37th Iranian Mathematical Olympiad

Lemma. If $1\le k_1<k_2<\cdots <k_t\le n-1$ be natural numbers, then $$\forall x\in \mathbb{R}:x^{2n}+\sum _{i=1}^t(x^{2k_i+1}+x^{2k_i})+1>0.$$ Proof. If $x>0$, it's trivial. So let's assume that $x<0$. If $-1\le x<0$, we write $$x^{2n}+\sum _{i=1}^t(x^{2k_i+1}+x^{2k_i})+1$$ as $$x^{2n}+1+(x+1)\sum _{i=1}^t{x}^{2k_i}.$$ Since $x+1\ge 0$ and ${\sum }_{i=1}^t{x}^{2k_i}\ge 0$, we have $$x^{2n}+1+(x+1)\sum _{i=1}^t{x}^{2k_i}>0.$$ If $x<-1$, then $x+1<0$ and $x^{2k+1}<x<1$. So, $x^{2n+1}(x^{2k+1}+1)>0$. Therefore, we can re-write $$x^{1398}+\sum _{i=1}^t(x^{2k_i+1}+x^{2k_i})+1$$ as $$(x^{1398}-(2k_t+1)+1)x^{2k_t+1}+\sum _{i=1}^{t-1}{x}^{2k_1+1}(x^{2k_{i+1}}-(2k_i+1))+2^{2k_1}+1$$ where $1\le k_1\le k_2\le \cdots \le \frac{1396}{2}$. So, it is positive. $\square$

Back to the problem. When Roozbeh chooses $x^k$, if $k$ is an even number, Keyvan should choose $x^k$ as well. But, if $k$ is an odd number, and if coefficient of $x^k$ is odd, Keyvan should choose $x^{k-1}$ and if coefficient of $x^k$ is even, Keyvan should play $x^{k+1}$. Obviously after Keyvan's move, the polynomial will be in the following form: $$x^{1398}\sum _{i=1}^m(x^{s_i}+x^{s_i-1}{)}^2+2\sum _{i=1}^n{x}^{2t_i}+\sum _{i=1}^l(x^{2k_i+1}+x^{2k_i})+1.$$ Where $$1\le s_1\le s_2\le \cdots \le s_m\le \frac{1398}{2},$$ $$0\le t_1\le t_2\le \cdots \le t_n\le \frac{1398}{2},$$ $$1\le k_1<k_2<\cdots <k_l\le 1398.$$ According to the lemma we proved, the polynomial should be always positive. $■$