LetsSolveMathProblems 77th Problem of the Week

Consider the function $P(x) = (x- \omega)(x- \omega^2)...(x- \omega^{2019})$ . All solutions of the equation $x^{2020} = 1$ except for $x = 1$ are the zeros of $P(x)$ , so $P(x) = \frac{x^{2020}-1}{x-1} = x^{2019} + x^{2018} + ... + 1$ . From the original expression for $P(x)$ , we find that $P'(x) = (x- \omega^2)(x- \omega^3)...(x- \omega^{2019}) + (x- \omega)(x- \omega^3)...(x- \omega^{2019}) + ... + (x- \omega)(x- \omega^2)...(x- \omega^{2018})$ Put concisely, P'(x) = \sum\limits_{n = 1}^{2019} \prod\limits_{j=1}_{j \neq n}^{2019} (x - \omega^j) . Note that when $P'(\omega ^ k)$ is evaluated, all terms of the sum become zero except for the term where $n = k$ . Therefore, P'(\omega^k) = \prod\limits_{j=1}_{j \neq k}^{2019} (\omega^k - \omega^j) . The expression we are solving for is now equal to $-\sum\limits_{k=1}^{2019} P'(\omega^k) = -\sum\limits_{k=1}^{2019} (2019\omega^{2018k} + 2018\omega^{2017k} + ... + 1)$ We can evaluate a similar expression:

$\sum\limits_{k=0}^{2019} (2019\omega^{2018k} + 2018\omega^{2017k} + ... + 1)$ $= 2019 \sum\limits_{k=0}^{2019} \omega^{2018k} + 2018 \sum\limits_{k=0}^{2019} \omega^{2017k} + ... + \sum\limits_{k=0}^{2019} 1$

$\sum\limits_{k=0}^{2019} \omega^{ck}$ is a geometric series that evaluates to $\frac{1-\omega^{2020c}}{1-\omega^c}$ when $c \neq 0$ . Because $\omega^{2020} = 1$ , this equals $\frac{1-1}{1-\omega^c} = 0$ . Therefore, the expression becomes $\sum\limits_{k=0}^{2019} 1 = 2020$ .

Finally, we can evaluate the original expression:

$-\sum\limits_{k=1}^{2019} P'(\omega^k) = - (-P'(\omega^0) + \sum\limits_{k=0}^{2019} P'(\omega^k))$ $= P'(1) - 2020 = 2019 + 2018 + ... + 2 + 1 - 2020 = (1010)(2019) - 2020$ $\equiv (10)(19) - 20 \pmod{1000} = \boxed{170}$ .