MHB Can Jessica Ensure Divisibility by x^2+1 in the Polynomial Game?

[anemone]

Here is this week's POTW:

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This week's high school POTW uses the same context as last week's problem (http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-188-november-3rd-2015-a-16779.html), but we have cranked up the difficulty level and we hope members will take a stab at it, and we're looking forward to receive your participation!

Consider a polynomial

$$P(x)=a_0+a_1x+\cdots+a_{2011}x^{2011}+x^{2012}$$
Nigel and Jessica are playing the following game. In turn, they choose one of the coefficients $a_0,\,\cdots,\,a_{2011}$ and assign a real value to it. Nigel has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values.

Jessica's goal is to make $P(x)$ divisible by a fixed polynomial $m(x)$ and Nigel's goal is to prevent this.

Which of the players has a winning strategy if $m(x)=x^2+1$?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. Fallen Angel
3. MarkFL

Solution from Fallen Angel:

Spoiler

Let $g(y)=a_{0}+\displaystyle\sum_{k=1}^{1005}a_{2k}y^{k}+y^{1006}$ and $h(y)=a_{1}+\displaystyle\sum_{k=1}^{1005}a_{2k+1}y^{k}$.

This way we have that $f(x)=g(x^{2})+xh(x^{2})$, thus the goal of the second player will be get $g(-1)=h(-1)=0$.

A simple strategy is to follow the other player, this is, if Nigel choose a coefficient of $g$, Jessica can choose a coefficient of $g$, and conversely, if Nigel choose a coefficiente of $h$, Jessica can choose a coefficient of $h$.

Since both $g$ and $h$ has an even number of coefficients Jessica will choose the last coefficient in both cases and it's clear that she can get $h(-1)=g(-1)=0$ just like in the last POTW (by solving a linear equation).