Find all polynomials $P$ such that $(X+2019) P(X)=X P(X+1)$.
0 is a root of $P(X)$ and -2019 is a root of $P(X+1)$, i.e., -2018 is a root of $P(X)$. We can therefore factorize $P: P(X)=X(X+2018)Q(X)$ for some polynomial $Q$, and the equation gives after simplification $(X+2018) Q(X)=(X+1) Q(X+1)$. Following this idea, we can show that $(X+k)(X+2018-k)$ divides $P(X)$ for all $k$ between 0 and 2018. We can therefore write $P(X)=\prod_{k=0}^{2018}(X+k) \times Q(X)$ and the equation gives $Q(X)=Q(X+1)$, which implies that $Q$ is constant (it takes the value $Q(0)$ an infinite number of times).