Polynomial problem - 2020

Since we want $P(\omega) = 0$ we must have $n \equiv 2 \pmod{3}$ , and so $n=2021$ and then $(x^3-1)Q(x) \; = \; (x-1)P(x) \; = \; (x - 1)(x^{2021} + x^{2020} + 1) \; = \; x^{2022} - x^{2020} + x - 1 \; =\; (x^3-1)\left(\sum_{n=0}^{673}x^{3n} - x\sum_{n=0}^{672}x^{3n}\right)$ so that $Q(x) \; = \; \sum_{n=0}^{673}x^{3n} - x\sum_{n=0}^{672}x^{3n}$ and hence $Q(\omega) \; = \; 674 - 673\omega$ so that $\mathrm{Re}\big[Q(\omega)\big] \; =\; 674 + \tfrac{673}{2}$ making the answer $2 \times 674 + 673 - 2 = \boxed{2019}$ .

Let us define the sequence of polynomials $P_n(x)=x^n+x^{n-1}+1$ . It is easy to see that $P_n(x)=(x^2+x+1)(x^{n-2}-x^{n-4})+P_{n-3}(x), \quad n\geq 5.\quad (*)$ From the previous recurrence relation, we have that $(x^2+x+1)| P_n(x)$ if and only if $(x^2+x+1)|P_{n-3}(x)$ for any integer $n\geq 5.$ Notice that $P_2(x)=x^2+x+1,$ and then it is divisible by $(x^2+x+1),$ but $P_3(x)$ and $P_4(x)$ are not divisible by $(x^2+x+1).$ Therefore, $(x^2+x+1)$ is a factor of $P_n(x)$ if and only if $n=3k+2,$ where $k$ is any non-negative integer. Then $P(x)=P_{2021}(x).$ Now for any $n$ integer greater than or equal to 2 we define $Q_n(x)=\frac{P_n(x)}{x^2+x+1}.$ Obviously, $Q(x)=Q_{2021}(x)$ and from the recurrence relation (*), we get that $Q_n(x)=x^{n-2}-x^{n-4}+Q_{n-3}(x),$
where $n$ is any integer number that is greater than or equal to 5. Then we have the following

$Q_5 (x)=x^3-x+Q_2(x)$ $Q_8(x)=x^6-x^4+Q_5(x)$ $Q_{11} (x)=x^9-x^7+Q_8(x)$ $...................................................$ $Q_{2021} (x)=x^{2019}-x^{2017}+Q_{2018}(x)$ Adding all these equalities and cancelling repeated terms, we get that $Q_{2021}(x)=(x^3+x^6+...+x^{2019})-x(1+x^3+...+x^{2016})+Q_2(x).$ Evaluating the last formula at $-\frac{1}{2}+i\frac{\sqrt{3}}{2}=e^{\frac{2\pi}{3}i},$ using that $(e^{\frac{2\pi}{3}i})^{3k}=1,$ and that $Q_2(e^{\frac{2\pi}{3}i})=1,$ we can conclude that $Q_{2021}(-\frac{1}{2}+i\frac{\sqrt{3}}{2})=674-673(-\frac{1}{2}+i\frac{\sqrt{3}}{2})=1010.5-673i\frac{\sqrt{3}}{2}.$ Then the answer for this question is $\boxed{2019}.$

Once you know that $P(x) = x^{2021} + x^{2020} + 1$ and realize that plugging in $\omega$ to the factorization gives $0 = 0$ , it is also possible to tease out $Q(\omega)$ using an algebraic derivative (which follows all the same power rules from calculus and are valid via complex analysis; FToA gets its proof there anyway) on the polynomial $P$ : $2021\omega^{2020} + 2020\omega^{2019} = (2\omega + 1)Q(\omega) + \underbrace{(\omega^2 + \omega + 1)}_{0}Q'(\omega)$ . From here it follows that $Q(\omega) = \dfrac{2021\omega + 2020}{2\omega + 1}$ , rationalize, take the real part to get the answer to the question which is $2019$ .