Periodic Continued Fractions

Definition 2.1   A periodic continued fraction is a continued fraction $[a_0, a_1, \ldots, a_n, \ldots]$ such that

$$a_n = a_{n+h}$$

for a fixed positive integer $h$ and all sufficiently large $n$. We call $h$ the period of the continued fraction.

Example 2.2   Consider the periodic continued fraction $[1,2,1,2,\ldots] = [\overline{1,2}]$. What does it converge to?

$$[\overline{1,2}] = 1+\frac{1}{2+\frac{1}{1+\frac{1}{2+ \frac{1}{1+\cdots}}}},$$

so if $\alpha=[\overline{1,2}]$ then

$$\alpha = 1 + \frac{1}{2+\alpha}.$$

Thus $2\alpha + \alpha^2 = 2 + \alpha + 1$, so

$$\alpha^2 + \alpha - 3 = 0$$    and $$\quad \alpha = \frac{-1+\sqrt{7}}{2}.$$

Theorem 2.3   An infinite integral continued fraction is periodic if and only if it represents a quadratic irrational.

Proof. ( $\Longrightarrow$) First suppose that

$$[a_0, a_1, \ldots, a_n, \overline{a_{n+1},\ldots, a_{n+h}}]$$

is a periodic continued fraction. Set $\alpha=[a_{n+1},a_{n+2}, \ldots]$. Then

$$\alpha = [a_{n+1},\ldots, a_{n+h}, \alpha],$$

so

$$\alpha = \frac{\alpha p_{n+h} + p_{n+h-1}}{\alpha q_{n+h} + q_{n+h-1}}.$$

(We use that $\alpha$ is the last partial convergent.) Thus $\alpha$ satisfies a quadratic equation. Since the $a_i$ are all integers, the number

$$[a_0, a_1, \ldots ] = [a_0, a_1, \ldots, a_n, \alpha]$$

$$= a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \cdots + \alpha}}$$

can be expressed as a polynomial in $\alpha$ with rational coefficients, so $[a_0, a_1, \ldots]$ also satisfies a quadratic polynomial. Finally, $\alpha\not\in\mathbb{Q}$ because periodic continued fractions have infinitely many terms.

( $\Longleftarrow$) This direction was first proved by Lagrange. The proof is much more exciting! Suppose $\alpha\in\mathbb{R}$ satisfies a quadratic equation

$$a \alpha^2 + b\alpha + c = 0$$

with $a, b, c\in\mathbb{Z}$. Let $[a_0, a_1, \ldots]$ be the expansion of $\alpha$. For each $n$, let

$$r_n = [a_n, a_{n+1}, \ldots],$$

so that

$$\alpha = [a_0, a_1, \ldots, a_{n-1}, r_n].$$

We have

$$\alpha = \frac{r_n p_n + p_{n-1}}{r_n q_n + q_{n-1}}.$$

Substituting this expression for $\alpha$ into the quadratic equation for $\alpha$, we see that

$$A_n r_n^2 + B_n r_n + C_n = 0,$$

where

$$A_n = a p_{n-1}^2 + b p_{n-1} q_{n-1} + c q_{n-1}^2,$$

$$B_n = 2a p_{n-1} p_{n-2} + b(p_{n-1} q_{n-2} + p_{n-2} q_{n-1}) + 2c q_{n-1} q_{n-2},$$

$$C_n = a p_{n-2}^2 + b p_{n-2} q_{n-2} + c p_{n-2}^2.$$

Note that $A_n, B_n, C_n\in\mathbb{Z}$, that $C_n = A_{n-1}$, and that

$$B^2 - 4A_n C_n = (b^2- 4ac)(p_{n-1}q_{n-2} - q_{n-1}p_{n-2})^2 = b^2 - 4ac.$$

Recall from the proof of Theorem 2.3 of the previous lecture that

$$\left\vert \alpha - \frac{p_{n-1}}{q_{n-1}}\right\vert < \frac{1}{q_n q_{n-1}}.$$

Thus

$$\left\vert \alpha q_{n-1} - p_{n-1}\right\vert < \frac{1}{q_n} < \frac{1}{q_{n+1}},$$

so

$$p_{n-1} = \alpha q_{n-1} + \frac{\delta}{q_{n-1}}$$   with $$\vert\delta\vert < 1.$$

Hence

$$A_n = a\left(\alpha q_{n-1} + \frac{\delta}{q_{n-1}}\right)^2 +b\left(\alpha q_{n-1} + \frac{\delta}{q_{n-1}}\right)q_{n-1} +c q_{n-1}^2$$

$$= (a\alpha^2 + b\alpha + c)q_{n-1}^2 + 2a\alpha\delta + a\frac{\delta^2}{q_{n-1}^2} + b\delta$$

$$= 2a\alpha\delta + a\frac{\delta^2}{q_{n-1}^2} + b\delta.$$

Thus

$$\vert A_n\vert = \left\vert 2a\alpha\delta + a\frac{\delta^2}{q_{n-1}^2} + b\delta\right\vert < 2\vert a\alpha\vert + \vert a\vert + \vert b\vert.$$

Thus there are only finitely many possibilities for the integer $A_n$. Also,

$$\vert C_n\vert = \vert A_{n-1}\vert$$    and $$\quad \vert B_n\vert = \sqrt{b^2 - 4(ac-A_n C_n)},$$

so there are only finitely many triples $(A_n, B_n, C_n)$, and hence only finitely many possibilities for $r_n$ as $n$ varies. Thus for some $h>0$,

$$r_n = r_{n+h}.$$

This shows that the continued fraction for $\alpha$ is periodic. $\qedsymbol$