purely periodic continued fractions

We know that periodic continued fractions represent quadratic irrationals; this article characterizes purely periodic continued fractions. We will use freely the results on convergents to a continued fraction.

Theorem 1.

(Galois) A quadratic irrational $t$ is represented by a purely periodic simple continued fraction if and only if $t>1$ and its conjugate $s$ under the transformation $\sqrt{d}\mapsto-\sqrt{d}$ satisfies $-1<s<0$ .

Proof.

Suppose first that $t$ is represented by a purely periodic continued fraction

t=[\overline{a_{0},a_{1},\ldots,a_{r-1}}].

Note that $a_{0}\geq 1$ since it appears again in the continued fraction. Thus $t>1$ . The $r^{\mathrm{th}}$ complete convergent is again $t$ , so that we have

t=\frac{p_{r-2}+tp_{r-1}}{q_{r-2}+tq_{r-1}}

so that

q_{r-1}t^{2}+(q_{r-2}-p_{r-1})t-p_{r-2}=0

Consider the polynomial $f(x)=q_{r-1}x^{2}+(q_{r-2}-p_{r-1})x-p_{r-2}$ . $f(t)=0$ , so the other root of $f(x)$ is the conjugate $s$ of $t$ . But $f(-1)=(p_{r-1}-p_{r-2})+(q_{r-1}-q_{r-2})>0$ since the $p_{i}$ and the $q_{i}$ are both strictly increasing sequences, while $f(0)=-p_{r-2}<0$ . Thus $s$ lies between $-1$ and $0$ and we are done.

Now suppose that $t>1$ and $-1<s<0$ , and let the continued fraction for $t$ be $[a_{0},a_{1},\ldots]$ . Let $t_{n}$ be the $n^{\mathrm{th}}$ complete convergent of $t$ , and $s_{n}=\overline{t_{n}}$ . Thus $s_{0}=s$ . Then

t=t_{0}=a_{0}+\frac{1}{t_{1}}

so that

s_{0}=\overline{t_{0}}=a_{0}+\frac{1}{\overline{t_{1}}}=a_{0}+\frac{1}{s_{1}}

and thus

\frac{1}{s_{1}}=-a_{0}+s_{0}<-a_{0}\leq-1

so that $-1<s_{1}<0$ . Inductively, we have $-1<s_{n}=\overline{t_{n}}<0$ for all $n\geq 0$ . Suppose now that the continued fraction for $t$ is not purely periodic, but rather has the form

t=[a_{0},a_{1},\ldots,a_{k-1},\overline{a_{k},a_{k+1},\ldots,a_{k+j-1}}]

for $k\geq 1$ . Then $t_{k}=t_{k+j}$ and so

t_{k-1}-t_{k+j-1}=\left(a_{k-1}+\frac{1}{t_{k}}\right)-\left(a_{k+j-1}+\frac{1% }{t_{k+j}}\right)=a_{k-1}-a_{k+j-1}

But $a_{k-1}\neq a_{k+j-1}$ , otherwise $a_{k-1}$ would have been the first element of the repeating period. Thus $t_{k-1}-t_{k+j-1}$ is a nonzero integer and thus $s_{k-1}-s_{k+j-1}$ is as well. But $-1<s_{k-1}-s_{k+j-1}<1$ , which is a contradiction. Thus $k=0$ and the continued fraction is purely periodic. ∎

References

1 A.M. Rockett & P. Szüsz, Continued Fractions, World Scientific Publishing, 1992.

Title	purely periodic continued fractions
Canonical name	PurelyPeriodicContinuedFractions
Date of creation	2013-03-22 18:04:44
Last modified on	2013-03-22 18:04:44
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 11Y65
Classification	msc 11A55