\documentstyle[12pt]{article}

\textwidth=17.0cm \textheight=630pt \evensidemargin=1mm \oddsidemargin=1mm


\newtheorem{theorem}{Theorem}

\title{Continued Fractions of Cubic Irrationalities}
\author{V.A.Malyshev}
\date{}

\begin{document}
\maketitle

\begin{abstract}
�{\small\it An integral relation for elements of the continued fraction of $\sqrt[3]{M}$
is presented.}
\end{abstract}

\begin{center}
\large{�Contents}
\end{center}
\bigskip

\noindent {Introduction}
\medskip


\noindent {0. An integral algorithm}

\noindent {1. A cubic Diophantine equation}

\noindent {2. Squares of numerators}

\noindent {3. A recurrence relation}

\noindent {4. Numbers $M=2,9,11,12,13,14,16,19,20,...$}

\noindent {5. A hypothesis about remainders}

\medskip\noindent {Bibliography}


%%%%%%%%%%%%%%%%%%%%%%%
\section*{Introduction}

In the paper we prove a recurrence relation for elements and present a reliable
hypothesis for remainders of the continued fraction of the cubic irrationality
$\sqrt[3]{M}$. From the hypothesis it follows that the sequence of elements is unbounded.
In particular, this gives the answer to the Kchinchin question~\cite{Hi1} about the
continued fractions of algebraic numbers.

For example, consider the continued fraction
$$
\sqrt[3]{2}\;=\;1
 +\frac{1}{\displaystyle 3
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 5
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 4
 +\frac{1}{\displaystyle 1+\cdots}}}}}}}
% +\frac{1}{\displaystyle 1
% +\frac{1}{\displaystyle 8
% +\frac{1}{\displaystyle 1
% +\frac{1}{\displaystyle 14+\cdots}}}}}}}}}}}
$$

\begin{center}
 \begin{picture}(400,20)
 \put(-40,30){\line(1,0){70}}
 \put(-25,15){{\footnotesize{\it 2000 Mathematics Subject Classification.} Primary 11-04, 11K50.}}
 \put(-25,0){{\footnotesize{\it Key words and phrases.} Algebraic numbers, continued fractions.}}
\end{picture}
\end{center}

\noindent Let $r_{0},r_{1},r_{2},\ldots$ be the sequence of remainders and let
$a_{0}=1,a_{1},a_{2},\ldots$ be the sequence of elements of the continued fraction.
Recall that
$$
�a_{k+1}=\left[\frac{1}{r_{k}}\right].
$$
Direct calculations show that the sequence
$$
�\xi_{k}=\log_{2}(1+r_{k})
$$
has the uniform distribution~\cite{Pol1}. In spite of the chaotic behavior of the
remainders all elements
$$
�3\;\;1\;\;5\;\;1\;\;1\;\;4\;\;1\;\;1\;\;8\;\;1\;\;14\;\;1\;\;10\;\;2\;\;1\;\;4\;\;12
 \;\;2\;\;3\;\;2\;\;1\;\;3\;\;4\;\;1\;\;1\;\;2\;\;14\;\;3\;\;12\;\;1\;\;\ldots
$$
may be calculated by the recurrence relation
$$
 a_{k}=\left[-\frac{(P_{k-2}P_{k-1}^{2}-2Q_{k-2}Q_{k-1}^{2})Q_{k-1}+(-1)^{k-1}2P_{k-1}^{2}}
 {(P_{k-1}^{3}-2Q_{k-1}^{3})Q_{k-1}}\right],
$$
where
\begin{eqnarray*} �
 P_{-1}=1,&P_{0}=1,&P_{k}=a_{k}P_{k-1}+P_{k-2},\\
 Q_{-1}=0,&Q_{0}=1,&Q_{k}=a_{k}Q_{k-1}+Q_{k-2}.\\
\end{eqnarray*}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{0. An integral algorithm}

