Kulikov E.Yu.

Find some polynomial with integer coefficients, for which the number

$\sqrt[5]{2-\sqrt{3}}$

is a root.
Let's introduce the notations $a=\sqrt[5]{2+\sqrt{3}}, b=\sqrt[5]{2-\sqrt{3}}, x=a+b$. Note that $a b=1$, which means $x=a+\frac{1}{a}$.

We will use the formulas

$$
\left(a+\frac{1}{a}\right)^{5}=a^{5}+\frac{1}{a^{5}}+5\left(a^{3}+\frac{1}{a^{3}}\right)+10\left(a+\frac{1}{a}\right), \quad\left(a+\frac{1}{a}\right)^{3}=a^{3}+\frac{1}{a^{3}}+3\left(a+\frac{1}{a}\right)
$$

Substituting the value of $a^{3}+1 / a^{3}$ from the second equality into the first, we get

$$
\left(a+\frac{1}{a}\right)^{5}=a^{5}+\frac{1}{a^{5}}+5\left(a+\frac{1}{a}\right)^{3}-5\left(a+\frac{1}{a}\right)
$$

From here, considering that $a^{5}+b^{5}=4$, we obtain $x^{5}=4+5\left(x^{3}-3 x\right)+10 x$, or $x^{5}-5 x^{3}+5 x-4=0$.

## Answer

## Problem