largest term in simple continued fraction of $\sqrt{n}$

Question

Let $n$ be a positive integer that is not a perfect square. It is well known that the simple continued fraction of $\sqrt{n}$ is of the form $$\sqrt{n} = [a_0;\overline{a_1,a_2,\ldots,a_{k-1},2a_0}],$$ where of course $a_0 = \lfloor \sqrt{n} \rfloor$. My question is, is $2a_0$ the maximum of all of the terms $a_i$ in the given continued fraction? I'm unable to find a counterexample.

Answer 1

That's correct. The partial quotients of a continued fraction expansion of the square root of a non square integer $n$, written $\sqrt{n}=[a_0;\overline{a_1, a_2, ..., a_{k-1},2a_0}]$, with primitive period $k_p$, have the property $a_i\leqslant a_0$ for $k_p\nmid i$. For non primitive period, $a_i\leqslant 2a_0$, when $i\geqslant 0$.

Proof. Let the complete quotients of the expansion of the continued fraction of $\sqrt{n}$ be denoted $r_h=\dfrac{\sqrt{n}+A_{h-1}}{\alpha_{h-1}}$, with $r_0=\sqrt{n}$. Using the following properties ($h\geqslant 0$, $j\geqslant -1$):

$\;\;\;\;\;\;\;\;(1)\;\;\;\;\;\;\;\;A_j\leqslant a_0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$(Perron p.76 formula 7)

$\;\;\;\;\;\;\;\;(2)\;\;\;\;\;\;\;\;A_{h-1}+A_h=a_h\alpha_{h-1}\;\;\;\;\;$ (Perron p.83 formula 4)

$\;\;\;\;\;\;\;\;(3)\;\;\;\;\;\;\;\;\alpha_{i-1}\geqslant 2 \text{ for } 0<i<k_p\;\;$ (Perron p.93 formula 5)

(Note:above indices are shifted for $A_h$ and $\alpha_h$ compared to Perron's notations)

(2) and (1) gives $a_h\leqslant \dfrac{2a_0}{\alpha_{h-1}}$, and using (3) we get $a_i\leqslant a_0$ for $0<i<k_p$ and by periodicity, for all $a_i$ with $k_p\nmid i$. And since all $\alpha_{h-1}$ for $h\geqslant0$ are positive integers, we have $a_h\leqslant 2a_0$ for $h\geqslant0$, the period being primitive or not.

Die Lehre von den Kettenbrüchen (Perron)