66th Belarusian Mathematical Olympiad

6. Answer : b) no, it does not. a) Let $b_0\le b_1\le \cdots \le b_{2n}$ denote the coefficients $a_0,a_1,\dots ,a_{2n}$ of $p(x)$, which are arranged in non-decreasing order. Consider the permutation $c_0,c_1,\dots ,c_{2n}$ of the numbers $a_0,a_1,\dots ,a_{2n}$ such that $c_{2n}=b_{2n},c_{2n-2}=b_{2n-1},\dots ,c_0=b_n$ and $c_{2n-1}=b_{n-1},c_{2n-3}=b_{n-2},\dots ,c_1=b_0$. For the set of $c_0,c_1,\dots ,c_{2n}$ we have $$c_0\ge c_1\le c_2\ge c_3\le \cdots \le c_{2n-2}\ge c_{2n-1}\le c_{2n}.$$ Show that the polynomial $q(x)=c_{2n}{x}^{2n}+c_{2n-1}{x}^{2n-1}+\cdots +c_{1}x+c_0$ has no real roots. The coefficients of $q(x)$ are positive, so $$q(x)>0(1)$$ for $x\ge 0$. Show that inequality (1) holds for negative $x$, too. Consider two cases: 1) $-1\le x<0$ and 2) $x<-1$. In case 1) we have $|x{|}^{2k+1}\le x^{2k}$ for $k\in \{0,1,\dots ,n-1\}$, hence, $c_{2k+1}{x}^{2k+1}+c_{2k}{x}^{2k}\ge (c_{2k}-c_{2k+1})|x{|}^{2k}\ge 0$. Therefore, $$q(x)=c_{2n}{x}^{2n}+(c_{2n-1}{x}^{2n-1}+c_{2n-2}{x}^{2n-2})+\cdots +(c_{1}x+c_0)\ge c_{2n}{x}^{2n}>0.$$ In case 2) we have $|x{|}^{2k}>|x{|}^{2k-1}$, hence, $$c_{2k}{x}^{2k}+c_{2k-1}{x}^{2k-1}>(c_{2k}-c_{2k-1})x^{2k}\ge 0$$ for $k\in \{1,2,\dots ,n\}$. Therefore, $$q(x)=(c_{2n}{x}^{2n}+c_{2n-1}{x}^{2n-1})+\cdots +(c_2{x}^2+c_{1}x)+c_0>c_0>0.$$ Thus, $q(x)>0$ for all real $x$, so $q(x)$ has no real roots.

b) Consider the polynomial $p(x)=x^{2n}+x^{2n-1}+\cdots +x-2n$. The sum of the coefficients of $p(x)$ is equal to 0, i.e., $p(1)=0$. Therefore, for any permutation $c_0,c_1,\dots ,c_{2n}$ of the numbers $-2n,1,\dots ,1$ we have $q(1)=0$, where $q(x)=c_{2n}{x}^{2n}+c_{2n-1}{x}^{2n-1}+\cdots +c_{1}x+c_0$. It follows that 1 is a root of the polynomial $q(x)$.