A polynomial product of the form
$$(1 – z)^{b_1} (1 – z^2)^{b_2} (1 – z^3)^{b_3} (1 – z^4)^{b_4} (1 – z^5)^{b_5} \dotsm (1 – z^{32})^{b_{32}},$$where the $b_k$ are positive integers, has the surprising property that if we multiply it out and discard all terms involving $z$ to a power larger than 32, what is left is just $1 – 2z.$ Determine $b_{32}.$
Your answer can be in exponential notation.
I tried letting this polynomial equal to $z^{32}q(z) + 1-2z$, but I don't see how this can help.
I also tried to solve simpler versions of the problem, such as finding $b_2$ if the product was $(1-z)^{b_1}(1-z^2)^{b_2}$. I found that the product $(1-z)^2(1-z^2)^1$ works. I then tried to solve the problem when the product was $(1-z)^{b_1}(1-z^2)^{b_2}(1-z^3)^{b_3}$. However, this got quite confusing, and I'm not sure what to do other than randomly test numbers. Now I'm not sure what I should do next.
Thanks in advance!!!
Best Answer
It turns out that, for any $n\in\mathbb{Z}^+$ and $a\in\mathbb{Z}$, the unique solution $(b_1,\ldots,b_n)\in\mathbb{Z}^n$ to $$\prod_{k=1}^{n}(1-z^k)^{b_k}\equiv 1-az\pmod{z^{n+1}}\tag{1}\label{eqtn}$$ is given by $$b_k=\frac1k\sum_{d\,\mid\,k}\mu(d)a^{k/d}\tag{2}\label{sltn}$$ using the Möbius function $\mu$; note the independence on $n$. Thus the answer is $\color{blue}{2^{11}(2^{16}-1)}$.
Here is a not-very-elementary proof of \eqref{sltn} using formal power series.
Let $\ell(z)=-\log(1-z)=\sum_{n=1}^\infty z^n/n$; then \eqref{eqtn} implies $$\sum_{k=1}^n b_k\ell(z^k)\equiv\ell(az)\pmod{z^{n+1}},$$ that is $$\sum_{m=1}^n\frac{a^m}{m}z^m=\sum_{k=1}^n b_k\sum_{d=1}^{\lfloor n/k\rfloor}\frac{z^{kd}}{d}=\sum_{m=1}^n z^m\sum_{d\,\mid\,m}\frac{b_{m/d}}{d},$$ i.e. $a^m=\sum_{d\,\mid\,m}(m/d)b_{m/d}=\sum_{d\,\mid\,m}db_d$, and \eqref{sltn} follows by Möbius inversion.