June 2023 Paper 2 Q7
7 A student wishes to prove that, for all positive integers \(a\) and \(b\), \(a^2 - 4b \neq 2\).
| Scheme | Marks | AO |
|---|---|---|
| \(a^2 = 4b + 2\) | M1 | 2.1 |
| Hence \(a^2\) is even. Hence \(a\) is even | A1 | 2.2a |
| [2] |
Notes
M1: Setting up so that the deduction \(a^2\) is even can be made.
A1: www, must see both statements and a convincing, correct, argument oe (e.g. \(a^2 = 2(2b + 1)\))
Alternative method
| Scheme | Marks |
|---|---|
| Assume that \(a\) is odd, then \(a^2\) is odd | M1 |
| \(4b\) is even, so \(a^2 - 4b\) is odd Hence 2 is odd (so contradiction) Hence \(a\) is even. | A1 |
M1: For setting up and stating that \(a\) is odd \(\Rightarrow a^2\) is odd
May see (not required) \(a = 2n + 1\), \(a^2 = 2(2n^2 + 2n) + 1\) Hence \(a^2\) is odd
A1: www, Must see both statements and a convincing, correct, argument
| Scheme | Marks | AO |
|---|---|---|
| Assume that \(a^2 - 4b = 2\) Let \(a = 2n\), (where \(n\) is an integer) | M1 | 2.1 |
| Either of: \(4n^2 - 4b = 2\) \(2n^2 - 2b = 1\), Hence 1 is even or: \(4n^2 - 4b = 2\) \(n^2 - b = 0.5\), \(n^2 - b\) is an integer | A1 | 2.1 |
| (Contradiction) Hence \(a^2 - 4b \neq 2\) | A1 | 2.2a |
| [3] |
Notes
M1: Setting up (must see assumption and use of \(a\) is even)
A1: Substituting in \(a = 2n\) and correctly reaching an equation which shows a contradiction.
Accept the equivalent in words if clear and correct.
Also accept: \(4n^2 - 4b\) is a multiple of 4, Hence \(a^2 - 4b\) is a multiple of 4, which is a contradiction
A1: www, Must see both statements and a convincing, correct, argument
Alternative method (1)
| Scheme | Marks |
|---|---|
| Assume that \(a^2 - 4b = 2\) \(\Rightarrow a^2 = 4b + 2 = 2(2b + 1)\) | M1 |
| For \(a\) to be an integer, \(2b + 1\) must be even But (\(2b\) is even, so) \(2b + 1\) is odd | A1 |
| (Which is a contradiction) hence \(a^2 - 4b \neq 2\) | A1 |
A1: Must see both statements and a convincing, correct, argument www
Alternative method (2)
| Scheme | Marks |
|---|---|
| \(a\) even \(\Rightarrow a^2 = 4n\) (\(n\) an integer) \(\Rightarrow a^2\) is congruent to 0 mod 4 | M1 |
| \(4b + 2\) is congruent to 2 mod 4 | A1 |
| Therefore \(a^2\) cannot equal \(4b + 2\) | A1 |
A1: Must see previous two lines and a convincing, correct, argument www
Alternative method (3)
| Scheme | Marks |
|---|---|
| Assume that \(a^2 - 4b = 2\), then \(a\) is even (and consider whether \(b\) is odd or even) | M1 |
| If \(b\) is odd then 2 is either 0 or a multiple of 4, (so contradiction) AND If \(b\) is even then 2 is either 0 or a multiple of 4, (so contradiction) | A1 |
| Therefore \(a^2\) cannot equal \(4b + 2\) | A1 |
M1: For setting up using part (a) and considering either case where \(b\) is odd or even (ignore any reference to the cases where \(a\) is odd as these are not required) May see (but not required) \(a = 2n\) so \(a^2 = 4n^2\)
Condone using the same letter (e.g. \(n\)) in \(a\) and \(b\) for this mark only.
A1: For correctly considering both cases either algebraically or in words.
May see (but not required) \(b = 2m + 1\), so \(a^2 - 4b = 4(n^2 - (2m + 1))\)
And \(b = 2m\), so \(a^2 - 4b = 4(n^2 - 2m)\)
Do not award this mark if same integer (e.g. \(n\)) used in both \(a\) and \(b\)
A1: A fully correct, convincing argument with conclusion, www.