Newton or Vieta?

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Since $r_k$ , for $k=1,2,3, ... , 20$ , is a root of $z^{20}-19z+2 = 0$ , then $r_k^{20}-19r_k+2 = 0$ , $\implies z^{20} = 19z - 2$ . Therefore, we have:

$\begin{aligned} \sum_{k=1}^{20} r_k^{20} & = \sum_{k=1}^{20} (19r_k - 2) \\ & = 19 \color{#3D99F6}{\sum_{k=1}^{20} r_k} - 2 \sum_{k=1}^{20} 1 \quad \quad \small \color{#3D99F6}{\text{Vieta's formula: }\sum_{k=1}^{20} r_k = 0} \\ & = \boxed{-40} \end{aligned}$

Define the following terms

$S_1 = r_1+r_2+\ldots+r_{20}\\ S_2 = r_1r_2+r_2r_3+\ldots+r_{19}r_{20}\\ S_3=r_1r_2r_3+r_1r_2r_4+\ldots+r_{18}r_{19}r_{20}\\ \vdots\\ S_{20}=r_1r_2r_3\ldots r_{18}r_{19}r_{20}$

From Vieta's formula, we know that:

$\color{#3D99F6}{S_1=S_2=\ldots=S_{18}=0}\\ \color{#D61F06}{S_{19}=19}\\ \color{#EC7300}{S_{20}=2}$

Next, we define

$P_1 = r_1+r_2+\ldots+r_{20}\\ P_2 = r_1^2+r_2^2+\ldots+r_{20}^2\\ P_3 = r_1^3+r_2^3+\ldots+r_{20}^3\\ \vdots\\ P_{20} = r_1^{20}+r_2^{20}+\ldots+r_{20}^{20}$

We are looking for the value of $P_{20}$ , and it can be calculated as:

$P_{20} = \color{#3D99F6}{S_1}P_{19}-\color{#3D99F6}{S_2}P_{18}+\color{#3D99F6}{S_3}P_{17}-\color{#3D99F6}{S_4}P_{16}+\color{#3D99F6}{S_5}P_{15}-\color{#3D99F6}{S_6}P_{14}+\color{#3D99F6}{S_{7}}P_{13}-\color{#3D99F6}{S_{8}}P_{12}+\color{#3D99F6}{S_{9}}P_{11}-\color{#3D99F6}{S_{10}}P_{10}\\+\color{#3D99F6}{S_{11}}P_{9}-\color{#3D99F6}{S_{12}}P_{8}+\color{#3D99F6}{S_{13}}P_{7}-\color{#3D99F6}{S_{14}}P_{6}+\color{#3D99F6}{S_{15}}P_{5}-\color{#3D99F6}{S_{16}}P_{4}+\color{#3D99F6}{S_{17}}P_{3}-\color{#3D99F6}{S_{18}}P_{2}+\color{#D61F06}{S_{19}}P_{1}-20\color{#EC7300}{S_{20}}$

Notice that the first $18$ terms are all $0$ . This leaves us with

$P_{20} = \color{#D61F06}{S_{19}}P_{1}-20\color{#EC7300}{S_{20}}$

Note that $P_1=\color{#3D99F6}{S_1}=0$ . Therefore,

$P_{20} = \color{#D61F06}{S_{19}}\color{#3D99F6}{S_1}-20\color{#EC7300}{S_{20}}\\ =\color{#D61F06}{19}(\color{#3D99F6}{0})-20(\color{#EC7300}{2})\\ =\boxed{-40}$