statement.html

problem 64 · 41 lines

<div class="problem_content" role="problem">
<p>All square roots are periodic when written as continued fractions and can be written in the form:</p>

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

<p>For example, let us consider $\sqrt{23}:$</p>
$$\sqrt{23} = 4 + \sqrt{23}-4=4 + \dfrac 1 {\dfrac 1 {\sqrt{23}-4}} = 4+\dfrac 1  {1 + \dfrac{\sqrt{23}-3}7}$$

<p>If we continue we would get the following expansion:</p>

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

<p>The process can be summarised as follows:</p>

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


<p>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.</p>
<p>The first ten continued fraction representations of (irrational) square roots are:</p>
<p>
$\quad \quad \sqrt{2}=[1;(2)]$, period=$1$<br/>
$\quad \quad \sqrt{3}=[1;(1,2)]$, period=$2$<br/>
$\quad \quad \sqrt{5}=[2;(4)]$, period=$1$<br/>
$\quad \quad \sqrt{6}=[2;(2,4)]$, period=$2$<br/>
$\quad \quad \sqrt{7}=[2;(1,1,1,4)]$, period=$4$<br/>
$\quad \quad \sqrt{8}=[2;(1,4)]$, period=$2$<br/>
$\quad \quad \sqrt{10}=[3;(6)]$, period=$1$<br/>
$\quad \quad \sqrt{11}=[3;(3,6)]$, period=$2$<br/>
$\quad \quad \sqrt{12}=[3;(2,6)]$, period=$2$<br/>
$\quad \quad \sqrt{13}=[3;(1,1,1,1,6)]$, period=$5$
</p>
<p>Exactly four continued fractions, for $N \le 13$, have an odd period.</p>
<p>How many continued fractions for $N \le 10\,000$ have an odd period?</p>
</div>