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CONTINUED FRACTIONS

INVERSE PERIODS AND A THEOREM OF GALOIS

\end{center}

Let $\alpha_0=[\overline{a_0,\ldots,a_{k-1}}]$ be a purely
periodic continued fraction of period $k$.

The $\alpha_n$ are also reduced, and satisfy

$$\alpha_0=a_0+\frac{1}{\alpha_1}\
\alpha_1=a_1+\frac{1}{\alpha_2}\ldots\
\alpha_{k-1}=a_{k-1}+\frac{1}{\alpha_0}$$

using periodicity.

Reversing the sequence, rearranging and taking conjugates gives

$$-\frac{1}{\overline{\alpha_0}}=a_{k-1}-\overline{\alpha_{k-1}
},\
-\frac{1}{\overline{\alpha_{k-1}}}=a_{k-2}-\overline{\alpha_{k-1
}}\ldots\ -\frac{1}{\overline{\alpha_1}}=a_0-\overline{\alpha_0}$$

Write $-\frac{1}{\overline{\alpha_n}}=\beta_n$. Then $\beta_n&gt;1$
and

%page 42

$$\beta_0=a_{k-1}+\frac{1}{\beta_{k-1}}\
\beta_{k-1}=a_{k-1}+\frac{1}{\beta_{k-1}},\ldots,\
\beta_1=a_0+\frac{1}{\beta_0}$$

From which we deduce

$$|beta_0\left(=-\frac{1}{\overline{\alpha_0}}\right)=[\overline{
a_{k-1},a_{k-2},\ldots,a_1,a_0}]$$

We can also investigate the complete quotients, developing
formulae of use later.

Let $\alpha_0=\frac{\sqrt{D}+P_0}{Q_0}$ where $Q_0|D-P_0^2$

then $\alpha_n=\frac{\sqrt{D}+p_n}{Q_n}$ and $Q_n|D-P_n^2$ so
$\frac{\sqrt{D}+P_n}{Q_n}=a_n+\frac{Q_{n+1}}{\sqrt{D}+P_{n+1}}$

Clearing of fractions and equating rational and irrational parts
gives

\begin{eqnarray*}
D+P_nP_{n+1}&amp;=&amp;a_nQ_nP_{n+1}+Q_nQ_{n+1}\\ P_n+P_{n+1}&amp;=&amp;a_nQ_n
\end{eqnarray*}

Multiplying the second equation by $P_{n+1}$ and subtracting gives

$$D-P_{n+1}^2=Q_nQ_{n+1}$$

which we have met before.

Now $\alpha_n=\frac{\sqrt{D}+P_n}{Q_n}$ so
$\overline{\alpha_n}=\frac{-\sqrt{D}+P_n}{Q_n}$

Thus

$$\beta_n=-\frac{1}{\overline{\alpha_n}}=\frac{Q_n}{\sqrt{D}-P_n}=
\frac{Q_n(\sqrt{D}+P_n)}{D-P_n^2}=\frac{\sqrt{D}-P_n}{Q_{n-1}}$$

This needs interpreting for $n=0$.

However

$$\beta_0=\beta_k\textrm{ by periodicity
}=\frac{\sqrt{D}+P_k}{Q_{k-1}}=\frac{\sqrt{D}+P_0}{Q_{k-1}}$$

This relates the complete quotients of $\alpha_0$ and
$-\frac{1}{\overline{\alpha_0}}$

%page 43

Finally we deduce

Theorem (Serret)

Two conjugate quadratic irrationals have inverse periods (not
necessarily reduced).

Proof

Let $\alpha_0=[a_0,\ldots a_{k-1}\overline{a_k,\ldots a_{m+k-1}}]$

$$\alpha_0=\frac{p_{k-1}\alpha_k+p_{k-2}}{q_{k-1}\alpha_k+q_{k-1}},$$

$\alpha_k$ purely periodic.

so $-\frac{1}{\overline{\alpha_k}}=[\overline{a_{m+k-1},\ldots
a_k}]$. However

\begin{eqnarray*}
\overline{\alpha_0}&amp;=&amp;
\frac{p_{k-1}\overline{\alpha_k}+p_{k-1}}{q_{k-1}\overline{\alpha_k}
+q_{k-2}}\\
&amp;=&amp;\frac{p_{k-2}\left(-\frac{1}{\overline{\alpha_k}}\right)+
(-p_{k-1})}{q_{k-2}\left(-\frac{1}{\overline{\alpha_k}}\right)
+(q_{k-1})}
\end{eqnarray*}

