Divisibility of $n^2+9$ by $n+3$ How would one find all integers $n$ such that $n+3 \vert n^2 +9$? I assume it is important that $(n+3)^2 - 6n = n+3$, but I am struggling to see how you can find all $n$, and confirm an upper bound such that there are no more such $n$.

Alternatively:
$n+3|n^2+9 \iff$
$n+3|n^2 + 9 - n(n+3)\iff$
$n+3|-3n + 9\iff$
$n+3|-3n + 9 + 3(n+3)\iff$
$n+3|18$
$n+3 = \pm 1; \pm 2;\pm 3;\pm 6;\pm 9;\pm 18$
$-21,-12,-9,-6,-5,-4,-2,-1,0,3,6,15$
