HMMT February

Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:
Let $b(x) = x^{2} + x + 1$ . The polynomial $x^{2015} + x^{2014} + \cdots + x + 1$ has a unique "base $b(x)$ " representation
$x^{2015} + x^{2014} + \cdots + x + 1 = \sum_{k=0}^{N} a_{k}(x) b(x)^{k}$
where
- $N$ is a nonnegative integer;
- each "digit" $a_{k}(x)$ (for $0 \leq k \leq N$ ) is either the zero polynomial (i.e. $a_{k}(x) = 0$ ) or a nonzero polynomial of degree less than $\operatorname{deg} b = 2$ ; and
- the "leading digit $a_{N}(x)$ " is nonzero (i.e. not the zero polynomial).
Find $a_{N}(0)$ (the "leading digit evaluated at 0").

Solution

Solution:
Answer: $-1006$

Comparing degrees easily gives $N = 1007$ . By ignoring terms of degree at most $2013$ , we see
$a_{N}(x)\left(x^{2} + x + 1\right)^{1007} \in x^{2015} + x^{2014} + O\left(x^{2013}\right)$
Write $a_{N}(x) = u x + v$ , so
$\begin{aligned} a_{N}(x)\left(x^{2} + x + 1\right)^{1007} & \in (u x + v)\left(x^{2014} + 1007 x^{2013} + O\left(x^{2012}\right)\right) \\ & \subseteq u x^{2015} + (v + 1007 u) x^{2014} + O\left(x^{2013}\right) \end{aligned}$
Finally, matching terms gives $u = 1$ and $v + 1007 u = 1$ , so $v = 1 - 1007 = -1006$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.