$5 \cdot 34$ 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 removed, it leaves exactly $1-2 z$. Determine $b_{32}$ (the answer can be expressed as the difference of two powers of 2).
\[ \text{Solution: Let } f(z)=(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}} \cdots\left(1-z^{32}\right)^{b_{32}} = 1-2 z+g(z) \]
where every term in $g(z)$ is of degree higher than 32.
By comparing the coefficients of $z$ on both sides, we get $b_{1}=2$.
\[ \begin{aligned}
f(-z)= & (1+\approx)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1+z^{3}\right)^{b_{3}} \cdots\left(1-z^{32}\right)^{b_{32}}=1+2 z+ \\
& g(-z).
\end{aligned} \]
\[ \begin{aligned}
f(z) \cdot f(-z)= & \left(1-z^{2}\right)^{b_{1}+2 b_{2}} \cdot\left(1-z^{4}\right)^{2 b_{4}} \cdot\left(1-z^{6}\right)^{b_{3}+2 b_{6}} \cdot(1- \\
& \left.z^{8}\right)^{2 b_{8}} \cdot\left(1-z^{10}\right)^{b_{5}+2 b_{10}} \cdots \cdot\left(1-z^{30}\right)^{b_{15}+2 b_{30}} \cdot(1 \\
& \left.-z^{32}\right)^{2 b_{32}} \cdots \\
= & 1-4 z^{2}+g_{1}(z),
\end{aligned} \]

where every term in $g_{1}(z)$ is of degree higher than 32 and is even.
Let $w=z^{2}, f_{1}(w)=f(z) \cdot f(-z)$, then
\[ \begin{array}{l}
f_{1}(w)=(1-w)^{b_{1}+2 b_{2}}\left(1-w^{2}\right)^{2 b_{4}}\left(1-w^{3}\right)^{b_{3}+2 b_{6}}\left(1-w^{4}\right)^{2 b_{8}}(1-
\end{array} \]
\[ \begin{array}{l}
=1-4 w+g_{2}(w),
\end{array} \]

where every term in $g_{2}(w)$ is of degree higher than 16.
By comparing the coefficients of $w$ on both sides, we get
\[ b_{1}+2 b_{2}=4 \]

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

where every term in $g_{3}(r)$ is of degree higher than 8.
By comparing the coefficients of $r$ on both sides, we get
\[ b_{1}+2 b_{2}+4 b_{4}=16 \]

From $b_{1}=2, b_{2}=1$, we get $b_{4}=3$.
Continuing the above process, we can sequentially get $b_{8}=30, b_{16}=2^{12}-2^{4}=4080$. Finally, from
\[ b_{1}+2 b_{2}+4 b_{4}+8 b_{8}+16 b_{16}+32 b_{32}=2^{32} \]

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