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Answer: six values of $n$ ($-3, -1, 0,2,3,5$)
Using factors
$n-1$ is a factor of $n+3$ means $n$ is quite small because otherwise $n-1$ is too close to $n+3$
| $n$ | $n-1$ | $n+3$ | fit? |
| 2 | 1 | 5 | yes |
| 3 | 2 | 6 | yes |
| 4 | 3 | 7 | no |
| 5 | 4 | 8 | yes |
| 6 | 5 | 9 | no and it won't work for larger numbers because 5 is more than half of 9 |
Or if $n$ can be negative:
| $n$ | $n-1$ | $n+3$ | fit? |
| 0 | $-$1 | 3 | yes |
| $-$1 | $-$2 | 2 | yes |
| $-$2 | $-$3 | 1 | no |
| $-$3 | $-$4 | 0 | yes |
| $-$4 | $-$5 | $-$1 | no |
| from now on, the 'size' of $n-1$ will be greater than the 'size' of $n+3$ so we won't get any more fits | |||
Using algebra
$\frac{n+3}{n-1} = \frac{n-1}{n-1} + \frac{4}{n-1} = 1 + \frac{4}{n-1}$. Thus $\frac{n+3}{n-1}$ is an integer if and only if $n-1$ divides exactly into $4$. The values of $n$ for which this is true are $-3, -1, 0,2,3,5$.