I came across the Wiki-proof sites and I wanted to find out a proof that the continued fraction of $\sqrt{N}$, where $N$ is a positive integer, not being a perfect square, has the form $[a,\overline {r_1,r_2,\cdots,r_n,2a}]$

In Wiki-proof, I only found the proof that the continued fraction expansion must eventually become periodic. But two things are missing :

How can I prove that the period starts immediately after the first entry $a=\lfloor\sqrt{N}\rfloor$ ?

How can I prove that the period ends with the entry $2a$ and that the corresponding convergent $\frac{p}{q}$ is the fundamental solution of the Pell-equation $p^2-Nq^2=\pm1$ ?

Please answer with a proof directly written down and not just with a link!


I will answer the first two parts of your question, i.e., $\sqrt N$ has a continued fraction of the form $[\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, 2\lfloor \sqrt N \rfloor}]$.

We will assume that we already know that a quadratic irrational number has a periodic continued fraction.

First, we need the following theorem of Galois, the proof of which I take from Theorem 7.20 in I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991:

Theorem of Galois. The continued fraction of the quadratic irrational number $\xi$ is purely periodic if $\xi > 1$ and $-1 < \xi' < 0$, where $\xi'$ is the conjugate of $\xi$.

Proof: Recall the inductive definition for expanding $\xi = \xi_0$ into a continued fraction: $$a_i = \lfloor \xi_i \rfloor, \qquad \xi_{i+1} = \frac{1}{\xi_i - a_i}.$$

Invert and take conjugates of the latter equation to obtain $$\frac{1}{\xi'_{i+1}} = \xi'_i - a_i.$$ Now $a_i \ge 1$. Hence, if $\xi'_i < 0$, then $1/\xi'_{i+1} < -1$, and we have $-1 < \xi'_{i+1} < 0$. Because $-1 < \xi'_0 < 0$ by hypothesis, we see by mathematical induction that $-1 < \xi'_i < 0$ is true for all $i$. Then by the previous equation, $$0 < -\frac{1}{\xi'_{i+1}} - a_i < 1, \qquad a_i= \left\lfloor -\frac{1}{\xi'_{i+1}} \right\rfloor.$$ Because $\xi$ is a quadratic irrational number, it has a periodic continued fraction, so $\xi_j = \xi_k$ for some integers $j$ and $k$ with $0 < j < k$. Then we also have $\xi'_j = \xi'_k$, and

\begin{align} a_{j-1} & = \left\lfloor -\frac{1}{\xi'_j} \right\rfloor = \left\lfloor -\frac{1}{\xi'_k}\right\rfloor = a_{k-1}\\ \xi_{j-1} & = a_{j-1} + \frac{1}{\xi_j} = a_{k-1} + \frac{1}{\xi_k} = \xi_{k-1}. \end{align} Hence, $\xi_j = \xi_k$ implies $\xi_{j-1} = \xi_{k-1}$. A $j$-fold iteration of that implication gives $\xi_0 = \xi_{k-j}$, and we have $\xi = \xi_0 = [\overline{a_0; a_1, \ldots, a_{k-j-1}}]$, which completes the proof of the theorem of Galois.

The rest of the answer comes from Proposition 32 of "Continued Fractions, Pell's Equation, and Transcendental Numbers" by Jeremy Booher:

Because $$\frac{1}{\sqrt N - \lfloor \sqrt N \rfloor} > 1 \qquad \text{and} \qquad -1 < \frac{1}{-\sqrt N - \lfloor \sqrt N \rfloor} < 0,$$ the continued fraction of $\upsilon_1 = 1/(\sqrt N - \lfloor \sqrt N \rfloor)$ is purely periodic by the theorem of Galois, so $\upsilon_0 = \sqrt N$ has the form $[\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, a_{n+1}}]$ where $n+1$ denotes the length of the shortest period in that expansion. (I am using the notation $\upsilon_0$ and $\upsilon_1$ because $\upsilon_0 = \sqrt N$ and $\upsilon_1 = 1/(\sqrt N - \lfloor \sqrt N \rfloor)$ are related by the inductive definition above for expanding $\upsilon_0$ into a continued fraction.) Also, $\sqrt N + \lfloor \sqrt N \rfloor = [2\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, a_{n+1}}]$. Moreover, $$\sqrt N + \lfloor \sqrt N \rfloor > 1, \qquad \text{and} \qquad -1 < \lfloor \sqrt N \rfloor - \sqrt N < 0,$$ so the continued fraction of $\sqrt N + \lfloor \sqrt N \rfloor$ is purely periodic by the theorem of Galois, which means that the period (of length $n + 1$) starts at the beginning of the expansion. We therefore have more ways to express the continued fraction for $\sqrt N + \lfloor \sqrt N \rfloor$: \begin{align} \sqrt N + \lfloor \sqrt N \rfloor & = [2\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, a_{n+1}}]\\ & = [\overline{2\lfloor \sqrt N \rfloor; a_1, a_2, \ldots, a_n}, a_{n+1}, \ldots]\\ & = [\overline{2\lfloor \sqrt N \rfloor; a_1, a_2, \ldots, a_n}]. \end{align} We see from the last two continued fractions that $a_{n+1}$ begins the second period, so $a_{n+1}$ has the same value as the first integer of the first period, i.e., $a_{n+1} = 2\lfloor \sqrt N \rfloor$. Substituting for $a_{n+1}$ in the continued fraction for $\sqrt N$ concludes the proof: $$\sqrt N = [\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, 2\lfloor \sqrt N \rfloor}].$$

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  • $\begingroup$ Nice answer (+1) $\endgroup$ – Peter May 30 '17 at 8:01

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