1
$\begingroup$

Let $\alpha$ be a real irrational algebraic number. The now-classic Thue-Siegel-Roth theorem asserts that for any $\varepsilon > 0$ there exists a positive number $c = c(\alpha, \varepsilon)$ such that for all rational numbers $p/q$ with $\gcd(p,q) = 1, q > 0$ we have

$$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert > \frac{c}{q^{2 + \varepsilon}}.$$

Equivalently, this says that for the linear form $x - \alpha y$ one has

$$\displaystyle \left \lvert x - \alpha y \right \rvert > c |y|^{-1 - \varepsilon}$$

provided that $x,y \in \mathbb{Z}, y \ne 0$.

My question is whether something similar can be said for other ranges. Fix $0 < \theta < 1$ (independent of $\varepsilon$) and $\beta$ another real algebraic number. Can one find infinitely many $x,y \in \mathbb{Z}$ such that

$$\displaystyle |x - \alpha y - \beta y^\theta| = o(1)?$$

Further, can one obtain a Thue-Siegel-Roth type result? That is, does there exist a $\psi = \psi(\theta) > 0$ such that for all $\varepsilon > 0$ there exists a positive number $c = c(\alpha, \beta, \theta, \varepsilon)$ such that

$$\displaystyle |x - \alpha - \beta y^\theta| > c q^{-\psi - \varepsilon}$$

for all $x,y \in \mathbb{Z}$ with $x,y \ne 0$?

$\endgroup$
3
  • $\begingroup$ In you statement of Roth's theorem, you need to specify that $\alpha$ is a non-rational real number (and for that matter, it could be a non-real complex number, but something stronger is trivially true in that case). $\endgroup$ Commented Oct 30, 2023 at 2:43
  • $\begingroup$ @JoeSilverman thank you, added some verbiage that avoids the rationality issue $\endgroup$ Commented Oct 30, 2023 at 2:46
  • $\begingroup$ $\{(x, y) | |y|\leq N+\frac12, |x - \alpha y - b y^\theta| \leq \frac1N\}$ isn't convex, right? You can't (directly) use Minkowski's theorem $\endgroup$ Commented Oct 30, 2023 at 3:18

0

You must log in to answer this question.

Browse other questions tagged .