Assume that $p(x)\in \Bbb{Q}[x]$ is irreducible of degree $n\ge3$.

Is it possible that $p(x)$ has three distinct zeros $\alpha_1,\alpha_2,\alpha_3$ such that $\alpha_1-\alpha_2=\alpha_2-\alpha_3$?

As also observed by Dietrich Burde a cubic won't work here, so we need $\deg p(x)\ge4$. The argument goes as follows. If $p(x)=x^3+c_2x^2+c_1x+c_0$, then $-c_2=\alpha_1+\alpha_2+\alpha_3=3\alpha_2$ implying that $\alpha_2$ 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 $[\Bbb{Q}(\alpha_1-\alpha_2):\Bbb{Q}]$. I had the idea that I want to find a case, where $\alpha_1-\alpha_2$ is fixed by a large number of elements of the Galois group $G=\operatorname{Gal}(L/\Bbb{Q})$, $L\subseteq\Bbb{C}$ the splitting field of $p(x)$. One way of enabling that would be to have a lot of repetitions among the differences $\alpha_i-\alpha_j$ of the roots $\alpha_1,\ldots,\alpha_n\in\Bbb{C}$ 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.


  • If it is possible for three zeros, what about longer arithmetic progressions?
  • Does the scene change, if we replace $\Bbb{Q}$ with another field $K$ of characteristic zero? (Artin-Schreier polynomials show that the assumption about the characteristic is relevant.)
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    $\begingroup$ I do not have time to think this through until tomorrow, but for your second variant, it pops to my mind that one should look at the quotient field of $\mathbb Q[x,y,z,w,u]/(x-y-u,y-z-u)$ and consider $(T-x)(T-y)(T-z)(T-w)$ over the smallest field, you can define this (i.e. the field generated by the elementary symmetric polynomials in x,y,z,w). $\endgroup$
    – MooS
    Jul 1, 2016 at 10:59
  • $\begingroup$ If this does not work, i.e. if this polynomial turns out to be reducible, I dont see that it could work with concrete numbers instead of indeterminants. Of course this does only cover the $n=4$-case. $\endgroup$
    – MooS
    Jul 1, 2016 at 11:01
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    $\begingroup$ On the other hand if this would work with indeterminates, couldn't we invoke Hilbert's irreducibility theorem to deduce that it is also possible over $\mathbb Q$? $\endgroup$
    – MooS
    Jul 1, 2016 at 11:12
  • $\begingroup$ A promising plan @MooS! Take your time :-) I will be gone for 30 hrs or so. $\endgroup$ Jul 1, 2016 at 13:36

1 Answer 1


The answer to your question is NO. Suppose by contradiction that such a polynomial $P$ exists, and denote by $S$ the set of roots. By hypothesis, some $a\in S$ can be written $a=\frac{b+c}{2}$ where $b,c$ are distinct elements of $S$. But since the Galois group acts transitively on $S$, this property holds for all $a\in S$. This motivates the following definition :

Definition. A (non-empty) set $S\subseteq {\mathbb C}$ is AP-extensive if any $a\in S$ can be written $a=\frac{b+c}{2}$ where $b,c$ are distinct elements of $S$.

Note that an AP-extensive $S\subseteq {\mathbb R}$ cannot have a largest element. In particular, any (non-empty) AP-extensive $S\subseteq {\mathbb R}$ is necessarily infinite. This still holds in $\mathbb C$ :

Lemma. If $S\subseteq {\mathbb C}$ is AP-extensive, then $S$ is infinite (or empty).

Proof of lemma. Suppose by contradiction that $S$ is finite and nonempty. Then the set $\lbrace b\in{\mathbb R}\ | \ \exists a, a+ib\in S\rbrace$ is finite also and therefore has a largest element $b_0$. Let $S_1=\lbrace z\in S \ | \ Im(z)=b_0 \rbrace$. It is easy to see that if $a=\frac{b+c}{2}$ with $a,b,c\in S$ and further $a\in S_1$, then $b$ and $c$ must be in $S_1$ also. So $S_1$ is AP-extensive as well. Next, let $S_2=S_1-ib_0$. Then $S_2$ is AP-extensive also, but by construction $S_2\subseteq {\mathbb R}$. So $S_2$ (and hence $S_1,S$ also) must be infinite which is impossible. This concludes the proof.

  • $\begingroup$ A beautiful argument :-) $\endgroup$ Jul 3, 2016 at 9:20
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    $\begingroup$ I got a lot of inspiration from the Krein-Milman theorem. $\endgroup$ Jul 3, 2016 at 9:29
  • $\begingroup$ Unless I misunderstood, the same argument works for polynomials over any characteristic zero field. The set of zeros spans (affinely or linearly) a finite dimensional space over $\Bbb{Q}$, and we can then work with one coordinate at the time much the same way. In other words, we don't need the set $S$ to be a subset of $\Bbb{C}$ in particular. $\endgroup$ Jul 3, 2016 at 9:41
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    $\begingroup$ Wow! The key idea to use the transitive action to deduce that any root could play the role of $\alpha_2$ is so simple but wonderful! $\endgroup$
    – MooS
    Jul 4, 2016 at 14:13
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    $\begingroup$ This is one of the best proofs I've ever read. Ordinarily I won't bump old things just to heap praise, but it is deserved here. $\endgroup$
    – The Count
    Dec 28, 2016 at 4:40

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