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Odd Period Square Roots

 Published on Friday, 27th February 2004, 06:00 pm; Solved by 25314;
Difficulty level: 3

Problem 64

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

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

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

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

If we continue we would get the following expansion:

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

The process can be summarised as follows:

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

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 $N \le 10\,000$ have an odd period?