Belarus competition problems

Suppose that such polynomial $p(x)$ with integer coefficients exists. It follows from the equality $p(\sqrt{2}) = \sqrt{2}$ that $p(-\sqrt{2}) = -\sqrt{2}$, i.e. the numbers $\sqrt{2}$ and $-\sqrt{2}$ are roots of the polynomial $p(x) - x$. According to Bezout's theorem, the polynomial $p(x) - x$ is divisible by $(x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2$ as polynomials with rational coefficients. Moreover, it follows from the Gauss lemma that in the equality $p(x) - x = (x^2 - 2)h(x)$ the rational coefficients of the polynomial $h(x)$ are integers. Substituting into this equality the numbers $2\sqrt{2}$ and $-2\sqrt{2}$ instead of $x$, we obtain the equalities
$$2 = 6 \cdot h(2 + \sqrt{2}) \quad \text{u} \quad 2 = 6 \cdot h(2 - \sqrt{2}).$$
Multiplying these equalities and reducing by 4, we obtain the equality
$$1 = 9 \cdot (h(2 + \sqrt{2})h(2 - \sqrt{2})).$$
Since the product of conjugate numbers in brackets is an integer, it implies that 1 is divisible by 9 — a contradiction.