4. Let the three sides of the right triangle $\triangle ABC$ be $a$, $b$, and $c$, with $a < b < c$. If $\frac{b}{c+a} + \frac{a}{c+b} = \frac{17}{20}$. Then $a: b: c$ $=$ $~$. $\qquad$
4.8:15:17.

Since $c^{2}-a^{2}=b^{2}, c^{2}-b^{2}=a^{2}$, we have
$$
\begin{array}{l}
\frac{17}{20}=\frac{b}{c+a}+\frac{a}{c+b}=\frac{b(c-a)}{c^{2}-a^{2}}+\frac{a(c-b)}{c^{2}-b^{2}} \\
=\frac{c-a}{b}+\frac{c-b}{a}=\frac{c(a+b)-\left(a^{2}+b^{2}\right)}{a b} \\
=\frac{c(a+b-c)}{a b} .
\end{array}
$$

Also, $a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2}$
$$
\begin{array}{l}
=\frac{(a+b)^{2}-c^{2}}{2} \\
=\frac{1}{2}(a+b+c)(a+b-c),
\end{array}
$$

Therefore, $\frac{17}{20}=\frac{2 c}{a+b+c}$, which means
$$
17(a+b)=23 c \text {. }
$$

Squaring both sides, we get
$$
289\left(a^{2}+2 a b+b^{2}\right)=529 c^{2}=529\left(a^{2}+b^{2}\right) \text {. }
$$

Rearranging, we have $(15 a-8 b)(8 a-15 b)=0$.
Thus, $\frac{a}{b}=\frac{8}{15}$ or $\frac{a}{b}=\frac{15}{8}$.
Since $8^{2}+15^{2}=17^{2}$, and $a<b<c$, we have $a: b: c=8: 15: 17$.