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An Algebraic View on Continued Fraction Factorization

Consider the field

\begin{displaymath}
\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b \in \mathbb{Z}\}.
\end{displaymath}

Looking at the determinant of

\begin{displaymath}
M=
\left
(\begin{array}{cc}
A & 2B\\
B & C
\end{array} \right)
\end{displaymath}

we see that $AC-2B^2=\pm 1$ and thus $AC=\pm 1+2B^2$. Let $\zeta=1+B\sqrt{2} \in
\mathbb{Z}[\sqrt{2}]$. Then $\zeta\bar{\zeta}=1-2B^2$. It is clear that there seems to be a correspondence between $\zeta\bar{\zeta}$ and $1+2B^2$. Thus one could consider the factorization of $M$ by considering the factorizatino of $AC$ in $\mathbb{Z}[\sqrt{2}]$.

A unit in $\mathbb{Z}[\sqrt{2}]$ is a number of the form $(1+\sqrt{2})^n$. A number $\beta$ in $\mathbb{Z}[\sqrt{2}]$ is said to be prime if $\beta=\gamma\epsilon$implies that $\gamma$ or $\epsilon$ is a unit. Let $\alpha_i's$ be such numbers. Then we can look at the factorization

\begin{displaymath}
AC=\alpha_0\alpha_1\ldots\alpha_n(1+\sqrt{2})^m.
\end{displaymath} (13)


This correspondence between prime factorization in $\mathbb{Z}[\sqrt{2}]$ and the factorization of continued fractions could be an area for further research.


scanez 2000-12-04