$5 \cdot 34$ Polynomial
$$
(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}} \cdots\left(1-z^{32}\right)^{b_{32}}
$$

In this polynomial, $b_{i}$ are positive integers $(i=1,2, \cdots, 32)$, and the polynomial has the following remarkable property: when expanded, and terms with powers of $z$ higher than 32 are deleted, exactly $1-2 z$ remains. Determine $b_{32}$ (the answer can be expressed as the difference of two powers of 2).
[Solution] Let $f(z)=(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3} \cdots}\left(1-z^{32}\right)^{b_{32}}$
$$
=1-2 z+g(z) \text {, }
$$

where each term of $g(z)$ is of degree higher than 32. By comparing the coefficients of $z$ on both sides, we get $b_{1}=2$.
$$
\begin{array}{l} 
f(-z)=(1+z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1+z^{3}\right)^{b_{3} \cdots}\left(1-z^{32}\right)^{b_{32}}=1+2 z+ \\
g(-z) . \\
f(z) \cdot f(-z)=\left(1-z^{2}\right)^{b_{1}+2 b_{2}} \cdot\left(1-z^{4}\right)^{2 b_{4}} \cdot\left(1-z^{6}\right)^{b_{3}+2 b_{6}} \cdot(1- \\
\left.z^{8}\right)^{2 b_{8}} \cdot\left(1-z^{10}\right)^{b_{5}+2 b_{10}} \cdots \cdot\left(1-z^{30}\right)^{b_{15}+2 b_{30}} \cdot(1 \\
-z^{32} 2^{2 b_{32}} \cdots \\
= 1-4 z^{2}+g_{1}(z),
\end{array}
$$

where each term of $g_{1}(z)$ is of degree higher than 32 and is even. Let $w=z^{2}, f_{1}(w)=f(z) \cdot f(-z)$, then
$$
\begin{array}{l}
=1-4 w+g_{2}(w), \\
\end{array}
$$

where each term of $g_{2}(w)$ is of degree higher than 16 with respect to $w$. By comparing the coefficients of $w$ on both sides, we get
$$
b_{1}+2 b_{2}=4 \text {, }
$$

thus $\quad b_{2}=1$.
Repeat the above steps. Let $r=w^{2}, f_{2}(r)=f_{1}(w) f_{1}(-w)$, then we get
$$
\begin{array}{l}
f_{2}(r)=(1-r)^{b_{1}+2 b_{2}+4 b_{4}} \cdot\left(1-r^{2}\right)^{4 b_{8}} \cdot\left(1-r^{3}\right)^{b_{3}+2 b_{6}+4 b_{12}} \cdot(1- \\
\left.r^{4}\right)^{4 b_{16}}\left(1-r^{5}\right)^{b_{5}+2 b_{10}+4 b_{20}} \cdot\left(1-r^{6}\right)^{4 b_{24}} \cdot(1- \\
\left.r^{7}\right)^{b} 7^{+2 b_{14}+4 b_{28}}\left(1-r^{8}\right)^{4 b_{32}} \ldots \\
=1-16 r+g_{3}(r) \text {, } \\
\end{array}
$$

where each term of $g_{3}(r)$ is of degree higher than 8 with respect to $r$. By comparing the coefficients of $r$ on both sides, we get
$$
b_{1}+2 b_{2}+4 b_{4}=16 \text {. }
$$

Given $b_{1}=2, b_{2}=1$, we have $b_{4}=3$.
Continuing the above process, we can sequentially obtain $b_{8}=30, b_{16}=2^{12}-2^{4}=4080$. Finally, from
$$
b_{1}+2 b_{2}+4 b_{4}+8 b_{8}+16 b_{16}+32 b_{32}=2^{32} \text {, }
$$

we get $\quad b_{32}=\frac{2^{32}-2^{16}}{32}=2^{27}-2^{11}$.