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

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

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

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

If we continue we would get the following expansion:

$$\sqrt{23}=4+{\displaystyle \frac{1}{1+{\displaystyle \frac{1}{3+{\displaystyle \frac{1}{1+{\displaystyle \frac{1}{8+\dots }}}}}}}}$$

The process can be summarised as follows:

$$\begin{array}{rl}{a}_0& =4,\frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}\\ a_1& =1,\frac{7}{\sqrt{23}-3}=\frac{7(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}\\ a_2& =3,\frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}\\ a_3& =1,\frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4\\ a_4& =8,\frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}\\ a_5& =1,\frac{7}{\sqrt{23}-3}=\frac{7(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}\\ a_6& =3,\frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}\\ a_7& =1,\frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4\end{array}$$

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:

$\sqrt{2}=[1;(2)]$, period=$1$
$\sqrt{3}=[1;(1,2)]$, period=$2$
$\sqrt{5}=[2;(4)]$, period=$1$
$\sqrt{6}=[2;(2,4)]$, period=$2$
$\sqrt{7}=[2;(1,1,1,4)]$, period=$4$
$\sqrt{8}=[2;(1,4)]$, period=$2$
$\sqrt{10}=[3;(6)]$, period=$1$
$\sqrt{11}=[3;(3,6)]$, period=$2$
$\sqrt{12}=[3;(2,6)]$, period=$2$
$\sqrt{13}=[3;(1,1,1,1,6)]$, 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?