Do all numbers of the form $\sqrt{n}$ have a periodic continued fraction in the form $[a_0; \overline{a_1, a_2, \cdots, a_k}]$??

Question

After trying out This question, I noticed that the continued fraction of $\sqrt{n}$ where $n$ is a non-square integer is in the form $[a_0; \overline{a_1, a_2, \cdots, a_k}]$, that is to say, that they are in the form $$a_0 + \cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{\ddots}{\quad\cfrac{\ddots}{\qquad a_k+\cfrac{1}{a_1+\cfrac{1}{a_2+\ddots}}}}}}$$

I was able to show this was true for $n\leq10000$. However, I'm not sure how to show this is true.

Answer 1

Taking $$ a_0 = \left\lfloor \sqrt n \right\rfloor \; . $$ There is a characterization of purely periodic continued fractions; this is due to Galois. Namely, if $P,Q,D$ are integers, $D>0$ is not a square, define $$ \zeta = \frac{P + \sqrt N}{Q} $$ and call its conjugate $$ \eta = \frac{P - \sqrt N}{Q} $$ The first result is that $\zeta$ has a purely periodic continued fraction if and only if it is reduced, which means both $$ \zeta > 1 \; , \; \; \; \; \; -1 < \eta < 0 $$ In your example you have $Q=1,$ $$ a_0 < \sqrt n < 1 + a_0, $$ so $$ a_0 - \sqrt n < 0 < 1 + a_0 - \sqrt n, $$ the right hand part giving $$ -1 < a_0 - \sqrt n, $$ The conclusion is that $a_0 + \sqrt n$ has a purely periodic continued fraction. As this differs in only the first entry, we see that the fraction for $\sqrt n$ begins being purely periodic after just the initial partial quotient.

Answer 2

The general result is usually referred to Lagrange. Your observation is due to Galois by writing $$\sqrt{n}=[n]+\frac 1 t,$$ where $$t=\frac 1{\sqrt{n}-[n]}>1$$ and its Galois conjugate $s=1/(-\sqrt{n}-[n])$ satisfies $$-1<s<0.$$ See a proof here: https://planetmath.org/purelyperiodiccontinuedfractions