Let $\omega=\sqrt[3]{M}$ be irrational. Consider the continued fraction
$$
\omega\;=\;a_{0}+r_{0}\;=\;a_{0}+\frac{1}{\displaystyle a_{1}+r_{1}}\;=\;
  a_{0}+\frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}+r_{2}}}
\;=\;
 a_{0}+\frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}
 +\frac{1}{\displaystyle a_{3}+r_{3}}}}\;=\;\cdots\;
$$
Here
$$
�\begin{array}{llr}
 a_{k}&=&\left[1/r_{k-1}\right],\\
 r_{k}&=&1/r_{k-1}-\left[1/r_{k-1}\right],
\end{array}
$$
where
$$
�\begin{array}{llr}
 a_{0}&=&[\omega],\\
 r_{0}&=&\omega-[\omega].
\end{array}
$$
We first give an integral algorithm for calculation $a_{0},a_{1},\ldots\;$.

Let us put
\begin{eqnarray*} �
 P_{k}&=&a_{k}P_{k-1}+P_{k-2},\\
 Q_{k}&=&a_{k}Q_{k-1}+Q_{k-2},\\
\end{eqnarray*}
where
$$
�\begin{array}{c}
  �P_{-1}=1 \\
  �Q_{-1}=0
\end{array}
\;\;\;\;{\rm and}\;\;\;\; �\begin{array}{c}
  �P_{0}=a_{0} \\
  �Q_{0}=1
\end{array}
$$
The identity
$$
�\omega=\frac{(a_{k}+r_{k})P_{k-1}+P_{k-2}}{(a_{k}+r_{k})Q_{k-1}+Q_{k-2}}\eqno(1.0)
$$
leads to the rational function
$$
�F_{k}(x)=\frac{P_{k-1}x+P_{k-2}}{Q_{k-1}x+Q_{k-2}}.
$$
The function $F_{k}(x)$ is strictly monotone on $[0,\infty)$. Hence $a_{k}$ is a unique
integer $\alpha\ge 1$ such that
$$
�(M-F^{3}_{k}(\alpha)(M-F^{3}_{k}(\alpha+1))<0. \eqno(2.0)
$$
We check the inequality $(2.0)$ for $\alpha=1,2\ldots\;$ and find $a_{k}$ .

Of course, the algorithm is rather bulky. By identity $(1.0)$, we have
$$
a_{k}+r_{k}=-\frac{P_{k-2}-\omega Q_{k-2}}{P_{k-1}-\omega Q_{k-1}}.
$$
Hence
$$
a_{k}+r_{k}=-\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})(P_{k-1}^{2}+\omega P_{k-1}Q_{k-1}+
 \omega^{2}Q_{k-1}^{2})}
{(P_{k-1}^{3}-MQ_{k-1}^{3})(P_{k-2}^{2}+\omega P_{k-2}Q_{k-2}+
 \omega^{2}Q_{k-2}^{2})}.
$$
For even $k$ we have
$$
�\frac{P_{k-2}}{Q_{k-2}}<\omega<\frac{P_{k-1}}{Q_{k-1}}
$$
and for odd $k$ we have
$$
�\frac{P_{k-1}}{Q_{k-1}}<\omega<\frac{P_{k-2}}{Q_{k-2}}.
$$
Hence for even $k$ we obtain
$$
 -\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})Q_{k-1}^{2}}{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-2}^{2}}
 <a_{k}+r_{k}<
 -\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})P_{k-1}^{2}}{(P_{k-1}^{3}-MQ_{k-1}^{3})P_{k-2}^{2}},
$$
and for odd $k$ we obtain
$$
 -\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})P_{k-1}^{2}}{(P_{k-1}^{3}-MQ_{k-1}^{3})P_{k-2}^{2}}
 <a_{k}+r_{k}<
 -\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})Q_{k-1}^{2}}{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-2}^{2}}.
$$
Let
$$
�b_{k}=\left[-\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})Q_{k-1}^{2}}
                   {(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-2}^{2}}\right]
\;\;\;\;{\rm and}\;\;\;\; c_{k}=\left[-\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})P_{k-1}^{2}}
                   {(P_{k-1}^{3}-MQ_{k-1}^{3})P_{k-2}^{2}}\right] .
$$
We see that $b_{k}\le a_{k}\le c_{k}$ for even $k$ and $c_{k}\le a_{k}\le k_{k}$ for odd
$k$. Therefore we may check the inequality $(2.0)$ only for $\alpha$ between $b_{k}$ and
$c_{k}$. In the cases $k=0$ and $k=1$ we put $b_{0}=0, c_{0}=\infty$ and $b_{1}=\infty,
c_{1}=0$.
\newpage


