Challenging +1.2 This is a proof by contradiction requiring students to assume there exists another prime p = n² - 1, then factor as p = (n-1)(n+1) to show p must be composite unless n = 2. While it requires understanding of proof structure and factorization insight, the algebraic manipulation is straightforward and the question explicitly tells students which proof method to use, making it moderately above average difficulty but not requiring deep mathematical insight.
Assume there is a prime number \(p\) which is one less than a square number
M1*
Setting up proof by contradiction
\(p = n^2 - 1\) for some positive integer \(n \geq 2\)
\(p = (n-1)(n+1)\)
M1*
Factorising
If \(n = 2\): \(p = 1 \times 3 = 3\) which is prime [\(p = 2\) is not 1 less than a square number]
E1
Considers the possibility that one factor might be 1
If \(n > 2\) 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)
Condone missing reference to \(n=2\) (or \(p=3\)) for this step. Conclusion must be clear. Allow SC1 where M1M0 or M0M0 awarded and \(3 = 2^2 - 1\) is established
[4]
## Question 12:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume there is a prime number $p$ which is one less than a square number | M1* | Setting up proof by contradiction |
| $p = n^2 - 1$ for some positive integer $n \geq 2$ | | |
| $p = (n-1)(n+1)$ | M1* | Factorising |
| If $n = 2$: $p = 1 \times 3 = 3$ which is prime [$p = 2$ is not 1 less than a square number] | E1 | Considers the possibility that one factor might be 1 |
| If $n > 2$ 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) | Condone missing reference to $n=2$ (or $p=3$) for this step. Conclusion must be clear. Allow SC1 where M1M0 or M0M0 awarded and $3 = 2^2 - 1$ is established |
| **[4]** | | |
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