Is it possible for an irreducible polynomial with rational coefficients to have three zeros in an arithmetic progression? Assume that p(x) in Q[x] is irreducible of degree n >= 3.

Is it possible for an irreducible polynomial with rational coefficients to have three zeros in an arithmetic progression?
Assume that $p(x)\in <!--\; \in \; -->\mathbb{Q}[x]$ is irreducible of degree $n\ge <!--\; \ge \; -->3$.
Is it possible that p(x) has three distinct zeros ${\mathrm{\alpha <!--\; \alpha \; -->}}_1,{\mathrm{\alpha <!--\; \alpha \; -->}}_2,{\mathrm{\alpha <!--\; \alpha \; -->}}_3$ such that ${\mathrm{\alpha <!--\; \alpha \; -->}}_1-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_2={\mathrm{\alpha <!--\; \alpha \; -->}}_2-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_3$?
As also observed by Dietrich Burde a cubic won't work here, so we need $\mathrm{deg}<!--\; \; -->p(x)\ge <!--\; \ge \; -->4$. The argument goes as follows. If $p(x)=x^3+c_2{x}^2+c_{1}x+c_0$, then $-<!--\; -\; -->c_2={\mathrm{\alpha <!--\; \alpha \; -->}}_1+{\mathrm{\alpha <!--\; \alpha \; -->}}_2+{\mathrm{\alpha <!--\; \alpha \; -->}}_3=3{\mathrm{\alpha <!--\; \alpha \; -->}}_2$ implying that ${\mathrm{\alpha <!--\; \alpha \; -->}}_2$ would be rational and contradicting the irreducibility of p(x).
This came up when I was pondering this question. There the focus was in minimizing the extension degree $[\mathbb{Q}({\mathrm{\alpha <!--\; \alpha \; -->}}_1-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_2):\mathbb{Q}]$. I had the idea that I want to find a case, where ${\mathrm{\alpha <!--\; \alpha \; -->}}_1-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_2$ is fixed by a large number of elements of the Galois group $G=\mathrm{Gal}<!--\; \; -->(L/\mathbb{Q})$, $L\subseteq <!--\; \subseteq \; -->\mathbb{C}$ the splitting field of p(x). One way of enabling that would be to have a lot of repetitions among the differences ${\mathrm{\alpha <!--\; \alpha \; -->}}_i-<!--\; -\; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_j$ of the roots ${\mathrm{\alpha <!--\; \alpha \; -->}}_1,\dots <!--\; \dots \; -->,{\mathrm{\alpha <!--\; \alpha \; -->}}_n\in <!--\; \in \; -->\mathbb{C}$ of p(x). For the purposes of that question it turned out to be sufficient to be able to pair up the zeros of p(x) in such a way that the same difference is repeated for each pair (see my answer).
But can we build "chains of zeros" with constant interval, i.e. arithmetic progressions of zeros.
Variants:
- If it is possible for three zeros, what about longer arithmetic progressions?
- Does the scene change, if we replace Q with another field K of characteristic zero? (Artin-Schreier polynomials show that the assumption about the characteristic is relevant.)