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Call a real number $\lambda>1$ special if it is a root of a polynomial $f(x)$ such that

  1. $f(x)$ is monic with integer coefficients,
  2. all roots of $f(x)$ are distinct,
  3. for all $z\in\mathbb{C}$, if $f(z)=0$ and $z\neq \lambda$ then $|z|<1$.

For example, the Golden Ratio $\frac{1+\sqrt{5}}{2}$ is special.

Question: Are there special numbers arbitrarily close to 1?

Some Remarks:

  1. After messing around on Wolfram Alpha, I was able to beat the Golden Ratio using the unique real root of $x^3-x-1$, which is approximately 1.3247.

  2. I have found lots of work on polynomials such that all roots are on the unit disc, and some things where all but two roots are on the unit disc. But nothing close to what I want: exactly one real root not on the unit disc, and the rest strictly inside the unit disc.

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    $\begingroup$ See Pisot–Vijayaraghavan_number. $\endgroup$ Commented Nov 7, 2016 at 16:14
  • 3
    $\begingroup$ Congratulations, you discovered an interesting problem, unfortunately already solved. $\endgroup$ Commented Nov 8, 2016 at 0:50

1 Answer 1

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This is a well-known topic. Such numbers are called Pisot numbers. Salem (1945) proved that they form a closed subset $S$ of $(1,+\infty)$, and Siegel (1944) proved that its lowest element is the positive root of $X^3-X-1$. Therefore the answer to your question is No.

The Golden ratio is however the smallest among the cluster points of $S$. The derived set $S'$ was described by Dufresnoy & Pisot (1953). A Pisot number $\theta$ (root of $f$ as above) belongs to $S'$ iff there exist another polynomial $g\in{\mathbb Z}[X]$ such that $|g|\le|f|$ over the unit circle, and $\pm g$ is neither $f$, nor its reciprocal polynomial $f^*(X)=X^nf(1/X)$.

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