Belarus2022

a) Consider any monomial $a_n{x}^n$ of $p(x)$ and substitute $x=\sqrt{2}+\sqrt{3}$ and $x=\sqrt{2}-\sqrt{3}$ into it. When opening the brackets in the expression $a_n(\sqrt{2}+\sqrt{3}{)}^n$, all terms will look like $a_n{C}_n^m{\sqrt{2}}^k{\sqrt{3}}^m$, where $k+m=n$. If $m$ is even then if $\sqrt{2}+\sqrt{3}$ is replaced by $\sqrt{2}-\sqrt{3}$, the sign of the term doesn't change, and the value of this term is a positive integer or a positive integer multiplied by $\sqrt{3}$. If $m$ is odd then with such a change the sign of the term changes and the value of this term is a positive integer multiplied by $\sqrt{3}$ or a positive integer multiplied by $\sqrt{6}$. Since these observations are true for every monomial, they are also true for the value of the entire polynomial, i.e. $p(\sqrt{2}-\sqrt{3})=\sqrt{2}+\sqrt{3}$.

b) It is easy to verify that $p(x)=x^3-10x$ satisfies the problem conditions. Let's show how to find this example. Arguing as in paragraph a) we can write two more equalities: $p(-\sqrt{2}+\sqrt{3})=\sqrt{2}+\sqrt{3}$ and $p(-\sqrt{2}-\sqrt{3})=\sqrt{2}+\sqrt{3}$. These equalities show that numbers $\pm \sqrt{2}\pm \sqrt{3}$ satisfy the equality $p(x)=\frac{1}{x}$, so they are roots of the polynomial $xp(x)-1=0$. By Bezout's theorem, a polynomial with these roots is divisible by a polynomial $$(x-\sqrt{2}-\sqrt{3})(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}-\sqrt{3})(x+\sqrt{2}+\sqrt{3})=x^4-10x^2+1.$$ It remains to find $p(x)$ from the equality $xp(x)-1=x^4-10x^2+1$.