identity is true for all values of $x$; $4x + 12 = 20$ is only true when $x = 2$.
Examiner tip: A formula links two or more different variables (like $A$ and $r$); an equation is only true for particular values, which you can solve to find – don't mix them up.

9Grade 4–5Non-calculator3 marks

$n$ is a positive integer.

For each statement, write down whether it is always true, sometimes true or never true. Give a reason or an example to support each answer.

(a) $3n + 1$ is an odd number. [1]

(b) $2(n + 3)$ is an even number. [1]

(a) 'sometimes true' with two examples, one of each kind, e.g. $n = 2$ gives 7 (odd) and $n = 1$ gives 4 (even). Accept a correct general reason, e.g. 'odd when $n$ is even, even when $n$ is odd'B1
(b) 'always true' with a reason, e.g. '$2(n + 3)$ is 2 multiplied by a whole number, so it is even'. Examples alone are not enough for 'always'B1
(c) 'never true' with a reason, e.g. '$n + (n + 1) = 2n + 1$, which is one more than an even number, so always odd'. Accept 'one of two consecutive numbers is odd and the other even, so the sum is odd'B1
Answer: (a) Sometimes true (e.g. $n=1$ gives 4, $n=2$ gives 7). (b) Always true: $2 \times$ an integer is even. (c) Never true: $2n + 1$ is always odd.
Examiner tip: One counter-example is enough to show a statement is not always true, but 'always' and 'never' need a general reason, not a list of examples.

10Grade 4–5Non-calculator4 marks

(a) Given that $$(x + 4)^2 - (x - 2)(x + 6) \equiv a(x + b)$$ find the value of $a$ and the value of $b$. [3]

(b) $x$ is an integer. Explain why $(x + 4)^2 - (x - 2)(x + 6)$ is always a multiple of 4. [1]

(a) for expanding at least one of the brackets correctly: $x^2 + 8x + 16$ or $x^2 + 4x - 12$M1
(a) for subtracting correctly to get $4x + 28$ (the $-12$ must become $+12$)M1
(a) for $a = 4$ and $b = 7$ caoA1
(b) for e.g. '$4(x + 7)$: $x + 7$ is an integer, so the expression is 4 times an integer, a multiple of 4'. Follow through their $4(x + b)$ from (a) if $b$ is an integerC1
Answer: (a) $a = 4$, $b = 7$. (b) It equals $4(x + 7)$ and $x + 7$ is an integer.
Examiner tip: Put the second product in a bracket before subtracting: $-(x^2 + 4x - 12)$ gives $+12$, not $-12$.

11Grade 6–7HigherNon-calculator3 marks

Sam tries to prove that the sum of any two odd numbers is even. Here is Sam's working.

Let the odd numbers be $2n + 1$ and $2n + 1$.
$(2n + 1) + (2n + 1) = 4n + 2 = 2(2n + 1)$, which is even.

(a) Explain why Sam's working does not prove the statement. [1]

(b) Prove that the sum of any two odd numbers is even. [2]

(a) for e.g. 'Sam has used the same odd number twice, so it only shows that an odd number added to itself is even' oe. Accept 'it doesn't work for two different odd numbers, e.g. 3 and 7'C1
(b) for two odd numbers with different letters added, e.g. $(2n + 1) + (2m + 1) = 2n + 2m + 2$ (or $2a - 1$ and $2b + 1$)M1
(b) for $2(n + m + 1)$ with a conclusion, e.g. 'a multiple of 2, so even'. Accept $2n + 2m + 2$ with 'every term is even, so the total is even'C1
Answer: (a) Sam uses the same odd number twice, so it only covers adding an odd number to itself. (b) $(2n+1)+(2m+1) = 2(n+m+1)$, a multiple of 2, so even.
Examiner tip: Two different unknown numbers need two different letters; using $n$ for both only proves a special case.

12Grade 6–7HigherNon-calculator3 marks

Priya shows her friends a number trick.

Think of a number.
Add 3.
Square your answer.
Subtract the square of the number you first thought of.
Subtract 9.
Divide by 6.

Priya says, 'Your final answer will always be the number you first thought of.'

Prove that Priya is correct. [3]

for using a letter for the starting number and writing the steps correctly as one expression, e.g. $\dfrac{(n + 3)^2 - n^2 - 9}{6}$ (or the steps in order)M1
for expanding $(n + 3)^2$ correctly as $n^2 + 6n + 9$M1
for $\dfrac{6n}{6} = n$ with a conclusion, e.g. 'which is the number first thought of'. Checking with numbers only scores 0C1
Answer: $\dfrac{(n+3)^2 - n^2 - 9}{6} = \dfrac{n^2 + 6n + 9 - n^2 - 9}{6} = \dfrac{6n}{6} = n$
Examiner tip: Trying a few numbers is not a proof; you must use a letter for 'the number you first thought of'.

13Grade 6–7HigherNon-calculator3 marks

The diagram shows a square and a rectangle. All lengths are in centimetres and $n$ is positive.

The square has sides of length $(n + 3)$ cm. The rectangle has length $(n + 4)$ cm and width $(n + 2)$ cm.

Prove that the area of the square is always 1 cm² more than the area of the rectangle. [3]

for the area of the square expanded correctly: $(n + 3)^2 = n^2 + 6n + 9$M1
for the area of the rectangle expanded correctly: $(n + 4)(n + 2) = n^2 + 6n + 8$M1
for $n^2 + 6n + 9 - (n^2 + 6n + 8) = 1$ with a conclusion, e.g. 'so the square is always 1 cm² bigger' (accept $n^2 + 6n + 9 = (n^2 + 6n + 8) + 1$)C1
Answer: $(n+3)^2 - (n+4)(n+2) = (n^2 + 6n + 9) - (n^2 + 6n + 8) = 1$
Examiner tip: Write both areas as full expansions before comparing; showing it for one value of $n$ earns no marks.

14Grade 8–9HigherNon-calculator4 marks

$n$ is an integer greater than 1.

Prove that $n^3 - n$ is always a multiple of 6. [4]

for taking out the common factor: $n(n^2 - 1)$M1
for factorising fully: $n(n - 1)(n + 1)$ or $(n - 1)n(n + 1)$M1
for stating that $n - 1$, $n$ and $n + 1$ are three consecutive integersC1
for a complete argument: 'at least one of them is even (a multiple of 2) and one of them is a multiple of 3, so the product is a multiple of $2 \times 3 = 6$' oe. Dep on previous C1C1
Answer: $n^3 - n = (n - 1)n(n + 1)$, the product of three consecutive integers, which contains a multiple of 2 and a multiple of 3, so it is a multiple of 6.
Examiner tip: After factorising, say why the product is a multiple of 6: you need both a factor of 2 and a factor of 3.

15Grade 8–9HigherNon-calculator4 marks

Prove that the square of any odd number is always 1 more than a multiple of 8. [4]

for writing an odd number algebraically and squaring, e.g. $(2n + 1)^2$M1
for expanding correctly: $4n^2 + 4n + 1$M1
for writing as $4n(n + 1) + 1$A1
for a complete argument: '$n$ and $n + 1$ are consecutive integers so one is even; $n(n + 1)$ is even, so $4n(n + 1)$ is a multiple of 8, and the square is 1 more than a multiple of 8' oeC1
Answer: $(2n + 1)^2 = 4n(n + 1) + 1$; $n(n + 1)$ is even, so $4n(n + 1)$ is a multiple of 8.
Examiner tip: $4n^2 + 4n$ only shows a multiple of 4; the last mark needs the reason $n(n + 1)$ is even.

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