Moderate -0.8 This is a straightforward proof by exhaustion requiring students to check the four cases n ≡ 0, 1, 2, 3 (mod 4) and show n² + 2 ≡ 2 or 3 (mod 4) in each case. The method is explicitly stated, the modular arithmetic is basic, and it's a routine application of a standard proof technique with minimal steps.
Given \(n \in \mathbb { N }\), prove, by exhaustion, that \(n ^ { 2 } + 2\) is not divisible by 4 .
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When \(n\) is odd, \(n^2\) is odd (odd \(\times\) odd = odd) so \(n^2 + 2\) is also odd
M1
So for all odd \(n\), \(n^2+2\) is also odd and so cannot be divisible by 4 (as all numbers in the 4 times table are even)
A1
When \(n\) is even, \(n^2\) is even and a multiple of 4, so \(n^2+2\) cannot be a multiple of 4
M1
Fully correct and exhaustive proof plus final statement "So for all \(n\), \(n^2+2\) cannot be divisible by 4"
A1*
Alternative (algebraic):
Answer
Marks
If \(n\) is even, \(n=2k\), so \(\frac{n^2+2}{4} = \frac{(2k)^2+2}{4} = \frac{4k^2+2}{4} = k^2 + \frac{1}{2}\)
M1
If \(n\) is odd, \(n=2k+1\), so \(\frac{n^2+2}{4} = \frac{(2k+1)^2+2}{4} = \frac{4k^2+4k+3}{4} = k^2+k+\frac{3}{4}\)
M1
Either \(k^2+\frac{1}{2}\) or \(k^2+k+\frac{3}{4}\) are not whole numbers, with valid reason why this means \(n^2+2\) is not a multiple of 4
A1
Full proof with no errors or omissions, including the conjecture, correct notation and algebra for both even and odd numbers, and full explanation for all \(n\)
A1*
## Question 4:
| When $n$ is odd, $n^2$ is odd (odd $\times$ odd = odd) so $n^2 + 2$ is also odd | M1 | |
|---|---|---|
| So for all odd $n$, $n^2+2$ is also odd and so cannot be divisible by 4 (as all numbers in the 4 times table are even) | A1 | |
| When $n$ is even, $n^2$ is even **and a multiple** of 4, so $n^2+2$ cannot be a multiple of 4 | M1 | |
| Fully correct and exhaustive proof plus final statement "So for all $n$, $n^2+2$ cannot be divisible by 4" | A1* | |
**Alternative (algebraic):**
| If $n$ is even, $n=2k$, so $\frac{n^2+2}{4} = \frac{(2k)^2+2}{4} = \frac{4k^2+2}{4} = k^2 + \frac{1}{2}$ | M1 | |
|---|---|---|
| If $n$ is odd, $n=2k+1$, so $\frac{n^2+2}{4} = \frac{(2k+1)^2+2}{4} = \frac{4k^2+4k+3}{4} = k^2+k+\frac{3}{4}$ | M1 | |
| Either $k^2+\frac{1}{2}$ or $k^2+k+\frac{3}{4}$ are not whole numbers, with valid reason why this means $n^2+2$ is not a multiple of 4 | A1 | |
| Full proof with no errors or omissions, including the conjecture, correct notation and algebra for both even and odd numbers, and full explanation for all $n$ | A1* | |
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Given $n \in \mathbb { N }$, prove, by exhaustion, that $n ^ { 2 } + 2$ is not divisible by 4 .\\
\includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-12_2658_1943_111_118}
\hfill \mbox{\textit{Edexcel P2 2018 Q4 [4]}}