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2023 Algebra NT #10 \zeta= e^{2\pi i/99}

Source:

February 28, 2024

algebracomplex numbers

Problem Statement

Let

\zeta= e^{2\pi i/99}

and

\omega e^{2\pi i/101}

. The polynomial

x^{9999} + a_{9998}x^{9998} + ...+ a_1x + a_0

has roots

\zeta^m + \omega^n

for all pairs of integers

(m, n)

with

0 \le m < 99

and

0 \le n < 101

. Compute

a_{9799} + a_{9800} + ...+ a_{9998}