Polynomial Game - Problem 4

Consider a polynomial f(x)=x2014+a2013x2013++a1x+a0f(x)=x^{2014}+a_{2013}x^{2013}+\dots+a_1x+a_0. Cody and Pi Han take turns choosing one of the coefficients a0,a1,,a2013a_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)f(x) divisible by a fixed polynomial m(x)m(x), and Cody's goal is to prevent this.

  • Which of the players has a strategy if m(x)=x2+x+1m(x)=x^2+x+1 and Cody goes first?
  • Which of the players has a strategy if m(x)=x41m(x)=x^4-1 and Cody goes first?

Note by Cody Johnson
5 years, 7 months ago

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Hardest question in the entire contest! :'(

Shaun Loong - 5 years, 7 months ago

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

Cody Johnson - 5 years, 7 months ago

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

Calvin Lin Staff - 5 years, 7 months ago

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By considering unassigned coefficients as they were zero, the polynomial f(x)f(x) changes move-by-move. For the first point, Pi Han's goal is to have f(ω)=0f(\omega)=0, with ω\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, argf(ω)\arg f(\omega) is not a multiple of π/3\pi/3. For the second point, if Cody plays in such a way that, just after his turns, argf(i)\arg f(i) is not a multiple of π/4\pi/4, Pi Han's doomed again, since f(i)=0f(i)=0 is a necessary condition in order to have (x41)f(x)(x^4-1)\mid f(x).

Jack D'Aurizio - 5 years, 7 months ago

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But Cody goes first.

Cody Johnson - 5 years, 7 months ago

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