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Sorry, this is quite a brief question, but I have no idea how to make progress on it, and there's not much to say in the question itself.

The minimal polynomial (over $\mathbb{Q}$) of the golden ratio is $x^2-x-1$, and this has exactly one root greater than one in absolute value, and all others less than one, i.e., it is a Pisot-Vijayaraghavan number. Just checking cases, all $n$-bonacci constants, the real solutions to $x^n-\sum_{i=0}^{n-1}x^i$ are PV. So, are they, and are their minimal polynomials in fact the ones given just previously?

Thomas Anton
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  • You can use Brauer's criterion to see it is irreducible (see this answer), and it's proof itself found in the referrenced book shows there are $n-1$ roots in the unit disc and one outside (it uses Rouche's theorem to do so). So I guess you only need to finish it by showing the one root outside is a real root. – Sil Sep 02 '21 at 09:16
  • Also it seems this was asked in past on MO: Roots of $x^n-x^{n-1}-\cdots-x-1$ – Sil Sep 02 '21 at 09:24
  • @Sil Any polynomial with real coefficients, one root with absolute value larger than 1, and all other roots with absolute values less than 1 must have the large root real so we are done. This is very related to the theory of PV numbers. Use the solution to the difference equation with the coefficients from the polynomial. – Thomas Anton Sep 02 '21 at 09:30
  • That's correct (it follows from complex conjugates having same modulus, thus must also lie ouside, so the root is equal to its conjugate). By the way you have a typo there, sum should be indexed from $i=0$ (constant coefficient $-1$ should be included) – Sil Sep 02 '21 at 09:46

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