7. Given a polynomial $P(x)$, not identically zero. It is known that for all $x$ the identity $(x-2020) \cdot P(x+1)=(x+2021) \cdot P(x)$ holds. How many roots does the equation $P(x)=0$ have? Answer: 4042
Solution:

First, we prove two auxiliary lemmas:

Lemma 1. If $x_{0} \neq 2020$ is a root of $P(x)$, then $x_{0}+1$ is also a root.

Proof: Substitute $\left(x_{0}-2020\right) \cdot P\left(x_{0}+1\right)=\left(x_{0}+2021\right) \cdot P\left(x_{0}\right)=0$ but $\left(x_{0}-2020\right) \neq 0$.

Lemma 2. If $x_{0} \neq -2021$ is a root of $P(x)$, then $x_{0}-1$ is also a root.

Proof: Substitute $0=\left(x_{0}-2020\right) \cdot P\left(x_{0}+1\right)=\left(x_{0}+2021\right) \cdot P\left(x_{0}\right)$ but $\left(x_{0}+2021\right) \neq 0$.

Substitute $x=2020$ into the identity, we get $0=4041 P(2020)$, hence 2020 is a root.

By sequentially applying Lemma 2, we obtain that $x=2019, 2018, \ldots, -2021$ are roots.

We will now prove that there are no other roots.

1) Suppose that the root $x_{0}$ is not an integer, then by Lemma 1, $x_{0}+1, x_{0}+2, \ldots, x_{0}+n, \ldots$ are also roots. But a polynomial (not identically zero) cannot have infinitely many roots - contradiction.
2) Suppose that $x_{0}$ is an integer and does not belong to the interval $[-2021 ; 2020]$. If $x_{0}>2020$, then (by Lemma 1) $x_{0}+1, x_{0}+2, \ldots, x_{0}+n, \ldots$ are also roots. If $x_{0}<-2021$, then (by Lemma 2) $x_{0}-1, x_{0}-2, \ldots, x_{0}-n, \ldots$ are also roots. In either case, a non-zero polynomial cannot have infinitely many roots - contradiction.

Thus, the roots are the integers from -2021 to 2020 inclusive, their number is 4042.