Belarus competition problems

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 $x p(x) - 1 = 0$. 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 $x p(x) - 1 = x^4 - 10x^2 + 1$.