HMMT-Feb 2015 guts, Problem 10

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 \begin{itemize} $N$ 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). \end{itemize} Find $a_{N}(0)$ (the "leading digit evaluated at 0 ").

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 \[ \begin{aligned} a_{N}(x)\left(x^{2}+x+1\right)^{1007} & \in(u x+v)\left(x^{2014}+1007 x^{2013}+O\left(x^{2012}\right)\right) \\ & \subseteq u x^{2015}+(v+1007 u) x^{2014}+O\left(x^{2013}\right) \end{aligned} \] Finally, matching terms gives $u=1$ and $v+1007 u=1$, so $v=1-1007=-1006$. Remark. This problem illustrates the analogy between integers and polynomials, with the nonconstant (degree $\geq 1$ ) polynomial $b(x)=x^{2}+x+1$ taking the role of a positive integer base $b>1$. $\fbox{-1006}$.

Answer: -1006