# Estimate number of solutions in the Roth's theorem

There is a fundamental theorem in Diophantine approximation :

For all algebraic irrational $\alpha$ $$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^{2 + \epsilon}}$$ with $\epsilon>0$, has finitely many solutions.

can we estimate number of solutions $N_{\alpha}(\epsilon)$?

for instance :

what is the upper bound of $N_{\sqrt{2}}(1)$?

number of solutions for $\sqrt{2}$, with $\epsilon=1$.

$$\displaystyle \left \lvert \sqrt{2}- \frac{p}{q} \right \rvert < \frac{1}{q^{3}}$$

• One can give an effective upper bound for $N_\alpha(\epsilon)$ but one cannot give an effective bound for the height of the solutions. The effective bound provided is extremely bad and very likely to be completely off the true count. An effective version of Roth's theorem that allows you to bound the height of the solutions remains one of the outstanding problems in diophantine approximation. Sep 6, 2016 at 15:12
• Since when is $\pi$ algebraic? Sep 6, 2016 at 15:36
• @Wojowu: The same finiteness result is expected for $\pi$, but has not yet been proved with $\epsilon = 1$ (only with larger $\epsilon$). Sep 6, 2016 at 15:38
• ...where $\varphi$ is the Golden Ratio? $N_{\varphi}(1) = 1$ as, obviously, $p/q = 1$ is the unique solution. This problem is primarily interesting for $\alpha$ of degree higher than $2$. Sep 6, 2016 at 16:14

For a fixed $\alpha$, the number $N_{\alpha}(\epsilon)$ is bounded by a polynomial function of $1/\epsilon$. The proof of this requires either Faltings's product theorem, or Esnault and Viehweg's multidimensional Dyson lemma. See section 6.5 of Bombieri and Gubler's book (Heights in Diophantine Geometry), with particular attention to point 6.5.8.

Roth's original proof gives a bound exponential in $\epsilon^{-2}$. An explicit form of this bound is worked out in Davenport and Roth's paper Rational approximations to algebraic numbers (Mathematika, 1955), and is also sketched in section 6.5 of Bombieri and Gubler's book.

You might be more interested in the dependence on the degree $d$ and absolute logarithmic height $h(\alpha)$ of $\alpha$ (I should also add the absolute value $|\alpha|$ for a cleaner dependence), and not so much on $\epsilon$. For that, the bound that comes out of these methods is $$\ll_{\epsilon} 1 + \log^+{h(\alpha)} + \log^+{|\alpha|} + (\log{d})^{O(1)}.$$

• This doesn’t look as bad as the other commentators suggest. Is $\log^+x=\max\{0,\log x\}$ or something? Sep 6, 2016 at 17:33
• @EmilJeřábek: Yes, so it is the logarithm basically. And yes, the polynomial in $1/\epsilon$ bound is actually quite a good estimate as far as it goes, i.e. for a fixed $\alpha$. One may compare this to the Gauss-Kuzmin statistics. The degree of the polynomial function is an absolute constant, but the coefficients depend on $d$ and $h(\alpha)$. It took quite a bit of time since Davenport and Roth's paper to obtain such an estimate, until finer tools became available in the mid 1980s. Roth's lemma only gives an exponential bound in $\epsilon^{-2}$. Sep 6, 2016 at 17:46

I believe that there is a completely explicit upper bound for $N_\alpha(\epsilon)$ (more generally counting in relative number fields and using more then one, possibly non-archimedean, absolute value) in the following paper:

Robert Gross, A note on Roth's theorem. J. Number Theory 36 (1990), no. 1, 127–132. MR1068678

Of course, it's not going to be pretty.