First Round of the 73rd Czech and Slovak Mathematical Olympiad (December 12th, 2023)

Denote the coefficients of $P$ as $P(x)=a{x}^2+bx+c$. By looking what happens for $|x|$ large, we conclude that $a\in \{1,2\}$. These two cases shall be treated separately:

Suppose that $a=1$. The lower bound on $P$ then tells us that the inequality $2x-2023<bx+c$ must hold for all $x$. Clearly, this happens if and only if $b=2$ and $c>-2023$. The upper bound is then equivalent to $$(x-1{)}^2>c+1,$$ which is satisfied for all $x$ precisely when $c<-1$. To summarize, we have shown that the only options here are $b=2$ and $-2023<c<-1$.

Suppose that $a=2$, then analogously to the previous case, the upper bound reduces to $bx+c<0$, which holds for all values of $x$ if and only if $b=0$ and $c<0$. The lower bound can then be rewritten as $$-c-2022<(x-1{)}^2,$$ which holds for all $x$ precisely when $c>-2022$. Therefore, we have shown that the only options for this case are $b=0$ and $-2022<c<0$.

Conclusion. There are $2\cdot 2021=4042$ such polynomials.