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Suppose $d$ is a positive integer that is not a perfect square such that the negative Pell equation, $x^{2}-dy^{2}=-1$ has no solution. Then we know the minimal period of the continued fraction expansion of $\sqrt{d}$ has even length, $2\ell$, and that the partial denominators, $a_{1},\dotsc,a_{2\ell-1}$, are symmetric about the $\ell$-th partial denominator, $a_{\ell}$. I.e., $a_{\ell-j}=a_{\ell+j}$ for $j=1,\dotsc,\ell-1$. It is this partial denominator, $a_{\ell}$, that I am referring to here as the middle partial denominator.

It is a classical result that $a_{2\ell}=2a_{0}$, but my question is what is known about the middle partial denominator, $a_{\ell}$, under the above conditions on $d$.

Of course, if $d=a_{0}^{2}+2$, where $a_{0}$ is a positive integer, for example, then we know that the continued fraction expansion of $\sqrt{d}$ takes the form $[a_{0}; \overline{a_{0},2a_{0}}]$, so the middle partial denominator is $a_{0}$ here. But are there more general results known that do not depend on $d$ satisfying such quadratic expressions that have “nice” continued fraction expansions?

Any known results with references, ideas,… would be greatly appreciated.

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  • $\begingroup$ What you are calling "partial denominator" is usually called "partial quotient". $\endgroup$ Commented Dec 6, 2018 at 11:08
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    $\begingroup$ @GerryMyerson I thought so too, but both wikipedia and wolfram seemed insistent across their continued-fraction-related pages that partial denominator was the term, so I just followed their lead. Happy to change it, especially if it helps get me a good answer here. $\endgroup$
    – user132145
    Commented Dec 6, 2018 at 18:26

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A quick search and you can find this very recent (working) paper which shows (theorem 3.4) that "the partial quotients $a_i$ for $0<i<n$, and by symmetry for $\ell-n<i<\ell$ can be given any positive integer value for a solution to exist. The only restriction, if any, is on the parity of the central quotient $a_n$". Note that here, the period is written $\ell$ and not $2\ell$.

It first recalls something shown by Muir, Perron, Friesen, that there is an integer solution $D$ to equation $\sqrt{D}=[a_0;\overline{a_1, a_2, \dotsc, a_{\ell-1},2a_0}]$ if and only if the quantity $q_{\ell-2}q_{\ell-3}^{+1}$ is even (where $q_i$ is the denominator of the $i^\text{th}$ convergent of $\sqrt{D}$).

Note from the paper: "The superscript notation $q_{v}\strut^{{}_{+\mu}}$ used in this paper is equivalent to Perron's notation $B_{v,\mu}$ (reference in the paper) where $\mu$ is what he calls an indice shift. They both translate to the simple continuant $K(a_{1+\mu},\ldots,a_{v+\mu})$"

It then uses lemma 3.2 and 3.3 that shows the following (which tell what are the parity constraint on $a_n$):

If $\ell=2n+1$, $n\in\mathbb{Z}^+$, then the following two assertions are equivalent

  1. $q_{\ell-2}q_{\ell-3}^{+1} \equiv 1 \pmod{2}$
  2. one of these statements is true:
    a) $q_{n-2} \equiv 0 \pmod{2}$ and $a_{n}\equiv 1 \pmod{2}$
    b) $q_{n-1}q_{n-2} \equiv 1 \pmod{2}$ and $a_{n}\equiv 0 \pmod{2}$

If $\ell=2n$, $n\in\mathbb{Z}^+$, then the following two assertions are equivalent

  1. $q_{\ell-2}q_{\ell-3}^{+1} \equiv 1 \pmod{2}$
  2. $q_{n-1} \equiv 0 \pmod{2}$ and $a_{n}\equiv 1 \pmod{2}$

and after pointing out that a) and b) are mutually exclusive, theorem 3.4 comes to the conclusion mentioned in the begining. Note: I know this is not a peer reviewed paper, but that part is pretty basic and easy to validate.

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  • $\begingroup$ I don't know what the $+1$ means in $q_{\ell-3}^{+1}$. $\endgroup$ Commented Aug 3 at 9:55
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    $\begingroup$ @Gerry Myerson Indeed, and that's not a common notation. Don't know why I forgot to mention that. $\endgroup$
    – Collag3n
    Commented Aug 3 at 13:37

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