jmc

Let $g(z)=(1-z{)}^{b_1}(1-z^2{)}^{b_2}(1-z^3{)}^{b_3}(1-z^4{)}^{b_4}(1-z^5{)}^{b_5}\cdots (1-z^{32}{)}^{b_{32}}.$ Since $g(z)$ reduces to $1-2z$ if we eliminate all powers of $z$ that are $z^{33}$ or higher, we write $$g(z)\equiv 1-2z(\mathrm{mod}{z}^{33}).$$Then $$\begin{array}{rl}g(-z)& =(1+z{)}^{b_1}(1-z^2{)}^{b_2}(1+z^3{)}^{b_3}(1-z^4{)}^{b_4}(1+z^5{)}^{b_5}\cdots (1-z^{32}{)}^{b_{32}}\\ & \equiv 1+2z(\mathrm{mod}{z}^{33}),\end{array}$$so $$\begin{array}{rl}g(z)g(-z)& =(1-z^2{)}^{b_1+2b_2}(1-z^4{)}^{2b_4}(1-z^6{)}^{b_3+2b_6}(1-z^8{)}^{2b_8}\cdots (1-z^{30}{)}^{b_{15}+2b_{30}}(1-z^{32}{)}^{2b_{32}}\\ & \equiv (1+2z)(1-2z)\equiv 1-2^2{z}^2(\mathrm{mod}{z}^{33}).\end{array}$$Let $g_1(z^2)=g(z)g(-z),$ so $$\begin{array}{rl}{g}_1(z)& =(1-z{)}^{c_1}(1-z^2{)}^{c_2}(1-z^3{)}^{c_3}(1-z^4{)}^{c_4}\cdots (1-z^{16}{)}^{c_{16}}\\ & \equiv 1-2^{2}z(\mathrm{mod}{z}^{17}),\end{array}$$where $c_i=b_i+2b_{2i}$ if $i$ is odd, and $c_i=2b_{2i}$ if $i$ is even. In particular, $c_{16}=2b_{32}.$

Then $$\begin{array}{rl}{g}_1(z)g_1(-z)& =(1-z^2{)}^{c_1+2c_2}(1-z^4{)}^{2c_4}(1-z^6{)}^{c_3+2c_6}(1-z^8{)}^{2c_8}\cdots (1-z^{14}{)}^{c_7+2c_{14}}(1-z^{16}{)}^{2c_{16}}\\ & \equiv (1-2^{2}z)(1+2^{2}z)\equiv 1-2^4{z}^2(\mathrm{mod}{z}^{17}).\end{array}$$Thus, let $g_2(z^2)=g_1(z)g_1(-z),$ so $$\begin{array}{rl}{g}_2(z)& =(1-z{)}^{d_1}(1-z^2{)}^{d_2}(1-z^3{)}^{d_3}(1-z{)}^{d_4}\cdots (1-z^7{)}^{d_7}(1-z^8{)}^{d_8}\\ & \equiv 1-2^{4}z(\mathrm{mod}{z}^9),\end{array}$$where $d_i=c_i+2c_{2i}$ if $i$ is odd, and $d_i=2c_{2i}$ if $i$ is even. In particular, $d_8=2c_{16}.$

Similarly, we obtain a polynomial $g_3(z)$ such that $$g_3(z)=(1-z{)}^{e_1}(1-z^2{)}^{e_2}(1-z^3{)}^{e_3}(1-z{)}^{e_4}\equiv 1-2^{8}z(\mathrm{mod}{z}^5),$$and a polynomial $g_4(z)$ such that $$g_4(z)=(1-z{)}^{f_1}(1-z^2{)}^{f_2}\equiv 1-2^{16}z(\mathrm{mod}{z}^3).$$Expanding, we get $$\begin{array}{rl}{g}_4(z)& =(1-z{)}^{f_1}(1-z^2{)}^{f_2}\\ & =\left(1-f_{1}z+\left(\genfrac{}{}{0ex}{}{f_1}{2}\right)z^2-\cdots \right)\left(1-f_2{z}^2+\cdots \right)\\ & =1-f_{1}z+\left(\left(\genfrac{}{}{0ex}{}{f_1}{2}\right)-f_2\right)z^2+\cdots .\end{array}$$Hence, $f_1=2^{16}$ and $\left(\genfrac{}{}{0ex}{}{f_1}{2}\right)-f_2=0,$ so $$f_2=\left(\genfrac{}{}{0ex}{}{f_1}{2}\right)=\left(\genfrac{}{}{0ex}{}{2^{16}}{2}\right)=\frac{2^{16}(2^{16}-1)}{2}=2^{31}-2^{15}.$$We have that $f_2=2e_4=4d_8=8c_{16}=16b_{32},$ so $$b_{32}=\frac{f_2}{16}=\boxed{2^{27}-2^{11}}.$$We leave it to the reader to find a polynomial that actually satisfies the given condition.