Example 10. In $\triangle A B C$, $A B$ $=36$ cm, $A C=21$ cm, $B C$ $=m$ cm, $m$ is an integer. Also, there is a point $D$ on $A B$ and a point $E$ on $A C$ such that $A D=D E=E C$ $=n$ cm, $n$ is an integer. Find the integer solutions for $m, n$.
In $\triangle A B C$ and $\triangle A D E$, according to the cosine rule, we have
$$
\begin{array}{l}
\frac{33^{2}+21^{2}-m^{2}}{2 \times 33 \times 21}=\cos A \\
=\frac{n^{2}+(21-n)^{2}-n^{2}}{2 \times n \times(21-n)}, \\
\frac{1530-m^{2}}{33 \times 21}=\frac{21-n}{n} . \\
n\left(2223-m^{2}\right)=3^{3} \times 7^{2} \times 11 .
\end{array}
$$

That is, $\frac{1530-m^{2}}{33 \times 21}=\frac{21-n}{n}$.

Since $m$ and $n$ are both prime numbers, 11 must be a factor of $3^{3} \times 7^{2} \times 11$. Therefore, $7 < n < 21$. So, $n=3^{2}$ or 11. When $n=9$, $m^{2}=606$ (which does not meet the condition, discard); when $n=11$, $m^{2}=900$, we get $m=30$.
The solution is $\left\{\begin{array}{l}m=30, \\ n=11 .\end{array}\right.$