| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Proof |
| Type | Proof by exhaustion with cases |
| Difficulty | Moderate -0.3 Part (a) requires finding a single counterexample to disprove a statement (e.g., m=3 gives 13, m=5 gives 29, m=7 gives 53, but m=9 gives 85=5×17), which is straightforward. Part (b) is a routine proof by cases using modular arithmetic: odd n gives n²≡1(mod 4) so n²+1≡2(mod 4), even n gives n²≡0(mod 4) so n²+1≡1(mod 4). Both parts are standard textbook exercises requiring minimal insight. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| eg \(9^2+4=85\) and \(85\) is multiple of \(5\) or \(85=5\times17\) or \(85\div5=17\) or \(85\) has a factor of \(5\) (or \(17\)) | B1 [1] | oe Not just "and 85 is not prime"; One correct example and one incorrect: B1; Condone eg \(9^2+4=85\div5=17\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((2k)^2+1=4k^2+1\) (\(k\) an integer) or eg \((2k+2)^2+1=4k^2+8k+5\) which is not a multiple of 4 | B1 | Allow any letter even \(n\); Allow omission of "(\(k\) an integer)" in both places |
| \((2k+1)^2+1=4k^2+4k+1+1\) (\(k\) an integer) \(= 4(k^2+k)+2\) or \(4k^2+4k+2\) | M1, A1* | Attempt expand & add 1, eg \(4k^2+1+1\): M1; Must see one of these forms; Same marks for \((2k-1)^2+1\) |
| Any sensible explanation why this is not mult of 4 eg This is of the form \(4\times\text{integer}+2\); \(4k^2\) & \(4k\) are mults of 4 but when \(+2\), not mult of 4 | A1 dep* [4] | eg \(4(k^2+k)\) is a multiple of 4; 2 is not a multiple of 4; \(k^2+k+\frac{1}{2}\) not an integer; Not just "This is not a mult of 4"; Numerical examples: no mks |
| Answer | Marks |
|---|---|
| \(n\) even \(\Rightarrow\) \(n^2\) is mult of 4, hence \(n^2+1\) is not mult of 4 or \(n^2\) is even so \(n^2+1\) is odd so not mult of 4 | B1 |
| \(n\) even, \(n+1\) is odd; \((n+1)^2+1 = n^2+2n+2\); \(n^2\) and \(2n\) are mults of 4, hence \(n^2+2n+2\) is not | M1, A1, A1 [4] |
## Question 5:
### Part (a):
eg $9^2+4=85$ and $85$ is multiple of $5$ or $85=5\times17$ or $85\div5=17$ or $85$ has a factor of $5$ (or $17$) | **B1** [1] | oe Not just "and 85 is not prime"; One correct example and one incorrect: B1; Condone eg $9^2+4=85\div5=17$
### Part (b):
$(2k)^2+1=4k^2+1$ ($k$ an integer) or eg $(2k+2)^2+1=4k^2+8k+5$ which is not a multiple of 4 | **B1** | Allow any letter even $n$; Allow omission of "($k$ an integer)" in both places
$(2k+1)^2+1=4k^2+4k+1+1$ ($k$ an integer) $= 4(k^2+k)+2$ or $4k^2+4k+2$ | **M1, A1*** | Attempt expand & add 1, eg $4k^2+1+1$: M1; Must see one of these forms; Same marks for $(2k-1)^2+1$
Any sensible explanation why this is not mult of 4 eg This is of the form $4\times\text{integer}+2$; $4k^2$ & $4k$ are mults of 4 but when $+2$, not mult of 4 | **A1** dep* [4] | eg $4(k^2+k)$ is a multiple of 4; 2 is not a multiple of 4; $k^2+k+\frac{1}{2}$ not an integer; Not just "This is not a mult of 4"; Numerical examples: no mks
**Alternative method:**
$n$ even $\Rightarrow$ $n^2$ is mult of 4, hence $n^2+1$ is not mult of 4 or $n^2$ is even so $n^2+1$ is odd so not mult of 4 | **B1** |
$n$ even, $n+1$ is odd; $(n+1)^2+1 = n^2+2n+2$; $n^2$ and $2n$ are mults of 4, hence $n^2+2n+2$ is not | **M1, A1, A1** [4] |
---5
\begin{enumerate}[label=(\alph*)]
\item Prove that the following statement is not true.\\
$m$ is an odd number greater than $1 \Rightarrow m ^ { 2 } + 4$ is prime.
\item By considering separately the case when $n$ is odd and the case when $n$ is even, prove that the following statement is true.\\
$n$ is a positive integer $\Rightarrow n ^ { 2 } + 1$ is not a multiple of 4 .
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q5 [5]}}