Question: Given a quadratic irrational $x = a + b\sqrt{D}$ ($a,b \in \Bbb{Q}$, $D \in \Bbb{N}_{> 0}$ square-free) and its Galois conjugate $x' = a - b\sqrt{D}$, is it true that the continued fraction expansions of $x$ and $x'$ have the same period?

Computations of a few random example seems to suggest that is indeed the case, and I think I even saw a statement to this effect somewhere, but can't remember where.

Motivation: The continued fraction expansions of two numbers $x,x' \in \Bbb{R}$ conjugate via an element of $\text{PGL}_2(\Bbb{Z})$ are 'almost the same', i.e. $a_{k+n} = a'_{k+m}$ for some $n,m$ and all $k \geq 0$, and I suspect that in the case of two conjugate quadratic-irrationals there always exists a 'generalized reflection' $s_H = wsw^{-1}$, $w \in \text{PGL}_2(\Bbb{Z})$, $s \in \left\{\begin{bmatrix} & 1 \\ 1& \end{bmatrix}, \begin{bmatrix} -1 & 1 \\ & 1\end{bmatrix}, \begin{bmatrix} -1 & \\ & 1\end{bmatrix}\right\}$ such that $s_H(x) = x'$.


The periodic part of the continued fraction for the Galois conjugate is always the mirror image of the periodic part for the original quadratic irrational. Here we are viewing the periodic part as a cycle arranged in a circle, and the mirror image is with respect to reflecting across a suitable diameter of this circle.

The periodic part of every irrational number $\sqrt{p/q}$ (with $p$ and $q$ integers) is always palindromic in the sense that its cyclic arrangement has mirror symmetry. This follows from the classical fact that the numbers $\sqrt{p/q}$ with $p/q>1$ are exactly the numbers whose continued fraction has the form $[a_0;\overline{a_1,\cdots,a_n}]$ with the terms $a_1,\cdots,a_{n-1}$ forming a palindrome and $a_n=2a_0$. Inserting the last term $a_n$ into the palindromic cycle formed by $a_1,\cdots,a_{n-1}$ then produces a cycle with mirror symmetry across the diameter through $a_n$.

All this becomes visually clear using Conway's notion of topographs for binary quadratic forms over the integers. One place to read about this viewpoint is in Chapter 4 of my unfinished book "Topology of Numbers", available on my webpage.

  • $\begingroup$ My computations semm to suggest that such cyclic mirror symmetry also occurs for irrationals of the form $a+b\sqrt{p}$ with $p$ prime. Either that, or I‘m an extremely bad RNG. $\endgroup$ May 25 '18 at 14:43
  • $\begingroup$ @ Nicolas Schmidt: Adding the integer $a$ just changes the initial term of the continued fraction and doesn't affect the periodic part. The term $b\sqrt p$ can be written as $\sqrt{b^2p}$ so this becomes a special case of the situation described in my answer. $\endgroup$ May 25 '18 at 23:14
  • $\begingroup$ $a$ can be any rational number, not necessarily an integer, as far as I tested. $\endgroup$ May 25 '18 at 23:43

No. For example:

$-\frac{2}{3} + \frac{5}{7}\sqrt{6} = [1; \overline{12, 18, 1, 32, 1, 1, 2, 171, 15, 3, 1, 1, 1, 18, 2, 2, 2, 3}]$


$-\frac{2}{3} - \frac{5}{7}\sqrt{6} = [-3; 1, 1, \overline{2, 2, 18, 1, 1, 1, 3, 15, 171, 2, 1, 1, 32, 1, 18, 12, 3, 2}]$

As it turned out, my "random" examples all had $D = p$ a prime, and in that case the periods are (always?) palindromic!

  • $\begingroup$ As Allen Hatcher wrote, "The periodic part of the continued fraction for the Galois conjugate is always the mirror image of the periodic part for the original quadratic irrational." Why your answer is NO? $\endgroup$ Feb 21 '20 at 12:45
  • $\begingroup$ My question (rephrased) was: are the expansion always palindromic? That's not the case as the above example shows; however, it's true for a large class of quadratic irrationals, namely all the pure square roots as Allen Hatcher wrote. $\endgroup$ Mar 26 '20 at 11:25

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