# 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.

###### 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, 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. Inductively, we have $-1 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, 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 PurelyPeriodicContinuedFractions 2013-03-22 18:04:44 2013-03-22 18:04:44 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 11Y65 msc 11A55