Saudi Arabian IMO Booklet

1) Since $\mathrm{deg}P=1$, put $P(x)=ax+b$ with $a,b\in \mathbb{Z}$ and $a\ne 0$. 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 $$2\le \mathrm{deg}P\le 2022\text{ and }{M}_{P(x)}<1011.$$ 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 $H(1)\ne 0$ then $M_{P(x)}>1011$. Similarly, $H(2020)\ne 0$ 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 $\mathrm{deg}P\le 2022$ 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 $M_{P(x)}\ge 1011$. $\square$