## Task B-4.4.

Determine all polynomials $p$ with real coefficients for which the equality

$$
x \cdot p(x-1)=(x-2021) \cdot p(x)
$$

is satisfied for all real numbers $x$.
## Solution.

By substituting the natural numbers $0,1, \ldots, 2020$ into the given equality, we obtain:

\[
\begin{aligned}
& x=0 \quad \Rightarrow \quad 0=-2021 \cdot p(0) \quad \Rightarrow \quad p(0)=0 \\
& x=1 \quad \Rightarrow \quad p(0)=-2020 \cdot p(1) \quad \Rightarrow \quad p(1)=0 \\
& x=2 \quad \Rightarrow \quad 2 p(1)=-2019 \cdot p(2) \quad \Rightarrow \quad p(2)=0 \\
& \vdots \\
& x=2020 \quad \Rightarrow \quad 2020 p(2019)=-p(2020) \quad \Rightarrow \quad p(2020)=0
\end{aligned}
\]

From the given equality, it is clear that if $p(x-1)=0$, then $p(x)=0$ for all natural numbers $x<2021$.

We conclude that $0,1,2, \ldots, 2020$ are the roots of the polynomial $p$, so

\[
p(x)=q(x) \cdot x(x-1)(x-2) \cdots(x-2020)
\]

for some polynomial $q(x)$.

Now, from the given equality $x \cdot p(x-1)=(x-2021) \cdot p(x)$, we have:

\[
x \cdot q(x-1) \cdot(x-1)(x-2) \cdots(x-2021)=(x-2021) \cdot q(x) \cdot x(x-1)(x-2) \cdots(x-2020)
\]

which implies that $q(x-1)=q(x)$ for all $x \in \mathbb{R}$, or $q(x)=c, c \in \mathbb{R}$.

The desired polynomials $p$ are of the form $p(x)=c \cdot x(x-1)(x-2) \cdots(x-2020), c \in \mathbb{R}$.