XIX Silk Road Mathematical Competition

Note that if for some real numbers $x_1,x_2,\dots ,x_m$ the following equality holds: $$|x_1|+|x_2|+\cdots +|x_m|=|x_1+x_2+\cdots +x_m|,$$ then they are of the same sign. Let $$Q(x)=P(x)(x-1{)}^t(x+1{)}^s=b_n{x}^n+b_{n-1}{x}^{n-1}+\cdots +b_0,$$ where $b_n=a_d>0$. From the problem statement it follows that $$|b_0|=|b_1|+|b_2|+\cdots +|b_n|.$$ Lemma: If $t\ge 1$, then $b_1,b_2,\dots ,b_n\ge 0$. Proof: $$\begin{array}{rl}Q(1)=0& ⟹b_0+b_1+\cdots +b_n=0\\ & ⟹|b_1+b_2+\cdots +b_n|=|b_0|=|b_1|+|b_2|+\cdots +|b_n|,\end{array}$$ hence, $b_1,b_2,\dots ,b_n$ are of the same sign. Since $b_n>0$, then $b_1,b_2,\dots ,b_{n-1}\ge 0$. We proved the lemma. Assume that $t\ge 2$. According to the lemma, $$b_1,b_2,\dots ,b_{n-1}\ge 0⟹b_1+2b_2+\cdots +n{b}_n>0.$$ On the other hand, let $R(x)=\frac{Q(x)}{(x-1{)}^2}$. Then $$Q^{\prime }(x)=2(x-1)R(x)+(x-1{)}^2{R}^{\prime }(x)⟹Q^{\prime }(1)=0⟹b_1+2b_2+\cdots +n{b}_n=0$$ — a contradiction. Thus, $t\le 1$.

Similarly, we can show that $s\le 1$. Let's consider three cases.

I) $t=1,s=1$ $$\begin{array}{rl}Q(x)=P(x)(x^2-1)& ⟹Q(1)=Q(-1)=0\\ & ⟹b_0+b_1+\cdots +b_n=b_0-b_1+b_2+\cdots +(-1{)}^n{b}_n=0\\ & ⟹b_1+b_2+b_3+\cdots =0.\end{array}$$ According to the lemma, $$b_1,b_2,\dots ,b_n\ge 0⟹b_1=b_2=b_3=\cdots =0,$$ — a contradiction, since $b_1=-a_1$ and by the problem statement $a_1\ne 0$.

II) $t=1,s=0$ $$Q(x)=P(x)(x-1)=-a_0+(a_0-a_1)x+\cdots +(a_{d-1}-a_d)x^d+a_d{x}^{d+1}.$$ By the lemma, $$a_0-a_1=b_1\ge 0,\dots ,a_{d-1}-a_d=b_d\ge 0⟹a_0\ge a_1\ge \cdots \ge a_d.$$

Therefore, $P(x)$ is non-increasing.

III) $t=0,s=1$. $$\begin{gathered} Q(-1)=0⟹b_0-b_1+\cdots +(-1{)}^n{b}_n=0⟹ \\ ⟹|b_1-b_2+\cdots +(-1{)}^n{b}_n|=|b_0|=|b_1|+|-b_2|+\cdots +|(-1{)}^n{b}_n|. \end{gathered}$$ Thus, $b_1,-b_2,\dots ,(-1{)}^n{b}_n$ are of the same sign, and since $b_n>0$, then $(-1{)}^{n-i}{b}_i\ge 0$ for each $1\le i\le n$. $$\begin{gathered} Q(x)=P(x)(x+1)=a_0+(a_0+a_1)x+\cdots +(a_{d-1}+a_d)x^d+a_d{x}^{d+1}⟹ \\ ⟹(-1{)}^{d+1-i}(a_{i-1}+a_i)\ge 0\text{ (for all }1\le i\le d)⟹ \\ ⟹a_d\le -a_{d-1}\le a_{d-2}\le \cdots \le (-1{)}^d{a}_0. \end{gathered}$$ It follows that $(-1{)}^{d}P(-x)=a_d{x}^d-a_{d-1}{x}^{d-1}+\cdots +(-1{)}^d{a}_0$ is non-increasing.