In $\triangle{ABC}$, $AB = 33, AC=21,$ and $BC=m$ where $m$ is an integer. There exist points $D$ and $E$ on $AB$ and $AC$, respectively, such that $AD=DE=EC=n$ where $n$ is also an integer. Find all the possible values of $m$.
Applying the law of cosine on $\triangle{ADE}$ and $\triangle{ABC}$, respectively: $$\cos{A}=\frac{n^2 + (21-n)^2 - n^2}{2\cdot n\cdot (21-n)}=\frac{33^2+21^2-m^2}{2\cdot 33\cdot 21}$$ Therefore $$\frac{21-n}{2n} = \frac{1530-m^2}{2\cdot 3^2 \cdot 7 \cdot 11} \implies n\cdot(2223 - m^2) = 3^3\cdot 7^2\cdot 11$$ Note that $$AE < AC\quad \text{and}\quad AD+DE > AE \implies 7 < n < 21$$ Based on the $2^{nd}$ question, it must be true that $n\mid(3^3\cdot 7^2\cdot 11)$. Hence, $n$ can only take $9$ or $11$. When $n=9$, no integer solution exists for $m$. When $n=11$, we have $m=30$. Therefore, the only answer is $m=\boxed{30}$.