Let $\beta\in(1,2)$ and $\gamma\in(1,2)$ be Galois conjugates of height 1. That is, there exists a polynomial $p$ with coefficients $-1,0,1$ such that $p(\beta)=p(\gamma)=0$ (not necessarily minimal).

Numerically, there appears to be an absolute constant $C>0$ such that $|\gamma-\beta|\ge C$. Is this true/known? If it is, what is the best known value for $C$?

I've looked into some literature on root separation but couldn't find this claim.

  • 2
    $\begingroup$ Chapter 9 deals with a similar problem, but the bound depends on $\beta$ (naturally, there are many results like that). The hight could be a red herring, as any polynomial with Mahler measure less than 2, always divides a polynomial of height 1. However, asking for an absolute bound might complicate the matter, and could potentially relate to Lehmer's conjecture. $\endgroup$
    – pavl0
    Nov 6, 2018 at 0:18

1 Answer 1


The polynomial

$$1 + x^n + x^{2n} - x^{3n} - x^{5n} - x^{6n} + x^{7n}$$

is irreducible and has two Galois conjugate roots $\beta_n$ and $\gamma_n$ in $(1,2)$ with

$$| \beta_n - \gamma_n| \sim \frac{\log(\beta_1/\gamma_1)}{n} \rightarrow 0.$$

  • $\begingroup$ Simple and beautiful! Maybe the irreducibilty (at least for infinitely many $n$, like the primes) deserves an explanation. $\endgroup$ Nov 8, 2018 at 10:46
  • 4
    $\begingroup$ Since the Galois group of the polynomial $p(x)$ for $n = 1$ is $S_7$, the irreducibility of $p(x^n)$ is immediate as long as a (any) root of $p(x)$ is not a perfect $n$th power for some $n > 1$. But any root is a fundamental unit in the corresponding degree $7$ field. $\endgroup$
    – user131093
    Nov 8, 2018 at 18:05
  • $\begingroup$ Thanks! I posted a follow-up question: mathoverflow.net/questions/314870/… $\endgroup$ Nov 8, 2018 at 20:03

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