# Question 9:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume $n^2-2$ is divisible by 4, so $n^2-2=4k$ | M1 | Sets up algebraic statement in terms of integer $k$ engaging with divisibility by 4; no need for explicit assumption statement |
| Then $n^2=4k+2=2(2k+1)$ (so $n^2$ is even) | A1 | Reaches $n^2=2(2k+1)$ |
| Hence $n^2$ is even so $n$ (=$2m$) is even, hence $n^2$ is a multiple of 4. As $n^2$ is a multiple of 4, $n^2-2=4m^2-2=2(2m^2-1)$ cannot be a multiple of 4 (as $2m-1$ is odd) — contradiction. | dM1 | Complete argument leading to full contradiction of initial statement; minor omissions allowed if overall argument is clear |
| So the original assumption is false. Hence "$n^2-2$ is never divisible by 4" is true for all $n$ | A1* | Draws contradiction and concludes statement true for all $n$; must have clear assumption at start that is contradicted; all working correct |

### Alternative (Alt 1):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume $n^2-2$ divisible by 4, so $\frac{n^2-2}{4}$ is an integer. If $n$ is even, $n=2m$, so $\frac{(2m)^2-2}{4}$ | M1 | |
| $= m^2 - \frac{1}{2}$ (which is not an integer) | A1 | |

## Question 9 (Proof by Contradiction that $n^2 - 2$ is never divisible by 4):

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### Main Method:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Since $m^2$ is an integer, $m^2 - \frac{1}{2}$ is not, hence $n$ cannot be even, but if $n$ is odd then $\frac{n^2-2}{4} = \frac{(2m+1)^2-2}{4} = m^2 + m - \frac{1}{4}$, which is again not an integer (since $m^2 + m$ is) | dM1 | For a complete argument leading to contradiction in both cases |
| Hence there is a contradiction (as $n$ cannot be an integer). Hence "$n^2 - 2$ is never divisible by 4" is true for all $n$ | A1* | Must draw contradiction to initial assumption and conclude true for all $n$; clear assumption must be contradicted; all working correct |

**Total: 4 marks**

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### Alt 2:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Assume that $n^2 - 2$ is divisible by 4 $\Rightarrow n^2 - 2 = 4k$ | M1 | Sets up algebraic statement engaging with divisibility by 4 |
| $\Rightarrow n^2 = 4k+2 \Rightarrow n = 2\sqrt{k+\frac{1}{2}}$ or $n = \sqrt{2}\sqrt{2k+1}$ | A1 | Reaches this expression |
| So for some integer $m$: $\sqrt{k+\frac{1}{2}} = \frac{m}{2} \Rightarrow 2k+1 = \frac{m^2}{2}$ but $m^2$ is odd if $m$ is odd so $\frac{m^2}{2}$ not an integer, or $m^2$ is a multiple of 4 if $m$ even, so odd=even. Or $2k+1$ is odd, so does not have a factor 2 to combine with the $\sqrt{2}$ outside, hence $n$ must be irrational | dM1 | Complete argument leading to contradiction |
| Hence we have a contradiction. So "$n^2 - 2$ is never divisible by 4" is true for all $n$ | A1* | Draws contradiction and concludes for all $n$ |

**Total: 4 marks**

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### Alt 3:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Assume $n^2-2$ is divisible by 4; for $n$ even: $n^2-2 = 4m^2-2$, or for $n$ odd: $n^2-2 = 4(m^2+m)-1$ | M1 | Uses $n$ odd or $n$ even to form expression for $n^2-2$ of the form $4\times\text{integer} \pm \text{non-multiple of 4}$ |
| $4m^2 - 2$ or $4(m^2+m)-1$ | A1 | Reaches $4m^2-2$ or $4(m^2+m)-1$ |
| Since 4 divides $n^2-2$ and $4m^2$, thus for $n$ even, 4 must divide 2, a contradiction; so $n$ cannot be even, and also 4 divides $4(m^2+m)$ so for $n$ odd, 4 divides 1, also a contradiction | dM1 | Both cases considered with reason for contradiction given |
| Hence contradiction for both cases (and $n$ must be either even or odd). So "$n^2-2$ is never divisible by 4" is true for all $n$ | A1* | Draws contradiction and concludes for all $n$ |

**Total: 4 marks**