\noindent{\bf Example.} We apply the algorithm for $\sqrt[3]{4}$ and obtain
\begin{center}
�\begin{tabular}{r||c|c|c|}
  �$k$  & �$b_{k}$  & �$a_{k}$ & �$c_{k}$  \\\hline\hline
  �0    & �0        & �1       & �$\infty$  \\\hline
  �1    & �$\infty$ & �1       & �0         \\\hline
  �2    & �0        & �1       & �3        \\\hline
  �3    & �3        & �2       & �1         \\\hline
  �4    & �2        & �2       & �2         \\\hline
  �5    & �1        & �1       & �1        \\\hline
  �6    & �3        & �3       & �3        \\\hline
  �7    & �2        & �2       & �2        \\\hline
  �8    & �3        & �3       & �3         \\\hline
\end{tabular}
\end{center}
From the table we see that
$
�a_{k}=b_{k}=c_{k}
$
for $k\ge 4$.
\bigskip

\noindent{\bf Hypothesis.} {\it Given $M$, there exists $n(M)$ such that
$
�a_{k}=b_{k}=c_{k}
$
or all $k\ge n(M)$}.
\bigskip

If the hypothesis is true, then
$$
�a_{k}\;=\;\left[-\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})Q_{k-1}^{2}}
                   {(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-2}^{2}}\right]
\;=\;\left[-\frac{(P_{k-2}^{3}-MQ_{k-2}^{3})P_{k-1}^{2}}
                   {(P_{k-1}^{3}-MQ_{k-1}^{3})P_{k-2}^{2}}\right]
$$
for all sufficiently large $k$.

In the paper we prove the analogous formula
$$
�a_{k}=\left[-\frac{(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})Q_{k-1}+(-1)^{k-1}2P_{k-1}^{2}}
{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-1}}\right].
$$
Unfortunately, the last formula is not symmetric.
\bigskip


\noindent{\bf Remark.} For $\sqrt[N]{M}$ we have
$$
�a_{k}\;=\;\left[-\frac{(P_{k-2}^{N}-MQ_{k-2}^{N})Q_{k-1}^{N-1}}
                   {(P_{k-1}^{N}-MQ_{k-1}^{N})Q_{k-2}^{N-1}}\right]
\;=\;\left[-\frac{(P_{k-2}^{N}-MQ_{k-2}^{N})P_{k-1}^{N-1}}
                   {(P_{k-1}^{N}-MQ_{k-1}^{N})P_{k-2}^{N-1}}\right].
$$





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. A cubic Diophantine equation}

\begin{theorem}
Let $A^{3}L\ne B^{3}N$. The Diophantine equation
$$
NX^{3}-LY^{3}=AX-BY
$$
has at most three relatively prime positive solutions.
\end{theorem}
{\bf Proof.} Let relatively prime $X\ge 1$ and $Y\ge 1$ satisfy the equation. From
$$
�(NX^{2}-A)X=(LY^{2}-B)Y
$$
it follows that
$$
�\begin{array}{rcl}
  �NX^{2}-A & �= & �\lambda Y,\\
  �LY^{2}-B & �= & �\lambda X
\end{array}
$$
for an integer $\lambda$. Hence
$$
�BNX^{2}-ALY^{2}=\lambda BY-\lambda AX.
$$
From
$$
�(BNX+\lambda A)X=(ALY+\lambda B)Y
$$
it follows that
$$
�\begin{array}{rcl}
  �BNX+\lambda A & �= & �\mu Y,\\
  �ALY+\lambda B & �= & �\mu X
\end{array}
$$
for an integer $\mu$. Hence
$$
B^{2}NX-A^{2}LY=\mu BY- \mu AX.
$$
From
$$
(B^{2}N+\mu A)X=(A^{2}L+\mu B)Y
$$
it follows that
$$
�\begin{array}{rcl}
  �X & �=  & �\gamma(A^{2}L+\mu B),\\
  �Y & �=  & �\gamma(B^{2}N+\mu A)
\end{array}
$$
for an integer $\gamma$. Since $X$ are $Y$ relatively prime, we have $\gamma=\pm 1$.
Since $X$ and $Y$ are positive, we have either $\gamma=-1$ or $\gamma=1$. From the input
Diophantine equation we obtain
$$
�\mu^{3}+1=3ABNL\mu+A^{3}NL^{2}+B^{3}N^{2}L.
$$
Hance the Diophantine equation has at most three relatively prime positive solutions.
$\bullet$
\bigskip

