June 2025 Paper 3 Q10
10 The magnitude of the vector \(\begin{pmatrix}4\\-1\\x\end{pmatrix}\) is an integer.
Determine all possible integer values of \(x\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(4^2 + 1^2 + x^2\) | B1 | 1.1 |
| \(17 + x^2 = n^2\) OR \(n^2 - x^2 = 17\) | B1 | 1.1 |
| \((n - x)(n + x) = 17\) | M1 | 3.1a |
| \(n - x = 1\) and \(n + x = 17\) AND \(n - x = -1\) and \(n + x = -17\) | A1 | 2.2a |
| 8 and \(-8\) | B2 | 1.1 |
| [6] |
Notes
B1: Use of Pythagoras implied by \(17 + x^2\)
B1: Understanding that \(17 + x^2\) is a square number soi e.g. by equating to at least one square number \(17 + x^2 = 25\)
Allow e.g. \(n = \sqrt{17 + x^2}\) and \(n\) integer soi e.g. by equating to an integer
M1: Rearranging to get their 17 on one side and factorising
A1: OR \(n - x = 17\) and \(n + x = 1\) AND \(n - x = -17\) and \(n + x = -1\)
Implies M1. If the final B2 is earned following only one pair of correct equations then award A1
B2: Both values and no others
B1 if only one value given (ignore extras)
Alternative method 1 for M1 A1
| Scheme | Marks |
|---|---|
| Evidence of adding square numbers to 17 to try to get a square number Or subtracting 17 from square numbers to get a square number Or \(x^2 = 64\) | M1 |
| Explanation of why there will be no further values | A1 |
M1: At least two different square numbers added to their 17 e.g. two from: \(17 + 1\), \(17 + 4\), \(17 + 9\), \(17 + 16\), \(17 + 25\), \(17 + 49\), \(17 + 64\), \(17 + 81\) and so on
Or tested two using \(17 + x^2 = 25\), \(17 + x^2 = 36\), \(17 + x^2 = 49\), \(17 + x^2 = 64\), \(17 + x^2 = 81\) and so on
A1: e.g. Since 64 and 17 + 64 are adjacent square numbers there will be no larger values of \(x^2\)
Alternative method 2 for M1 A1
| Scheme | Marks |
|---|---|
| \((x + 1)^2 - x^2 = 17\) | M1 |
| \(2x + 1 = 17\) and \((x - 1)^2 - x^2 = 17\) | A1 |
M1: Uses adjacent square numbers
Implied by \(2x + 1 = 17\)
Condone not testing earlier squares for this M1
A1: If the final B2 is earned following M1 then award A1