A polynomial $f(x)$ of degree $2000$ is given. It's known that $f(x^2-1)$ has exactly $3400$ real roots while $f(1-x^2)$ has exactly $2700$ real roots. Prove that there exist two real roots of $f(x)$ such that the difference between them is less that $0.002$. (А. Солынин)
HIDE: Thanks Thanks to the user Vlados021 for translating the problem.Problem
Source: 2019 Saint Petersburg
Tags: algebra
14.04.2019 17:41
Cool problem. Let $r_1,r_2,\dots,r_{2000}$ be (not necessarily real) roots of $f(x)$. Notice first that, $f(x^2-1)=(x^2-1-r_1)\cdots(x^2-1-r_{2000})$. Now, if $-1-r_i >0$, then the factor $x^2-1-r_i$ is irreducible. But since $f(x^2-1)$ has $3400$ real roots, it then holds that, for at least $1700$ values of $r_i$, it holds that, $r_i\geq -1$. Call this set of $1700$ values as $S_1$. Similarly, $f(1-x^2)=(1-x^2-r_1)\cdots(1-x^2-r_{2000})$. If $1-r_i<0$, then the factor, $1-x^2-r_i$ is irreducible. Thus, for at least $1350$ values of $r_i$, it holds that, $r_i\leq 1$. Call this set of $r_i$'s as $S_2$. Now, since $|S_1\cup S_2|=|S_1|+|S_2|-|S_1\cap S_2|\leq 2000$, it holds that, $|S_1\cap S_2|\geq 1100$. In particular, there are at least $1100$ real roots in the interval, $(-1,1)$. Now, divide this interval into $1000$ equal pieces, and conclude by Pigeonhole.
28.02.2021 16:04
Just a minor correction @grupyorum $$|S_1\cap S_2| \geq |S_1| + |S_2| - 2000 = 1700 + 1350 - 2000 = 1050.$$