Exercise 7.9.1 (Continued fraction expansion of $\sqrt{73}$)

Question

Continue the calculations started above for $\sqrt{73}$, and verify the continued fraction expansion given in Example 3 in the preceding section.

Richard Ganaye · Accepted Answer

\begin{proof} 
Using the method ``continued\_fraction'' given in Problem 7.7.3, and adding some instructions ``print'', we obtain
\begin{align*}
m_0 = 0, \quad q_0 = 1, \quad &a_0 = 8,\
m_1 = 8, \quad q_1 = 9, \quad &a_1 = 1,\
m_2 = 7, \quad q_2 = 3, \quad &a_2 = 5,\
m_3 = 8, \quad q_3 = 3, \quad &a_3 = 5,\
m_4 = 7, \quad q_4 = 8, \quad &a_4 = 1,\
m_5 = 1, \quad q_5 = 9, \quad &a_5 = 1,\
m_6 = 8, \quad q_6 = 1, \quad &a_6 =1 ,\
 &a_7 = 16.
\end{align*}
(The algorithm is terminated when $q_6 = 1$.)

So
$$\sqrt{73} = \langle 8, \overline{1,1,5,5,1,1,16} angle.$$

The convergents are given by formulas (7.6):
\begin{center}
\begin{tabular}{c|cccccccccc}
    $i$& -2 & -1 & 0 & 1 & 2&3 &4 &5& 6&7\
\hline
$a_i$ &    &    &8&  1  & 1 & 5 & 5 & 1 & 1 & 16\
$h_i$ & 0 & 1 & 8&  9&17 &94&487&581&1068& $\cdots$ \
$k_i$ & 1 & 0  & 1& 1&  2 &11&57&68&125&$\cdots$
\end{tabular}
\end{center}
So by Theorem 7.22, $h_6^2 - 73 k_6^2 = -1$, that is
$$1068^2 - 73 \cdot 125^2 = -1.$$
(This solution is also obtained by the method ``pell\_fermat'' given in Problem 7.8.9.)

\end{proof}

Exercise 7.9.1 (Continued fraction expansion of $\sqrt{73}$)

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$i$	-2	-1	0	1	2	3	4	5	6	7

$a_{i}$			8	1	1	5	5	1	1	16
$h_{i}$	0	1	8	9	17	94	487	581	1068	$\dots$
$k_{i}$	1	0	1	1	2	11	57	68	125	$\dots$

Exercise 7.9.1 (Continued fraction expansion of $\sqrt{73}$)

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