Belarus2022

$a=7k-2$ where $k$ is an arbitrary positive integer.

Suppose that the integer $a$ and the polynomial $p(x)$ satisfy the condition. For the polynomial $q(x)=p(x+1)$, the equalities from the condition have the form $q(\sqrt{2})=2-\sqrt{2}$ and $q(1+\sqrt{2})=a$. Since the coefficients of the polynomial $q(x)$ are integers, the equality $q(-\sqrt{2})=2+\sqrt{2}$ is true. Therefore the numbers $\sqrt{2}$ and $-\sqrt{2}$ satisfy the equality $q(x)=2-x$, i.e. they are roots of the polynomial $q(x)+x-2$. According to Bezout's theorem, the polynomial $q(x)+x-2$ is divisible by $(x-\sqrt{2})(x+\sqrt{2})=x^2-2$. Let $q(x)+x-2=(x^2-2)h(x)$, then the Gauss lemma implies that the rational coefficients of the polynomial $h(x)$ are integers. Substitute into the resulting equality $x=1+\sqrt{2}$ and take into account that $q(1+\sqrt{2})=a$, after transformations we obtain the equality $$a+\sqrt{2}-1=(1+2\sqrt{2})\cdot h(1+\sqrt{2}).(1)$$ Since the coefficients of the polynomial $h(x)$ are integers, $$a-\sqrt{2}-1=(1-2\sqrt{2})\cdot h(1-\sqrt{2}).$$ Let's multiply this equality with (1): $$(a-1{)}^2-2=-7\cdot (h(1+\sqrt{2})\cdot h(1-\sqrt{2})).$$ Since the expression in brackets is a product of conjugate numbers, it is an integer, so $(a-1{)}^2-2$ is divisible by $7$. This is equivalent to saying that $a$ is congruent to $4$ or $5$ modulo $7$.

For the numbers $a=7k+4$, $k\in \mathbb{Z}$, equality (1) takes the form $$7k+3-\sqrt{2}=(1+2\sqrt{2})\cdot h(1+\sqrt{2}),$$ which is equivalent to $$h(1+\sqrt{2})=\frac{7k+3-\sqrt{2}}{1+2\sqrt{2}}=\frac{1-7k}{7}+\frac{14k+5}{7}\cdot \sqrt{2},$$ which is impossible, since the coefficients $h(x)$ are integers.

For numbers $a=7k+5$, $k\in \mathbb{Z}$, equality (1) takes the form $$7k+4-\sqrt{2}=(1+2\sqrt{2})\cdot h(1+\sqrt{2}),$$ which is equivalent to $$h(1+\sqrt{2})=\frac{7k+4-\sqrt{2}}{1+2\sqrt{2}}=(2k-1)\sqrt{2}-k,$$ therefore, one can choose the polynomial $h(x)=(2k-1)x-(3k-1)$. Therefore, all such numbers satisfy the condition.