\noindent{\bf Remark.} Let $N\ge 1$ and $L\ge 1$. By the inequality
$$
�|\mu|\le |A|^{3}NL^{2}+|B|^{3}N^{2}L+1,
$$
we easily estimate $X$ and $Y$.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. Squares of numerators}

\begin{theorem}
�For any $N\ge 1$ and $A$ the congruence
$$
�NP_{k}^{2}\equiv A ({\rm mod} Q_{k})
$$
holds for a finite number of $k=1,2,\ldots$.
\end{theorem}
{\bf Proof.} Assume that $k_{1}<k_{2}<\cdots$ and that
$$
�NP_{k}^{2}=V_{k}Q_{k}+A\eqno(1.2)
$$
for all $k=k_{1},k_{2},\ldots\;$. It is known that
$$
\omega=\frac{P_{k}}{Q_{k}}+\frac{\theta_{k}}{Q_{k}Q_{k+1}},
$$
where $-1<\theta_{k}<0$ for odd $k$ and $0<\theta_{k}<1$ for even $k$. From
$$
�N\frac{P_{k}^{2}}{Q_{k}^{2}}=\frac{V_{k}}{Q_{k}}+\frac{A}{Q_{k}^{2}}
$$
and
$$
�\omega^{2}=\frac{P_{k}^{2}}{Q_{k}^{2}}+2\frac{P_{k}\theta_{k}}{Q_{k}^{2}Q_{k+1}}+
 \frac{\theta_{k}^{2}}{Q_{k}^{2}Q_{k+1}^{2}}
$$
it follows that
$$
�N\omega^{2}Q_{k}=V_{k}+\frac{A}{Q_{k}}+2N\frac{P_{k}\theta_{k}}{Q_{k}Q_{k+1}}+
 N\frac{\theta_{k}^{2}}{Q_{k}Q_{k+1}^{2}}.\eqno(2.2)
$$
Multiplying $(2.2)$ by
$$
�\omega Q_{k}=P_{k}+\frac{\theta_{k}}{Q_{k+1}},
$$
we obtain
$$
�NMQ_{k}^{2}=V_{k}P_{k}+B_{k},\eqno(3.2)
$$
with an integer
$$
�B_{k}=A\frac{P_{k}}{Q_{k}}
      +\frac{3NP_{k}^{2}}{Q_{k}Q_{k+1}}\theta_{k}
      +\frac{3NP_{k}}{Q_{k}Q_{k+1}^{2}}\theta_{k}^{2}
      +\frac{N}{Q_{k}Q_{k+1}^{3}}\theta_{k}^{3}.
$$
Since
$$
�\frac{P_{k}}{Q_{k}}\le \omega+\frac{1}{Q_{k}Q_{k+1}}\le \omega+1,
$$
we have
$$
�|B_{k}|\le |A|(\omega+1)+N(3\omega^{2}+9\omega+7)
$$
for all $k$. Multiplying $(1.2)$ by $P_{k}$, we obtain
$$
�NP_{k}^{3}=V_{k}P_{k}Q_{k}+AP_{k}.\eqno(4.2)
$$
Multiplying $(3.2)$ by $Q_{k}$, we obtain
$$
�NMQ_{k}^{3}=V_{k}P_{k}Q_{k}+B_{k}Q_{k}.\eqno(5.2)
$$
Subtracting $(5.2)$ from $(4.2)$, we obtain
$$
�NP_{k}^{3}-NMQ_{k}^{3}=AP_{k}-B_{k}Q_{k}.
$$
The sequence $B_{k_{1}},B_{k_{2}},\ldots$ is bounded. Hence there exists an integer $B$
such that
$$
�NP_{k}^{3}-NMQ_{k}^{3}=AP_{k}-BQ_{k}
$$
for infinitely many $k$. Since $A^{3}\ne B^{3}M$, and since $P_{k}\ge 1$ and $Q_{k}\ge 1$
are relatively prime, we have a contradiction with Theorem 1. The theorem is proved.
$\bullet$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. A recurrence relation}


