Problem 64

Odd period square roots

All square roots are periodic when written as continued fractions and can be written in the form:

$\quad \quad \sqrt{N}=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+…}}}$

For example, let us consider $\sqrt{23}$:

$\quad \quad \sqrt{23}=4+\sqrt{23}-4=4+\frac{1}{\frac{1}{\sqrt{23}-4}}=4+\frac{1}{1+\frac{\sqrt{23}-3}{7}}$

If we continue we would get the following expansion:

$\quad \quad \sqrt{23}=4+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{8+…}}}}$

The process can be summarised as follows:

$\quad \quad a_0=4, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7$

$\quad \quad a_1=1, \frac 7 {\sqrt{23}-3}=\frac {7(\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2$

$\quad \quad a_2=3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7$

$\quad \quad a_3=1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} 7=8+\sqrt{23}-4$

$\quad \quad a_4=8, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7$

$\quad \quad a_5=1, \frac 7 {\sqrt{23}-3}=\frac {7 (\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2$

$\quad \quad a_6=3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7$

$\quad \quad a_7=1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} {7}=8+\sqrt{23}-4$

It can be seen that the sequence is repeating. For conciseness, we use the notation $\sqrt{23} = [4;(1,3,1,8)]$, to indicate that the block $(1,3,1,8)$ repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:

$\quad \quad \sqrt{2}=[1;(2)]$, period=$1$

$\quad \quad \sqrt{3}=[1;(1,2)]$, period=$2$

$\quad \quad \sqrt{5}=[2;(4)]$, period=$1$

$\quad \quad \sqrt{6}=[2;(2,4)]$, period=$2$

$\quad \quad \sqrt{7}=[2;(1,1,1,4)]$, period=$4$

$\quad \quad \sqrt{8}=[2;(1,4)]$, period=$2$

$\quad \quad \sqrt{10}=[3;(6)]$, period=$1$

$\quad \quad \sqrt{11}=[3;(3,6)]$, period=$2$

$\quad \quad \sqrt{12}=[3;(2,6)]$, period=$2$

$\quad \quad \sqrt{13}=[3;(1,1,1,1,6)]$, period=$5$

Exactly four continued fractions, for $N \le 13$, have an odd period.

How many continued fractions for $N \le 10000$ have an odd period?

奇周期平方根

所有的平方根写成如下连分数表示时都是周期性重复的：

$\quad \quad \sqrt{N}=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+…}}}$

例如，让我们考虑$\sqrt{23}$：

$\quad \quad \sqrt{23}=4+\sqrt{23}-4=4+\frac{1}{\frac{1}{\sqrt{23}-4}}=4+\frac{1}{1+\frac{\sqrt{23}-3}{7}}$

如果我们继续展开，会得到：

$\quad \quad \sqrt{23}=4+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{8+…}}}}$

这个过程可以总结如下：

可以看出序列正在重复。我们将其简记为$\sqrt{23} = [4;(1,3,1,8)]$，表示在此之后$(1,3,1,8)$无限循环。

前$10$个（无理数）平方根的连分数表示是：

$\quad \quad \sqrt{2}=[1;(2)]$，周期为$1$

$\quad \quad \sqrt{3}=[1;(1,2)]$，周期为$2$

$\quad \quad \sqrt{5}=[2;(4)]$，周期为$1$

$\quad \quad \sqrt{6}=[2;(2,4)]$，周期为$2$

$\quad \quad \sqrt{7}=[2;(1,1,1,4)]$，周期为$4$

$\quad \quad \sqrt{8}=[2;(1,4)]$，周期为$2$

$\quad \quad \sqrt{10}=[3;(6)]$，周期为$1$

$\quad \quad \sqrt{11}=[3;(3,6)]$，周期为$2$

$\quad \quad \sqrt{12}=[3;(2,6)]$，周期为$2$

$\quad \quad \sqrt{13}=[3;(1,1,1,1,6)]$，周期为$5$

在$N \le 13$中，恰好有$4$个连分数表示的周期是奇数。

在$N \le 10000$中，有多少个连分数表示的周期是奇数？