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I have read a wikipedia article on Pisot number or PV number: https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number

I define a Pisot number $\alpha$ is "small" iff $\alpha<2$.(It is not a qualified definition; is there another reasonable bound for the problem below?)

Is there a small Pisot number $\alpha$ such that every (Galois) conjugate of it is real, except for the golden ratio $\frac{1+\sqrt{5}}{2}$? The conjugates of any quadratic Pisot number are real, but it is larger than 2 other than $\frac{1+\sqrt{5}}{2}$.

If then, how many? The nonexistence may be an exercise on Pisot numbers, but I can't even guess how I can approach it.

(The same post in Math Stackexchange was deleted.)

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  • $\begingroup$ Simul-posted to m.se, math.stackexchange.com/questions/2049412/… with no notice to either site. Very bad form. $\endgroup$
    – Gerry Myerson
    Commented Dec 8, 2016 at 11:40
  • $\begingroup$ Sorry, I didn't know anything about the rule... $\endgroup$
    – Setare G.
    Commented Dec 8, 2016 at 12:28
  • $\begingroup$ It's not a rule – it's common sense, and common courtesy. $\endgroup$
    – Gerry Myerson
    Commented Dec 8, 2016 at 21:55
  • $\begingroup$ Toufik Zaimi has a couple of papers that may be useful to you: Sur les nombres de Pisot totalement réels, Arab J Math Sci 5 (1999) 19-32, MR1734304 (2001b:11098) and On small Pisot numbers in a number field, Maghreb Math Rev 8 (1999) 163-167, MR1871538 (2002i:11103). $\endgroup$
    – Gerry Myerson
    Commented Dec 8, 2016 at 22:30

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The following theorem of Kronecker seems to (almost) answer your question.

Let $\beta$ be an algebraic integer such that all its algebraic conjugates are real (i.e. $\beta$ is totally real). Then either $\beta$ is of the form $\beta=2\cos(\pi q)$ with rational $q$, or some algebraic conjugate of $\beta$ has absolute value strictly greater than 2.

Due to this theorem, we only need to find for which $n>3$ the number $\beta=2\cos (\pi/n)$ is a Pisot number; in other words, there should be no integers coprime with $2n$ in the intervals $[2,n/3]$ and $[2n/3,n]$. (The golden ratio correspods to $n=5$).

Thus, if $p$ is the smallest prime not dividing $n$ then $p>n/3$. This leaves us with only small cases which one may deal with by hands.

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  • $\begingroup$ Thank you! Could I ask whether there is a useful related result on a bound bigger than $2$, say, $3$, $1+\sqrt{2}$ or $2+\epsilon$? $\endgroup$
    – Setare G.
    Commented Dec 8, 2016 at 15:52

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