So $\overline{\alpha_0}$ and $-\frac{1}{\overline{\alpha_k}}$ are
equivalent and so their continued fractions agree from some point
on. Thus $\overline{\alpha_0}$ has the reverse period of
$\alpha_0$

Examples

\begin{eqnarray*}
\frac{14-\sqrt{37}}{3}&amp;=&amp;[2,1,1,\overline{1,3,2}]\\
\frac{14+\sqrt{37}}{3}&amp;=&amp;[6,\overline{1,2,3}]=[6;1,\overline{2,3,1}]
\end{eqnarray*}

If $\alpha_0$ and $\overline{\alpha_0}$ are equivalent then they
agree from some point onward. This means that the have the same
period. They also have inverse periods, but this does not mean
that the period is symmetric, because of the shift noted above.

%page 44

Examples

$\alpha_0=\frac{\sqrt{7}+3}{2}\
\overline{\alpha_0}=\frac{\sqrt{7}-3}{-2}$

\begin{tabular}{ccc|cccc|c}
$\alpha_0$&amp;$k$&amp;0&amp;1&amp;2&amp;3&amp;4&amp;5\\ &amp;$P_k$&amp;3&amp;1$2$2$1$1\\
&amp;$Q_k$&amp;2&amp;3&amp;1&amp;3&amp;2&amp;3\\ &amp;$a_k$&amp;3&amp;1&amp;4&amp;1&amp;1
\end{tabular}

$\alpha_0=[3;\overline{1,4,1,1}]$

$\alpha_0-\overline{\alpha_0}=3$ so $\overline{\alpha_0}$ and
$\alpha_0$ are equivalent.
$\left(\overline{\alpha_0}=\frac{1.\alpha_0+3}{0.\alpha_0+1}\right)$

\begin{tabular}{ccc|ccccc|c}
$\alpha_0$&amp;$k$&amp;0&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\ &amp;$P_k$&amp;$-3$&amp;3&amp;2&amp;1&amp;1&amp;2&amp;2\\
&amp;$Q_k$&amp;$-2$&amp;1&amp;3&amp;2&amp;3&amp;1&amp;3\\ &amp;$a_k$&amp;0&amp;5&amp;1&amp;1&amp;1&amp;4&amp;1
\end{tabular}

\begin{eqnarray*}
\overline{\alpha_0}&amp;=&amp;[0,5,\overline{1,1,1,4}]\\
&amp;=&amp;[0;5,1,\overline{1,1,4,1}]\textrm{ - reverse period of
}\alpha\\ &amp;=&amp;[o;5,1,1,\overline{1,4,1,1}]\textrm{ - same period as
}\alpha
\end{eqnarray*}

because this shift starts somewhere, the period can be slit into 2
symmetric parts. In this case 1 and 1,4,1.

Square roots of rationals

Let $d\in Q$, not the square of a rational, and $d&gt;1$. Then
$-\sqrt{d}&lt;-1$ and the continued function for $\sqrt{D}$ has one
term before the period.

$\sqrt{d}=[a_0;\overline{a_1,\ldots,a_k}]$ so
$\frac{1}{\sqrt{d}-a_0}=[\overline{a_1,\ldots,a_k}]=\alpha_0$

$-\frac{1}{\overline{\alpha_0}}=\sqrt{d}+a_0=[\overline{a_k,\ldots,a_1}]$

but $\sqrt{d}+a_0=[2a_0,\overline{a_1,\ldots,a_k}]$

%page 45

Thus by uniqueness

$$a_k=2a_0,\ a_{k-1}=a_1,\ldots$$

so $\sqrt{d}=[a_0;\overline{a_1,a_2,\ldots,a_2,a_1,2a_0}]$

(If $k=1$ the symmetric part is empty)

Conversely we argue as follows

Suppose $\alpha_0=[a_o;\overline{a_1,a_2,\ldots,a_2,a_1,2a_0}].\
2a_0\neq0$ so $\alpha_0&gt;1$

$\frac{1}{\alpha_--a_0}=[\overline{a_1,a_2,\ldots,a_2,a_1,2a_0}]$
and so
$a_0-\overline{\alpha_0}=[\overline{2a_0,a_1,a_2,\ldots,a_1}]$ so
$-\overline{\alpha_0}=[a_0,\overline{a_1,a_2,\ldots,a_2,a_1,2a_0}]$

Thus $\alpha_0=-\overline{\alpha_0}$ and so $\alpha_0$ is the
square root of a rational number.


\end{document}
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