IMO Team Selection Test 2, June 2020

If $x=\alpha$ is a root of $P_n$, then $x=-\alpha$ is a root of $P_n$ as well, as all terms of $P_n$ have even degree. The minimal root of $P_n$ therefore cannot be positive. Therefore $a_n\le 0$ for all $n>2020$.

We have $P_{n+1}(x)=x^2\cdot P_n(x)+a_{n+1}$. Substitute $x=a_{n+1}$; as that is a root of $P_n$, we have $P_{n+1}(a_{n+1})=0+a_{n+1}\le 0$.

As the maximal degree term in $P_n(x)$ is $x^{2n}$, there exists an $N<0$ such that $P_n(x)>0$ for all $x<N$. Taking for example $-N=\max (2,|a_1|+|a_2|+\cdots +|a_n|)$, we see for $x<N$ that $x^{2i-2}\le x^{2n-2}$ for all $1\le i\le n$ and therefore that $$\begin{array}{rl}& |a_1{x}^{2n-2}+a_2{x}^{2n-4}+\cdots +a_{n-1}{x}^2+a_n|\\ & \le |a_1{x}^{2n-2}|+|a_2{x}^{2n-4}|+\cdots +|a_{n-1}{x}^2|+|a_n|\\ & \le |a_1{x}^{2n-2}|+|a_2{x}^{2n-2}|+\cdots +|a_{n-1}{x}^{2n-2}|+|a_n|x^{2n-2}\\ & \le (|a_1|+|a_2|+\cdots +|a_n|)x^{2n-2}\\ & \le -N\cdot x^{2n-2}\\ & <x^{2n},\end{array}$$ so $x^{2n}+a_1{x}^{2n-2}+a_2{x}^{2n-4}+\cdots +a_{n-1}{x}^2+a_n>0$. Hence for $n\ge 2021$ there exists an $N<0$ with $P_n(x)>0$ for all $x<N$, whereas $P_n(a_n)\le 0$. Therefore $P_n(x)$ has a root smaller than $a_n$. As $a_{n+1}$ is the minimal root, we have $a_{n+1}\le a_n$. $\square$