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We know that any positive real number $x$ can be represented as a simple continued fraction $$x = a_{0} + \dfrac{1}{a_{1} + \dfrac{1}{a_{2} + \dfrac{1}{a_{3} + \cdots}}} = [a_{0}, a_{1}, a_{2}, a_{3}, \ldots]$$ where $a_{0}$ is a non-negative integer and $a_{1}, a_{2}, a_{3}, \ldots$ are positive integers. Moreover this representation is unique when $x$ is an irrational number. The integers $a_{i}$ will be called terms of the simple continued fraction.

Of special interest are the numbers whose simple continued fraction consists of terms in arithmetic progression (after a certain point). Two examples are $$\frac{e - 1}{2} = [0, 1, 6, 10, 14, \dots]$$ and $$\frac{e^{2} - 1}{2} = [3, 5, 7, 9, \dots]$$

Hurwitz proved in 1891 that:

If $x, y$ are two positive numbers whose simple continued fractions have terms in arithmetic progression (after a certain point) then there is no non-trivial bi-linear relation of the form $$y = \frac{Ax + B}{Cx + D}$$ with integer coefficients $A, B, C, D$ unless the terms of both the continued fractions belong to the same arithmetic progression (i.e. after a certain point the terms in the continued fraction are equal).

I need a proof of the above result. Full context of this problem is in this answer.

Update: It is easy to observe that if $x = [a_{0}, a_{1}, a_{2}, \dots], y = [b_{0}, b_{1}, b_{2}, \dots]$ and $a_{n + i} = b_{n + j}$ for all $n$ and some fixed values of $i, j$ then we do have a simple bi-linear relation between $x, y$. I believe the theorem says that only in such case (i.e. when the terms in their continued fraction are ultimately same) there can be a bi-linear relation between $x, y$. I also believe that the theorem holds in general and it is not necessary for the terms in the continued fraction to be in arithmetic progression, but I am not sure.

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Paramanand Singh
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  • Hurwitz's article in german, 31 pages. As you might expect, handling $|AD-BC|\neq1$ requires detailed elaboration, but that should be familiar to those who have dealt with cosets of $\Gamma_0(N)$. After a first glance, I doubt that I could compress the proof to something appropriate for MSE posts, however. – ccorn Aug 18 '15 at 11:15
  • @ccorn: I would be more than happy if you could post the details somewhere and maybe an outline with links here. – Paramanand Singh Aug 18 '15 at 11:17
  • Hurwitz also points out that for $|AD-BC|=1$ (equivalent numbers $x,y$), Lagrange already proved that the tails of their simple continued fractions have to match, without restriction to arithmetic progressions. – ccorn Aug 18 '15 at 14:41
  • Hurwitz then handles $|AD-BC|=n\neq 0$ and considers tails formed by interleaved sequences of values of polynomials. Those he calls arithmetische Reihen der Ordnung $m$, where $m$ is the maximum degree among the involved polynomials. He later specializes on $m\in{0,1}$ which is more in accordance with what we would today call an arithmetic progression, but still allows interleavings of several such progressions. Naive translations would therefore be misleading today. – ccorn Aug 18 '15 at 14:50
  • Another translation issue: More modern literature, e.g. Thron et al., speaks of regular continued fractions instead of simple ones. – ccorn Aug 18 '15 at 15:17

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