Note that :

$\frac{n+3}{n-1} = \frac{n-1}{n-1} + \frac{4}{n-1} = 1 + \frac{4}{n-1}$. 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 $-3, -1, 0,2,3,5$.

*This problem is taken from the UKMT Mathematical Challenges.*