New Year's Algebra

Let $f(x)=g(x)+h(x)$ , where $g(x)$ contains only the odd powers of $x$ in $f(x)$ and $h(x)$ contains only the even powers of $x$ in $f(x)$ . $\Rightarrow f(-x)=-g(x)+h(x)$ .

Plugging this into the functional equation:

$g(x)+4033h(x)=4033\left( 1+x+x^{2}+x^{3}+...+x^{2016}+x^{2017} \right)$

We can now assign the even/odd powers of $x$ to $g(x)$ and $h(x)$ based on their definitions:

$\begin{array}{c}&g(x)=4033\left( x+x^{3}+x^{5}+...+x^{2015}+x^{2017} \right) \\ h(x)=1+x^{2}+x^{4}+x^{6}+...+x^{2014}+x^{2016} \end{array}$

Solving for $f(x):$

$\begin{array}{c}& f(x)=4033 \left( x+x^{3}+x^{5}+...+x^{2015}+x^{2017} \right) + 1+x^{2}+x^{4}+x^{6}+...+x^{2014}+x^{2016} \\ f(x)= \left( 4033x+1 \right) \left(1+x^{2}+x^{4}+x^{6}+...+x^{2014}+x^{2016} \right) \end{array}$

We now have $f(x)$ in the form $f(x)=i(x)j(x)$ where $i(x)=4033x+1$ and $j(x)=1+x^{2}+x^{4}+x^{6}+...+x^{2014}+x^{2016}$

• $i(x) \rightarrow x=-\frac{1}{4033}$ is a root.

• $j(x) \rightarrow j(x)\left( 1-x^{2}\right)=1-x^{2018}$ so the roots of $j(x)$ are the $2018^{th}$ roots of unity excluding $1$ and $-1$ , all of which are non-real, distinct, and sum to $0$ by Vieta's Formula .

$\Rightarrow C=0, R=-\frac{1}{4033} \Rightarrow C-R=\frac{1}{4033} \Rightarrow p+q=\boxed{4034}$