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# 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
3 years, 1 month ago

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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 $$(x^4-1)\mid f(x)$$. · 3 years, 1 month ago

But Cody goes first. · 3 years, 1 month ago

Hardest question in the entire contest! :'( · 3 years, 1 month ago

Hint: look at the roots of $$m(x)$$. · 3 years, 1 month ago