\begin{theorem}
�Given $\omega=\sqrt[3]{M}$, there exists $n(M)$ such that
$$
�a_{k}=\left[-\frac{(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})Q_{k-1}+(-1)^{k-1}2P_{k-1}^{2}}
{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-1}}\right]
$$
for all $k\ge n(M)$.
\end{theorem}
{\bf Proof.} From
$$
�\omega=\frac{(a_{k}+r_{k})P_{k-1}+P_{k-2}}{(a_{k}+r_{k})Q_{k-1}+Q_{k-2}}
$$
it follows that
$$
�a_{k}+r_{k}=-\frac{P_{k-2}-\omega Q_{k-2}}{P_{k-1}-\omega Q_{k-1}}.
$$
Hence
$$
a_{k}+r_{k}= -\frac{(P_{k-2}-\omega Q_{k-2})(P_{k-1}^{2}+\omega
P_{k-1}Q_{k-1}+\omega^{2}Q_{k-1}^{2})} {P_{k-1}^{3}-MQ_{k-1}^{3}}.
$$
From
$$
�Q_{k-1}P_{k-2}-Q_{k-2}P_{k-1}=(-1)^{k-1}
$$
we obtain
$$
�a_{k}+r_{k}=-\frac{P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2}+(-1)^{k-1}(\omega
P_{k-1}+\omega^{2}Q_{k-1})} {P_{k-1}^{3}-MQ_{k-1}^{3}}.
$$
From
$$
\omega=\frac{P_{k-1}}{Q_{k-1}}+\frac{\theta_{k-1}}{Q_{k-1}Q_{k}},
$$
where  $-1<\theta_{k}<0$ for odd $k$ and $0<\theta_{k}<1$ for even $k$, we obtain
$$
\omega P_{k-1}+\omega^{2}Q_{k-1}=
 \frac{2P_{k-1}^{2}}{Q_{k-1}}+\frac{3P_{k-1}\theta_{k-1}}{Q_{k-1}Q_{k}}+
 \frac{\theta_{k-1}^{2}}{Q_{k-1}Q_{k}^{2}}.
$$
Hence
$$
�a_{k}+r_{k}=
-\frac{(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})Q_{k-1}+(-1)^{k-1}(2P_{k-1}^{2}+\Delta_{k})}
{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-1}},\eqno(1.3)
$$
where
$$
\Delta_{k}=\frac{3P_{k-1}\theta_{k-1}}{Q_{k}}+
 \frac{\theta_{k-1}^{2}}{Q_{k}^{2}}.
$$
Let us put
$$
C=3\omega+4.
$$
Then
$$
|\Delta_{k}|\le C
$$
for all $k$. Let us write
$$
�2P_{k}^{2}=V_{k}Q_{k}+A_{k},
$$
where $0\le A_{k}\le Q_{k}-1$. By Theorem 2, there exists $n(M)$ such that
$$
�C+1\le A_{k-1}\le Q_{k-1}-(C+1)
$$
for all $k\ge n(N)$. Let us rewrite $(1.3)$ in the form
$$
�a_{k}+r_{k}=
\frac{(-1)^{k-1}(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})Q_{k-1}+2P_{k-1}^{2}+\Delta_{k}}
{(-1)^{k}(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-1}}.
$$
We note that the denominator is positive for all $k$. Consider
$$
�a_{k}+r_{k}=
\frac{(-1)^{k-1}(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})+V_{k-1}+(A_{k-1}+\Delta_{k})/Q_{k-1}}
{(-1)^{k}(P_{k-1}^{3}-MQ_{k-1}^{3})}.
$$
It is clear that
$$
�1\le A_{k-1}+\Delta_{k}\le Q_{k-1}-1
$$
for all $k\ge n(M)$. If $P$ and $Q\ge 1$ are integers, then
$$
\left[\frac{P+\delta}{Q}\right]=\left[\frac{P}{Q}\right]
$$
for all $0\le\delta<1$. Hence for all $k\ge n(M)$ the integer
$$
�a_{k}= \left[\frac{(-1)^{k-1}(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})+
V_{k-1}+(A_{k-1}+\Delta_{k})/Q_{k-1}}{(-1)^{k}(P_{k-1}^{3}-MQ_{k-1}^{3})}\right]
$$
is equal to the integer
$$
\left[\frac{(-1)^{k-1}(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})+V_{k-1}+A_{k-1}/Q_{k-1}}
{(-1)^{k}(P_{k-1}^{3}-MQ_{k-1}^{3})}\right].
$$
The theorem is proved. $\bullet$
\bigskip

