Hong Kong competition problems, 2021

Yes. We shall prove that $P(x) = (x^2+1)(x^2-2)(x^2+2)(x^2+7)$ satisfies all properties. It is not hard to see that $P$ has no rational root. It remains to show it satisfies the second condition.

Consider any odd prime $p$. Using facts about quadratic residues, we have the following.
* If $p \equiv 1 \pmod 4$, then $x^2+1 \equiv 0 \pmod p$ is solvable.
* If $p \equiv 7 \pmod 8$, then $x^2-2 \equiv 0 \pmod p$ is solvable.
* If $p \equiv 3 \pmod 8$, then $x^2+2 \equiv 0 \pmod p$ is solvable since
$$\left(\frac{-2}{p}\right) = \left(\frac{-1}{p}\right) \left(\frac{2}{p}\right) = (-1)(-1) = 1.$$
This shows there exists $x$ and $c \in \{1, -2, 2\}$ such that $x^2 + c \equiv 0 \pmod p$. Note that $p \nmid x$. Therefore, we have
$$(x^2 + c)' = 2x \not\equiv 0 \pmod p.$$
By Hensel's lifting lemma, $x^2+c \equiv 0 \pmod{p^k}$ is solvable for any positive integer $k$.

Next, we want to show for any $k \ge 1$, there exists an integer $b_k$ such that $2^k \mid (b_k^2+7)$. Indeed, the statement is true for $k=1,2,3$, by simply taking $b_k=1$. Assume now the statement is valid for some $k \ge 3$. For the case $k+1$, take $b_{k+1} = b_k + t2^{k-1}$, with $t$ to be determined. Now, since $b_k$ is odd,
$$b_{k+1}^2 + 7 = (b_k + t2^{k-1})^2 + 7 \equiv (b_k^2 + 7) + t2^k \pmod{2^{k+1}}.$$
By the inductive hypothesis, $2^k \mid (b_k^2 + 7)$. So we just take $t \equiv \frac{b_k^2 + 7}{2^k} \pmod 2$ to get $2^{k+1} \mid (b_{k+1}^2 + 7)$.

Now, let $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_s^{\alpha_s}$ be the prime factorization of $n$. From above we know that for every $1 \le j \le s$, there exists an integer $m_j$ such that $p_j^{\alpha_j} \mid P(m_j)$. By the Chinese remainder theorem, we can find $m \equiv m_j \pmod{p_j^{\alpha_j}}$ for $1 \le j \le s$. Then $P(m) \equiv P(m_j) \equiv 0 \pmod{p_j^{\alpha_j}}$. So $n \mid P(m)$. Therefore, the polynomial $P(x)$ satisfies the second condition, and we are done.