Belarus2022

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 $$2=6\cdot h(2+\sqrt{2})\text{u}2=6\cdot h(2-\sqrt{2}).$$ 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.