0%

Problem 64


Problem 64


Odd period square roots

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

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

For example, let us consider √23:

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

If we continue we would get the following expansion:

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

The process can be summarised as follows:

$$a_0=4 \text{, } \frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}$$
$$a_1=1 \text{, } \frac{7}{\sqrt{23}-3}=\frac{7(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}$$
$$a_2=2 \text{, } \frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}$$
$$a_3=1 \text{, } \frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4$$
$$a_4=8 \text{, } \frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}$$
$$a_5=1 \text{, } \frac{7}{\sqrt{23}-3}=\frac{7(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}$$
$$a_6=3 \text{, } \frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}$$
$$a_7=1 \text{, } \frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4$$

It can be seen that the sequence is repeating. For conciseness, we use the notation √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:

√2=[1;(2)], period=1
√3=[1;(1,2)], period=2
√5=[2;(4)], period=1
√6=[2;(2,4)], period=2
√7=[2;(1,1,1,4)], period=4
√8=[2;(1,4)], period=2
√10=[3;(6)], period=1
√11=[3;(3,6)], period=2
√12= [3;(2,6)], period=2
√13=[3;(1,1,1,1,6)], period=5

Exactly four continued fractions, for N ≤ 13, have an odd period.

How many continued fractions for N ≤ 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:

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

例如,让我们考虑√23:

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

如果我们继续这个过程,我们会得到如下的展开:

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

这个过程可以总结如下:

$$a_0=4 \text{, } \frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}$$
$$a_1=1 \text{, } \frac{7}{\sqrt{23}-3}=\frac{7(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}$$
$$a_2=2 \text{, } \frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}$$
$$a_3=1 \text{, } \frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4$$
$$a_4=8 \text{, } \frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}$$
$$a_5=1 \text{, } \frac{7}{\sqrt{23}-3}=\frac{7(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}$$
$$a_6=3 \text{, } \frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}$$
$$a_7=1 \text{, } \frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4$$

可以看出序列正在重复。我们将其简记为√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:

√2=[1;(2)],周期=1
√3=[1;(1,2)],周期=2
√5=[2;(4)],周期=1
√6=[2;(2,4)],周期=2
√7=[2;(1,1,1,4)],周期=4
√8=[2;(1,4)],周期=2
√10=[3;(6)],周期=1
√11=[3;(3,6)],周期=2
√12= [3;(2,6)],周期=2
√13=[3;(1,1,1,1,6)],周期=5

在N ≤ 13中,恰好有4个连分数表示的周期是奇数。

在N ≤ 10000中,有多少个连分数表示的周期是奇数?