AMC12 2021 12B, Problem 20

Let $Q(z)$ and $R(z)$ be the unique polynomials such that\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]and the degree of $R$ is less than $2.$ What is $R(z)?$

Solution: Let $z=s$ be a root of $z^2+z+1$ so that $s^2+s+1=0.$ It follows that \[(s-1)\left(s^2+s+1\right)=s^3-1=0,\] from which $s^3=1,$ but $s\neq1.$ Note that \begin{align} s^{2021}+1 &= s^{3\cdot673+2}+1 \\ &= (s^3)^{673}\cdot s^2+1 \\ &= s^2+1 \\ &= \left(s^2+s+1\right)-s \\ &= -s. \end{align} Since $z^{2021}+1=-z$ for each root $z=s$ of $z^2+z+1,$ the remainder when $z^{2021}+1$ is divided by $z^2+z+1$ is $R(z)=\fbox{-z}.$

Answer: -z