Kanel-Belov A.Ya.

Given a polynomial of degree 2022 with integer coefficients and leading coefficient 1. What is the maximum number of roots it can have in the interval $(0,1)$?
Evaluation. All 2022 roots of the polynomial cannot lie on the interval $(0,1)$: otherwise, the free term of the polynomial, equal to the product of the roots by Vieta's theorem, would lie on the interval $(0,1)$ and could not be an integer. Example 1. Consider the polynomial $\$ \mathrm{P}(\mathrm{x})=\mathrm{x}^{\wedge}\{2022\}+(1-4042 \mathrm{x})(3-4042 \mathrm{x}) \ldots(4041-4042 \mathrm{x}) \$$$. Note that for all $\$ k=0,1, \ldots, 2021$ the number

$\$ P \backslash \operatorname{Bigl}\left((\mid f r a c\{2 k\}\{4042\} \backslash \operatorname{Bigr})=\backslash \operatorname{Bigl}\left((f \operatorname{frac}\{2 \mathrm{k}\}\{4042\} \backslash \operatorname{Bigr})^{\wedge}\{2022\} \$+(-1)^{k}(2 k-1)!!(4041-2 k)!!\right.\right.$ is positive for even $\$ \mathrm{k} \$$ and negative for odd. Thus, on the interval $(0,1)$, the polynomial $\$ \mathrm{P}(\mathrm{x}) \$$ changes sign at least 2021 times and, therefore, has at least 2021 roots.

Example 2. Assume that $\$ \mathrm{P}(0) \$ \neq 0$. Consider the polynomial $\$ \mathrm{Q}(\mathrm{x})=\mathrm{x} \wedge \mathrm{n} P \backslash b i g($ frac1 $1 \mathrm{x} \backslash b i g) \$$ of degree $\$ \mathrm{n} \$ \mathrm{c}$ with integer coefficients and a free term equal to 1 (its coefficients are the coefficients of the polynomial \$P\$, arranged in reverse order). For each root \$x_0\$ of the polynomial \$P\$, lying on the interval $(0,1)$, there corresponds a root $\$ \mid$ frac $\{1\}\left\{x \_0\right\} \$$ of the polynomial $\$ Q \$$, lying on the ray $(1,+\infty)$. The converse is also true: for each root of the polynomial $\$ Q \$$, lying on the ray $(1,+\infty)$, there corresponds a root of the polynomial $\$ \$ \$$, which lies on the interval $(0,1)$.

Consider the polynomial $\$ Q(x)=1+x(x-10)(x-20) \ldots(x-10(n-1)) \$$. Since $\$ Q \$(5)0, \ldots$, $\$ \mathrm{Q}(10(\mathrm{n}-2)+5)0 \$$, in the considered $\$ \mathrm{n} \$$ points, the polynomial $\$ \mathrm{Q}(\mathrm{x}) \$$ takes
values of alternating signs, so it has $\$ n \$-1$ roots on the ray $(1,+\infty)$. These roots are located on the intervals $(5,15),(15,25), \ldots, \$(10(n-2)+5,10(n-1)+5) \$$. Therefore, the polynomial $\$ \mathrm{P} \$$ corresponding to the constructed polynomial $\$ \mathrm{Q} \$$ has exactly $\$ \mathrm{n} \$-1$ roots on the interval $(0,1)$.

## Answer

2021 roots.