Belarus competition problems

$a = 7k - 2$ 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
$$a + \sqrt{2} - 1 = (1 + 2\sqrt{2}) \cdot h(1 + \sqrt{2}). \quad (1)$$
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.