$\text{Ta có }5n+9∈B_{(2n+4)}$
$⇒5n+9\vdots2n+4$
$⇒2(5n+9)\vdots2n+4$
$⇒10n+18\vdots2n+4$
$⇒10n+20-2\vdots2n+4$
$⇒5(2n+4)-2\vdots2n+4$
$\text {vì } 5(2n+4)\vdots2n+4$
$⇒-2\vdots2n+4⇒2n+4∈Ư_{(-2)}=\text{{±1;±2}}$
lập bảng
$\left[\begin{array}{ccc}2n+4&-1&1&-2&2\\n&\text{(loại)}&\text{(loại)}&-3&-1\end{array}\right]$
$\text{Vậy n={-1;-3}}$