Prove the irreducibility of $P(x)$ which satisfies: $xP(x-1)=(x-2022)P(x)+2022$

Prove the irreducibility in $\mathbb{Z}[x]$ of $P(x)$ which satisfies: $xP(x-1)=(x-2022)P(x)+2022, \forall x\in \mathbb{R}$

My attempts I thought of substituting $x$ for $0$ and $x$ for $2022$, but that yielded no results. I have also attempted the Eisenstein criterion but that does not work as well.

Solution 1:

So I got $xP(x-1)=(x-2022)P(x)+2022$ equals $x(P(x-1)-1)=(x-2022)[P(x)-1]$,

I substitute $P(x)-1=Q(x)$ then I got $xQ(x-1)=(x-2022)Q(x)$

Substitute $x=2022$ we have $Q(2021)=0$ thus $x=2021$ is a root of $Q(x)$

This implies $x(x-2022)Q_1(x-1)=(x-2022)(x-2021)Q_1(x)$

Thus $xQ_1(x-1)=(x-2021)Q_1(x)$

Do the same thing for $Q_1, Q_2,...,Q_{2021}$ you will have \begin{align} P(x)= 1+\prod_{i=0}^{2021}(x-i) \end{align}

Using Eisenstein criterion for $p=2$ yields the result!