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Hurwitz's Theorem in Number Theory states that for every irrational number $\xi$, the equation $$\left|\xi-\frac{p}{q}\right|<\frac{1}{\sqrt{5}q^2}$$ has infinitely many solutions $(p,q)$ such that $p$ and $q$ are relatively prime, and that this is not true of $$\left|\xi-\frac{p}{q}\right|<\frac{1}{(\sqrt{5}+\varepsilon)q^2}$$ has finitely many relative prime solutions for any $\varepsilon>0$.

There is a meaningful sense in which this is a statement about $\phi=\frac{1+\sqrt{5}}{2}$, because the proof of the second part is a property of the number $\phi$, and if $\phi$ (plus a specific set of numbers that are equivalent to $\phi$ in a certain sense), then the bound changes to $2\sqrt{2}$. We might only care about some other irrational number, and so we might form the following problem:

Let $\mu(\xi)$ be the unique number (should it exist) that has the property that $$\left|\xi-\frac{p}{q}\right|<\frac{1}{\mu(\xi)q^2}$$ has infinitely many solutions $(p,q)$ such that $p$ and $q$ are relatively prime, and that $$\left|\xi-\frac{p}{q}\right|<\frac{1}{(\mu(\xi)+\varepsilon)q^2}$$ has finitely many relative prime solutions for any $\varepsilon>0$.

I am interested in when $\mu(\xi)$ exists (I conjecture that this is equivalent to asking when does $\xi$ have bounded coefficients in its continued fraction expansion), and when it does exist, how to calculate it (presumably via solving a recurrence). Proofs, references, and commentary on the current state of the problem (if it is unsolved) are all welcome.

For certain instances of the problem, I have found the value of $\mu(\xi)$ via recurrences. For example, I have proven that $\mu(\sqrt{d})=2\sqrt{d}$ for square-free $d$. I know that this is related to Markov Numbers and the Markov Spectrum, however neither of those answer my question.

This is the same problem as here but I'm looking for the general case.

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Stella Biderman
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2 Answers2

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Little unclear, it is very likely that, given the larger root of $$ A x^2 + B xy + Cy^2, $$ (root meaning either $x/y$ or $y/x$ so it becomes a function of one variable) of discriminant $$ \Delta = B^2 - 4AC > 0, $$ you want $$ \sqrt \Delta $$ Notice that the discriminant of $x^2 - d y^2$ is $4d,$ and the square root of that is $2 \sqrt d.$

Will Jagy
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  • I edited the question which hopefully makes things clearer.

    Is this post a conjecture as to the answer of my problem? Or something else?

     – Stella Biderman Dec 25 '15 at 00:42
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    @StellaBiderman, I think it is the answer. Note that the Golden Ratio solves $x^2 + xy - y^2$ of discriminant $5;$ divide through by $x^2,$ and call the ratio $r,$ we have $1 + r - r^2 = 0.$ – Will Jagy Dec 25 '15 at 00:45
  • This seems interesting, as it also satisfies the small number of other examples I have. Unfortunately, my set of solved examples is pretty small due to a heavy reliance on the continued fraction expansion. – Stella Biderman Dec 25 '15 at 00:49
  • @StellaBiderman, do me a favor, solve your problem for the larger root of $1 + 3 r - 2 r^2 = 0.$ The form $x^2 + 3 xy - 2 y^2$ is Lagrange reduced, meaning the continued fraction is purely periodic. Discriminant is $17.$ – Will Jagy Dec 25 '15 at 00:52
  • Sorry I lost Internet connection. Numerically I'm getting 4.12426, though I don't have access to a CAS right now and I wrote a program to solve it on my brother's TI-84 (ain't the holidays marvelous) – Stella Biderman Dec 25 '15 at 01:49
  • @Stella $\sqrt {17} \approx 4.123.$ Meanwhile, the continued fraction for $\frac{3 + \sqrt {17}}{4}$ is $1,1,3,1,1,3,1,1,3,1,1,3$ and so on – Will Jagy Dec 25 '15 at 02:09
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    @StellaBiderman I think you can do it. Be aware that this is all special to "quadratic irrationals." An inexpensive modern book that covers what else is known http://www.cambridge.org/us/academic/subjects/mathematics/number-theory/neverending-fractions-introduction-continued-fractions?format=PB – Will Jagy Dec 25 '15 at 04:26
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Because I already think this is what you are dealing with: given a binary form $A x^2 + B xy + C y^2$ with positive $\Delta = B^2 - 4 AC,$ not a square but permitted square factors.

If you have a single solution to $A x^2 + B xy + C y^2 = W,$ there are infinitely many, and $|x|,|y|$ get arbitrarily large while $W$ stays fixed.

Take any of the infinitely many solutions to $$ \tau^2 - \Delta \sigma^2 = 4. $$ Create the matrix $$ \left( \begin{array}{rr} \frac{\tau - B \sigma}{2} & -C \sigma \\ A \sigma & \frac{\tau + B \sigma}{2} \end{array} \right) $$

Given a solution $(x,y)$ to $A x^2 + B xy + C y^2 = W,$ we get infinitely many with $$ (x,y) \mapsto \left(\frac{\tau - B \sigma}{2} x -C \sigma y, \; \; A \sigma x + \frac{\tau + B \sigma}{2} y \right) $$

Oh, if we take the smallest $\sigma > 0$ in $ \tau^2 - \Delta \sigma^2 = 4, $ and call the resulting matrix $M,$ then all those matrices can be written as $M^n$ for $n$ a positive or negative integer.

Will Jagy
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