Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Find the answer

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 each "digit" $a_{k}(x)$ is either the zero polynomial or a nonzero polynomial of degree less than $\operatorname{deg} b=2$ ; and the "leading digit $a_{N}(x)$ " is nonzero. Find $a_{N}(0)$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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 $a_{N}(x)\left(x^{2}+x+1\right)^{1007} \in(u x+v)\left(x^{2014}+1007 x^{2013}+O\left(x^{2012}\right)\right) \subseteq u x^{2015}+(v+1007 u) x^{2014}+O\left(x^{2013}\right)$ . Finally, matching terms gives $u=1$ and $v+1007 u=1$ , so $v=1-1007=-1006$ .

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