55rd Ukrainian National Mathematical Olympiad - Third Round

Without loss of generality assume $x_1,x_2,\dots ,x_{2015}<0$. Vieta's theorem implies $x_1{x}_2{x}_3\dots x_{2015}{x}_{2016}=1$, that is, $x_{2016}$ is also a real negative number. Again, by Vieta's theorem: $$\{\begin{array}{l}{x}_1+x_2+\cdots +x_{2016}=-2016,\\ x_1{x}_2\dots x_{2016}=1,\end{array}\Rightarrow \{\begin{array}{l}|x_1|+|x_2|+\cdots +|x_{2016}|=2016,\\ |x_1|\cdot |x_2|\cdot \cdots \cdot |x_{2016}|=1.\end{array}$$ From the inequality between arithmetic and geometric means: $$1=\frac{|x_1|+|x_2|+\cdots +|x_{2016}|}{2016}\ge 2016\sqrt[2016]{|x_1|\cdot |x_2|\cdot \cdots \cdot |x_{2016}|}=1.$$ The equality implies $$|x_1|=|x_2|=\cdots =|x_{2016}|=1,\text{ or }{x}_1=x_2=\cdots =x_{2016}=-1,$$ thus, $P(x)=(x+1{)}^{2016}$, and $a_k=C_{2016}^k$, $k=0,\dots ,2016$.