Example 4 As shown in Figure 2, in the right triangle $\triangle ABC$, it is known that $\angle BCA=90^{\circ}$, $CD$ is the altitude, the three sides of the right triangle $\triangle ABC$ are all integers, and $BD=11^{3}$. Find the ratio of the perimeters of the right triangles $\triangle BCD$ and $\triangle ACD$.
(2002, National Junior High School Mathematics Competition)
Let $B C=a, C A=b, A B=c$.
From Rt $\triangle B C D \backsim \mathrm{Rt} \triangle B A C$, we get
$$
\frac{B C}{B A}=\frac{B D}{B C} \text {, }
$$

which means $B C^{2}=B D \cdot B A$.
Therefore, $a^{2}=11^{3} c$.
Since $a^{2}$ is a perfect square and 11 is a prime number, let $c=11 k^{2}$ (where $k$ is a positive integer).
Thus, $a=11^{2} k$.
Then, by the Pythagorean theorem, we have
$$
b=\sqrt{c^{2}-a^{2}}=11 k \sqrt{k^{2}-11^{2}} \text {. }
$$

Since $b$ is an integer, $k^{2}-11^{2}$ must be a perfect square.
Let $k^{2}-11^{2}=m^{2}$. Then
$$
(k+m)(k-m)=11^{2} \text {. }
$$

Since $k+m>k-m>0$ and 11 is a prime number, we have
$$
\left\{\begin{array} { l } 
{ k + m = 1 1 ^ { 2 } , } \\
{ k - m = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k=61, \\
m=60 .
\end{array}\right.\right.
$$

Thus, $a=11^{2} \times 61, b=11 \times 61 \times 60$.
Since Rt $\triangle B C D \backsim \mathrm{Rt} \triangle C A D$, the ratio of their perimeters is equal to their similarity ratio, i.e.,
$$
\frac{a}{b}=\frac{11^{2} \times 61}{11 \times 61 \times 60}=\frac{11}{60} .
$$