Saudi Arabian IMO Booklet

1) Since $\deg P = 1$, put $P(x) = ax + b$ with $a, b \in \mathbb{Z}$ and $a \neq 0$. Note that
$$|P(2021) - P(0)| = |2021a| \ge 2021 \Rightarrow \max\{|P(2021)|, |P(0)|\} \ge \frac{2021}{2}.$$
Thus, $|P(2021) - P(0)| = |2021a| \ge 2021$ which implies that
$$\max\{|P(2021)|, |P(0)|\} \ge \frac{2021}{2} \Rightarrow \max\{|P(2021)|, |P(0)|\} \ge 1011.$$
On the other hand, $T(x) = x - 1011$ satisfies the condition $|T(x)| \le 1011, \forall x \in [0; 2021]$. Hence, the minimum value of $M_{P(x)}$ is $1011$, the equality holds when $P(x) = x - 1011$.

2) Suppose that there is some integer polynomial $P(x)$ with
$$2 \le \deg P \le 2022 \text{ and } M_{P(x)} < 1011.$$
One can see that $|P(x)| < 1011, \forall x \in [0; 2021]$ thus $|P(x)| \le 1010, \forall x \in [0; 2021] \cap \mathbb{Z}$. By the property of integer polynomials, $2021|P(2021) - P(0)$ so $P(2021) = P(0)$ which leads to
$$|P(2021) - P(0)| \le |P(0)| + |P(2021)| \le 2020.$$
From that, let's put $P(x) = x(x - 2021)Q(x) + c$ with $c \in \mathbb{Z}$ and $Q(x) \in \mathbb{Z}[x]$. So $x(x - 2021) \ge 2022, \forall x \in \{2, 3, \dots, 2019\}$. If there exists $x_0 \in \{2, 3, \dots, 2019\}$ such that $Q(x_0) \ne 0$ then
$$\max |x_0(x_0 - 2021)Q(x_0) - c| \ge \frac{1}{2} |x_0(x_0 - 2021)Q(x_0)| \ge 1011.$$
From this, we can conclude that $Q(2) = Q(3) = \cdots = Q(2019) = 0$ so let's write
$$P(x) = x(x-2)(x-3)\cdots(x-2019)(x-2021)H(x) + c$$
with $H(x) \in \mathbb{Z}[x]$ then $P(1) = -2020 \cdot 2018!H(1) + c$. Note that if $H(1) \neq 0$ then $M_{P(x)} > 1011$. Similarly, $H(2020) \neq 0$ also leads to another contradiction, so $H(1) = H(2020) = 0$ then there is some $R(x) \in \mathbb{Z}[x]$ for which
$$Q(x) = (x-1)(x-2)\cdots(x-2020)R(x).$$
Since $\deg P \leq 2022$ then $R(x) \equiv c \in \mathbb{Z}$. On the other hand, $P\left(\frac{1}{2}\right) > 1011$ so $c = 0$, which contradicts $P(x)$ being non-constant. Therefore, the contrary hypothesis is false and it follows that $M_{P(x)} \geq 1011$. $\square$