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 $R(z)?$

Question

Problem Statement : 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 $R(z)=-z.$

My quesitons:

The solution shows the following: If $s$ is a root of $s^2+s+1$, then $R(s)=-s$. But how does this imply that $R(z)=-z$ for all $z$, instead of just those $z$ for which $z^2+z+1=0$?

Answer 1

Note that $R(z)$ is a polynomial of degree $1$ or less. So then is $R(z)+z$. Also note that a polynomial $p$ of degree $d$, has at most $d$ distinct roots, or is $0$ everywhere. For example, if say a polynomial $p(y)$ of the form $p(y)=ay+b$ is $0$ for $2$ distinct values of $y$, then both $a$ and $b$ must be $0$ and so $p(y)$ must be $0$ for all values of $y$.

So if $R(s)+s$ is $0$ for more than deg$(R(z)+z) =1$ values of $s$, [and it is for the $2$ roots $s$ of $z^2+z+1$], then as no nozero polynomial of degree $1$ can have $2$ roots, the only possibility is that $R(z)+z$ must be $0$ everywhere, or equivalently, $R(z)$ must be $-z$ everywhere.