Note that

\frac{n^{2}}{n + 4} = \frac{n^{2} + 16}{n + 4} - \frac{16}{n + 4} = \frac{(n + 4) (n - 4)}{n + 4} - \frac{16}{n + 4} = n - 4 - \frac{16}{n + 4}

So when $n > 12$ , the remainder when $n^{2}$ is divided by $n + 4$ is always 16.

For $1 \leq n \leq 12$ , the remainder when $n^{2}$ is divided by $n + 4$ is shown in the table below.

$n$     1     2     2     4     5     6     7     8     9     10     11     12

$n + 4$     5     6     7     8     9     10     11     12     13     14     15     16

remainder     1     4     2     0     7     6     5     4     3     2     1     0

So there are 9 different remainders, namely 0, 1, 2, 3, 4, 5, 6, 7, 16.

$n$	1	2	2	4	5	6	7	8	9	10	11	12
$n + 4$	5	6	7	8	9	10	11	12	13	14	15	16
remainder	1	4	2	0	7	6	5	4	3	2	1	0