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, $a_{1},\ldots,a_{2\ell-1}$, are symmetric about the $\ell$-th partial denominator, $a_{\ell}$. I.e., $a_{\ell-j}=a_{\ell+j}$ for $j=1,\ldots,\ell-1$. 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.

  • $\begingroup$ What you are calling "partial denominator" is usually called "partial quotient". $\endgroup$ – Gerry Myerson Dec 6 '18 at 11:08
  • $\begingroup$ @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 Dec 6 '18 at 18:26

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