Saudi Arabian Mathematical Competitions

1. Consider the polynomial
$$f(x) = x(x+2)(x+4)\dots(x+4040) - x - (2m-1)$$
for some $m \in \mathbb{Z}^+$ then $\lim_{x \to +\infty} f(x) = +\infty$ and $f(0) = 2^{2020} \cdot 2020! - (2m-1)$. Take $m$ big enough to get $f(0) < 0$ then by the continuous property of $f(x)$, there exists some real number $x_0$ such that $f(x_0) = 0$. Then take the set $T$ as follows
$$T = \{x_0, x_0 + 2, x_0 + 4, \dots, x_0 + 4040\}$$
Then $p(T) = x_0(x_0+2)(x_0+4)\dots(x_0+4040)$ and $p(T) - x_0 = 2m-1$ is an odd integer. Since every element in $T$ has form $x_0+2k$ with $k \in \mathbb{Z}$ then $p(T) - a$ is an odd integer for all $a \in T$. Finally, take $m = 2^{2019} \cdot 2020!$ then $f(0) = 1$, suppose on the contrary that $x_0 \in \mathbb{Q}$. Note that $f(x)$ is monic then $x_0 \in \mathbb{Z}$ and $x_0 \mid f(0) = 1$, implies that $x_0 \in \{-1; 1\}$, but this is contradiction since it is easy to check that $f(x)$ cannot have odd integer root.

2. We will proof that the answer is negative. Note that for all $x, y \in T$ then $x - y$ is even. Suppose that $T$ contains some rational number $b$. Then we can list elements of $T$ as follows $b, b + a_1, b + a_2, \dots, b + a_{2020}$ in which $a_1, a_2, \dots, a_k$ are even.

Put $c = b(b + a_1)(b + a_2)\dots(b + a_k) - b$ then $c$ is odd. Consider polynomial
$$g(x) = x(x + a_1)(x + a_2)\dots(x + a_k) - x - c$$
then $g(b) = 0$ and $x = b$ is the rational root of $g(x)$. Note that $g$ is monic then $b \in \mathbb{Z}$. From this we conclude that all the elements in $T$ are integers that share the same parity, so does $p(T)$. Thus $p(T) - b$ is even, contradiction.