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In this question on math.stackexchange.com I have made two conjectures the first of which I have proved. The second has not been settled. I post it here to seek a proof.


Given a quadratic surd $\sqrt d$ where $d$ is a natural number and not a perfect square. $(c_i)_{i=1}^\infty$ is the sequence of convergents of the continued fraction of $x$. Let $r_i:=\frac{c_{i+1}-\sqrt d}{c_i-\sqrt d},\,\forall i\in\mathbf N$. Let $n$ be the period of the continued fraction. It has been shown $\exists \,l_r:=\lim_\limits{i\rightarrow\infty}r_{in+r},\, \forall r\in\{0,1,\cdots,n-1\}$. Is the following statement, suggested by a numerical experiment, true?

  1. For $n\le 2$, $l_0=l_1$.

  2. For $n\ge 3$, there exists at least two distinct $l_r$'s.

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Proof to conjecture $1.$:

Define $$P(x):=\frac{\lfloor\sqrt d\rfloor x+d}{x+\lfloor\sqrt d\rfloor}.$$ Obviously $P_d(\sqrt d)=\sqrt d$.

By Theorem Plus of Square Root, Continued Fractions and the Orbit of $\frac10$ on $\partial H^2$ by Julian Rose, Krishna Shankar and Justin Thomas. The continued fraction of $\sqrt d$ is of period at most $2$ iff $P^i(\infty)=c_i$ for all convergents $\{c_i\}_{i=1}^\infty$.

$$\frac{P(c_i)-P(\sqrt d)}{c_i-\sqrt d}=-\frac{\sqrt d-\lfloor\sqrt d\rfloor}{c_i+\lfloor\sqrt d\rfloor}.$$ So $$l_0=l_1=-\frac{\sqrt d-\lfloor\sqrt d\rfloor}{\sqrt d+\lfloor\sqrt d\rfloor}.$$

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