A6: Algebraic Proof
Know the difference between an equation and an identity; argue mathematically to show algebraic expressions are equivalent
Know the difference between an equation and an identity; argue mathematically to show algebraic expressions are equivalent
Prove that (x + 3)² + (x - 3)² ≡ 2(x² + 9)
Solution:
LHS = (x + 3)² + (x - 3)²
= (x² + 6x + 9) + (x² - 6x + 9)
= x² + 6x + 9 + x² - 6x + 9
= 2x² + 18
= 2(x² + 9)
= RHS ∎
Prove that (n + 5)² - (n - 5)² ≡ 20n
Solution:
LHS = (n + 5)² - (n - 5)²
= (n² + 10n + 25) - (n² - 10n + 25)
= n² + 10n + 25 - n² + 10n - 25
= 20n
= RHS ∎
| Phrase | Expression |
|---|---|
| Any integer n | n |
| Even number | 2n |
| Odd number | 2n + 1 or 2n - 1 |
| Consecutive integers | n, n + 1, n + 2... |
| Consecutive even numbers | 2n, 2n + 2, 2n + 4... |
| Consecutive odd numbers | 2n + 1, 2n + 3, 2n + 5... |
| Multiple of 3 | 3n |
| Square of n | n² |
Prove that the sum of two consecutive even numbers is always a multiple of 4.
Solution:
Let the consecutive even numbers be 2n and 2n + 2
Sum = 2n + (2n + 2)
= 4n + 2
= 2(2n + 1)
This is NOT a multiple of 4 - the conjecture is false!
Counter-example: 4 + 6 = 10, which is not divisible by 4
Prove that the sum of three consecutive integers is always a multiple of 3.
Solution:
Let the integers be n, n + 1, n + 2
Sum = n + (n + 1) + (n + 2)
= 3n + 3
= 3(n + 1)
Since 3(n + 1) is divisible by 3 for any integer n, the statement is proven ∎
Prove that the difference between the squares of two consecutive integers is always odd.
Solution:
Let the consecutive integers be n and n + 1
Difference = (n + 1)² - n²
= (n² + 2n + 1) - n²
= 2n + 1
Since 2n is even, 2n + 1 is always odd ∎
Prove or disprove: "All square numbers are even."
Solution:
Counter-example: 9 = 3², but 9 is odd.
The statement is false.
Prove or disprove: "n² + n + 41 is always prime for any integer n."
Solution:
Try n = 41:
41² + 41 + 41 = 41(41 + 1 + 1) = 41 × 43
This is NOT prime. The conjecture is false.
Algebraic proof: name your variable ('let n be any integer'), build the expression, expand/simplify, and show the required form (even = 2n, odd = 2n + 1, multiple of k).
Prove that the sum of two consecutive even numbers is a multiple of 4... adjust: is always even.
Consecutive evens: 2n and 2n + 2. Sum = 4n + 2 = 2(2n + 1) — a multiple of 2 → always even ∎
M1 for 2n, 2n + 2, A1 for 2(2n + 1). The final line must END in the target form — factor out the 2.
Disprove: 'the square of any number is odd'.
Counter-example: 4² = 16, which is even. One counter-example is enough to disprove a universal claim ∎
B1 for a valid counter-example. Disproof needs only ONE example; proof needs the general argument.
Exam wording to expect: "Prove that …", "show that … is always …", "disprove by counter-example".
Q1: Prove that (x + 4)² - (x + 2)² ≡ 4(2x + 3)
Q2: Write an expression for two consecutive odd numbers and find their sum.
Q3: Prove that the sum of three consecutive even numbers is divisible by 6.
Q4: Prove that (2n + 1)² - (2n - 1)² = 8n
Q5: Find a counter-example to disprove: "All prime numbers are odd."
Q6: Prove that the product of two consecutive integers is always even.
Q6: Prove that the difference between two consecutive square numbers is always odd
Q7: Prove that n(n + 1) + n + 1 factors, and state what that shows
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. (x + 4)² − (x + 2)² = (x² + 8x + 16) − (x² + 4x + 4) = 4x + 12 = 4(2x + 3) ✓
Q2. 2n + 1 and 2n + 3 → sum 4n + 4 = 4(n + 1) — also always even
Q3. 2n, 2n + 2, 2n + 4 → sum 6n + 6 = 6(n + 1) → multiple of 6 ∎
Q4. (4n² + 4n + 1) − (4n² − 4n + 1) = 8n ✓
Q5. n = 2: 2 is prime and even — one counter-example disproves
Q6. n and n + 1: one is even, so the product n(n + 1) has a factor 2 → always even ∎
Q7. n(n + 1) + n + 1 = (n + 1)(n + 1) = (n + 1)² — a perfect square
Algebraic proof: 3–4 marks on Paper 1 (non-calculator). The marking rewards the final factored form — an expanded but unfinished argument loses the last mark.
Links to: expanding (A4), geometric proof (G6), types of numbers (N4).
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
Prove that the product of two odd numbers is always odd.
Solution:
Let the two odd numbers be 2m + 1 and 2n + 1
Product = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1
Since 2(2mn + m + n) is even (it's 2 × something), adding 1 makes it odd. ∎
1. Wrong: Using n and n + 1 for consecutive odd numbers Correct: Consecutive odd numbers are 2n + 1 and 2n + 3
2. Wrong: Proving an identity by substituting one value Correct: Must show both sides are algebraically identical, or test multiple values
3. Wrong: Saying "it works for n = 1, 2, 3 so it's proven" Correct: This only shows a pattern — a proof must work for ALL values using algebra
6 marks: Prove that the difference between the squares of any two consecutive even numbers is always a multiple of 4.
Let the consecutive even numbers be 2n and 2n + 2.
Their squares are (2n)² = 4n² and (2n + 2)² = 4n² + 8n + 4
Difference = (2n + 2)² - (2n)² = 4n² + 8n + 4 - 4n² = 8n + 4 = 4(2n + 1)
Since 4(2n + 1) has a factor of 4 for any integer n, the difference is always a multiple of 4. ∎
Mark scheme: 1 mark for correct expressions, 1 mark for squaring, 2 marks for correct subtraction, 1 mark for factorising, 1 mark for conclusion.
A student claims: "The sum of any three consecutive integers is always even."
(a) Write an expression for the sum of n, n + 1 and n + 2.
(b) Prove or disprove the claim.
(c) What type of number is the sum always a multiple of?
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