# Polynomial Game - Problem 4

Consider a polynomial $f(x)=x^{2014}+a_{2013}x^{2013}+\dots+a_1x+a_0$. Cody and Pi Han take turns choosing one of the coefficients $a_0,a_1,\dots,a_{2013}$ and assigning a real value to it. Once a value is assigned to a coefficient, it cannot be changed anymore. The game ends after all of the coefficients have been assigned values. Pi Han's goal is to make $f(x)$ divisible by a fixed polynomial $m(x)$, and Cody's goal is to prevent this.

• Which of the players has a strategy if $m(x)=x^2+x+1$ and Cody goes first?
• Which of the players has a strategy if $m(x)=x^4-1$ and Cody goes first? Note by Cody Johnson
5 years, 9 months ago

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Hardest question in the entire contest! :'(

- 5 years, 9 months ago

Hint: look at the roots of $m(x)$.

- 5 years, 9 months ago

I felt that remainder-factor theorem was a recurring theme in the questions about polynomials.

Staff - 5 years, 9 months ago

By considering unassigned coefficients as they were zero, the polynomial $f(x)$ changes move-by-move. For the first point, Pi Han's goal is to have $f(\omega)=0$, with $\omega$ being a third root of unity, with its last move. Cody can prevent this by playing in such a way that, just after his turns, $\arg f(\omega)$ is not a multiple of $\pi/3$. For the second point, if Cody plays in such a way that, just after his turns, $\arg f(i)$ is not a multiple of $\pi/4$, Pi Han's doomed again, since $f(i)=0$ is a necessary condition in order to have $(x^4-1)\mid f(x)$.

- 5 years, 9 months ago

But Cody goes first.

- 5 years, 9 months ago