Suppose $\theta>1$ is an irrational algebraic integer, i.e. $\theta\not\in\mathbb Z$ but satisfies a monic polynomial with integer coefficients, and $\{a_n\}_{n\ge0}$ is a sequence of nonzero rational integers, i.e. $0\neq a_n\in\mathbb Z$. Is it true that there's a constant $C>1$ independent of $\theta$ and choice of $\{a_n\}$ such that $a_0^2+\sum_{n>0}(a_n-\theta a_{n-1})^2\ge C$?


It's a result needed for estimation of roots of a monic polynomial. Suppose the sum $\sum_{n>0}(a_n-\theta a_{n-1})^2$ converges, then $a_n/a_{n-1}$ is a Diophantine approximation of $\theta$, so maybe a characterization of the speed of Diophantine approximation will work.

And maybe the condition that $\theta$ is an algebraic integer is irrelevant. I mean, maybe it holds for every irrational number $\theta>1$.

Any idea? Thanks!


As Robert Israel pointed out, the original version, as was written, $\sum_n(a_n-\theta a_{n-1})^2\ge C$ is wrong (that was not a typo, but my shortage of understanding the problem), for $a_j=F_{n+j}$ and $\theta=(1+\sqrt 5)/2$. I've edited it to a form closer to where it arises.

  • 1
    $\begingroup$ What is a "rational integer"? A rational number? An integer? $\endgroup$
    – Théophile
    Jul 17, 2015 at 14:48
  • $\begingroup$ It is an integer that is also a rational number. Let $A$ be the set of rational integers, then $A=\mathbb Z \cap \mathbb Q$ $\endgroup$
    – Asinomás
    Jul 17, 2015 at 14:50
  • 3
    $\begingroup$ @Théophile I found number theorists usually call algebraic integers as integers, thus to disambiguate, I added the adjective rational. $\endgroup$
    – Yai0Phah
    Jul 17, 2015 at 14:52
  • $\begingroup$ Ah, I see. Thanks for clarifying. $\endgroup$
    – Théophile
    Jul 17, 2015 at 14:52
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    $\begingroup$ @dREaM Technically true but completely missing the point: $\mathbb Z \cap \mathbb Q$ is a redundant intersection since every integer is rational. The correct interpretation is "integral element (integer) over the field of rational numbers", so $A = {\mathcal O}_{\mathbb Q}$, which happens to also be $\mathbb Z$. See en.wikipedia.org/wiki/Ring_of_integers $\endgroup$
    – Erick Wong
    Jul 17, 2015 at 15:50

1 Answer 1


No. If $\varphi = (\sqrt{5}+1)/2$ (which is a root of $x^2 - x - 1$) and $F_n$ are the Fibonacci numbers, $|F_n - \varphi F_{n-1}|$ goes to $0$ exponentially. Take $a_n = F_{n+k}$ for sufficiently large $k$.

EDIT: For the revised problem, it's still no. Consider the polynomial $P(x) = x^2 - m x - 1$ where $m$ is a positive integer. The roots of $P$ are $\theta = \dfrac{m + \sqrt{m^2+4}}{2}$ and $\xi = -1/\theta$, with $\theta \sim m$ and $\xi \sim -1/m$ as $m \to \infty$. Let $a_n$ be the solution of the recurrence $a_{n+1} = m a_n + a_{n-1}$ with $a_0 = 1$, $a_{-1} = 0$:

$$ a_n = \dfrac{\theta}{m \theta + 2} (\theta^{n+1} - \xi^{n+1})$$ Note that $a_n - \theta a_{n-1} = \xi^n$, so $$a_0^2 + \sum_{n=1}^\infty (a_n - \theta a_{n-1})^2 = 1 + \sum_{n=1}^\infty \xi^{2n} = 1 + \dfrac{\xi^2}{1 - \xi^2} = 1 + O(1/m^2)$$


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