June 2022 Paper 1 Q12
12 Prove by contradiction that 3 is the only prime number which is 1 less than a square number. [4]
| Scheme | Marks | AO |
|---|---|---|
| Assume there is a prime number \(p\) which is one less than a square number \(p = n^2 - 1\) for some positive integer \(n \geqslant 2\) | M1* | 2.1 |
| \(p = (n-1)(n+1)\) | M1* | 2.1 |
| If \(n = 2\), \(p = 1 \times 3 = 3\) which is prime [\(p = 2\) is not 1 less than a square number] | E1 | 2.1 |
| If \(n \gt 2\) then \(p\) has two [proper] factors so is not prime which is a contradiction. So there are no prime numbers other than 3 which are 1 less than a square number | E1 (dep) | 2.1 |
| [4] |
Notes
M1*: Setting up proof by contradiction
M1*: factorising
E1: Considers the possibility that one factor might be 1
E1: Condone missing reference to \(n = 2\) (or \(p = 3\)) for this step.
Conclusion must be clear.
SC1: Allow SC1 where M1M0 or M0M0 has been awarded and \(3 = 2^2 - 1\) is established