## Problem 5

Let $p(x)$ be the polynomial $(1-x)^{a}\left(1-x^{2}\right)^{b}\left(1-x^{3}\right)^{c} \ldots\left(1-x^{32}\right)^{k}$, where $a, b, \ldots, k$ are integers. When expanded in powers of $x$, the coefficient of $x^{1}$ is -2 and the coefficients of $x^{2}, x^{3}, \ldots, x^{32}$ are all zero. Find $\mathrm{k}$.


## Solution

Answer: $2^{27}-2^{11}$

We have $p(x)=1-2 x+O\left(x^{33}\right)$. Hence $p(-x)=1+2 x+O\left(x^{33}\right)$. Multiplying $p(x) p(-x)=1-2^{2} x^{2}+O\left(x^{33}\right)$. Now $p(x) p(-x)$ cannot have any odd terms, so we can write it as a polynomial in $x^{2}, q\left(x^{2}\right)$. Hence $q\left(x^{2}\right)=1-$ $2^{2} x^{2}+O\left(x^{34}\right)$. Similarly, $r\left(x^{4}\right)=q\left(x^{2}\right) q\left(-x^{2}\right)=1-2^{4} x^{4}+O\left(x^{36}\right), s\left(x^{8}\right)=r\left(x^{4}\right) r\left(-x^{4}\right)=1-2^{8} x^{8}+O\left(x^{40}\right)$, and $t\left(x^{16}\right)=1-2^{16} x^{16}+O\left(x^{48}\right)$.

Now go back to the definition of $p(x)$. When we take $p(x) p(-x)$, the term $(1-x)^{a}$ becomes $\left(1-x^{2}\right)^{a}$. All the even terms just double their exponent, so $\left(1-x^{2}\right)^{b}$ becomes $\left(1-x^{2}\right)^{2 b},\left(1-x^{4}\right)^{d}$ becomes $\left(1-x^{4}\right)^{2 d}$ and so on. The odd terms all keep the same exponent, so $\left(1-\mathrm{x}^{3}\right)^{\mathrm{c}}$ becomes $\left(1-\mathrm{x}^{6}\right)^{\mathrm{c}}$ and so on. Thus we get $\mathrm{t}\left(\mathrm{x}^{16}\right)=(1-$ $\left.x^{16}\right)^{n}\left(1-x^{32}\right)^{16 k} \ldots$. The first exponent is a sum of several exponents from $p(x)$, but the details are unimportant. We know that $\mathrm{t}\left(\mathrm{x}^{16}\right)=1-2^{16} \mathrm{x}^{16}+\mathrm{O}\left(\mathrm{x}^{48}\right)$. The $\mathrm{x}^{16}$ term can only come from $\left(1-\mathrm{x}^{16}\right)^{\mathrm{n}}$, so $\mathrm{n}=2^{16}$. Now there is no $\mathrm{x}^{32}$ term, so putting $\mathrm{N}=2^{16}$ we have $\mathrm{NC} 2=16 \mathrm{k}$, were $\mathrm{NC} 2$ is the binomial coefficient $\mathrm{N}(\mathrm{N}-$ $1) / 2=2^{31}-2^{15}$. Hence $\mathrm{k}=2^{27}-2^{11}$.

## 18th USAMO 1989

## 19th USAMO 1990