\noindent{\bf Hypothesis.} {\it Given $\sqrt[N]{M}$, there exists $n(N,M)$ such that
$$
�a_{k}=\left[-\frac{(P_{k-2}P_{k-1}^{N-1}-MQ_{k-2}Q_{k-1}^{N-1})Q_{k-1}+
(-1)^{k-1}(N-1)P_{k-1}^{N-1}}{(P_{k-1}^{N}-MQ_{k-1}^{N})Q_{k-1}}\right]
$$
for all $k\ge n(N,M)$.}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{4. Numbers $M=2,9,11,12,13,14,16,19,20,...$}

Let $n(M)$ be a minimal natural such that
$$
�a_{k}=\left[-\frac{(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})Q_{k-1}+(-1)^{k-1}2P_{k-1}^{2}}
{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-1}}\right]
$$
for all $k\ge n(M)$.

\begin{theorem}
Let $n$ satisfy the inequality $P_{n-1}\ge 10^{8} M^{4}$. Assume that
$$
�a_{k}=\left[-\frac{(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})Q_{k-1}+(-1)^{k-1}2P_{k-1}^{2}}
{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-1}}\right]
$$
for all $1\le k\le n-1$. Then $n(M)=1$.
\end{theorem}
{\bf ������⥫��⢮.} Let us write
$$
�2P_{k}^{2}=V_{k}Q_{k}+A_{k},
$$
where $0\le A_{k}\le Q_{k}-1$. By the proof of Theorem 3, the identity
$$
�a_{k}=\left[-\frac{(P_{k-2}P_{k-1}^{2}-MQ_{k-2}Q_{k-1}^{2})Q_{k-1}+(-1)^{k-1}2P_{k-1}^{2}}
{(P_{k-1}^{3}-MQ_{k-1}^{3})Q_{k-1}}\right]
$$
follows from the inequality
$$
3\omega+5 \le A_{k-1}\le Q_{k-1}-(3\omega+5),
$$
where $k\ge n$.

Assume that
$$
0\le A_{k-1}\le 3\omega +5\;\;\;\; {\rm or}\;\;\;\; Q_{k-1}-(3\omega +5)\le A_{k-1}\le
Q_{k-1}
$$
for some $k\ge n$. Then
$$
�2P_{k-1}^{2}=U_{k-1}Q_{k-1}+A,
$$
where
$$
|A|\le 3\omega +5.
$$
By the proof of Theorem 2, there exists an integer $B$ such that
$$
�|B|\le (3\omega +5)(\omega+1)+2(3\omega^{2}+9\omega+7)=9\omega^{2}+26\omega+19
$$
and
$$
�2P_{k-1}^{3}-2MQ_{k-1}^{3}=AP_{k-1}-BQ_{k-1}.
$$
Hence $P_{k-1}$ and $Q_{k-1}$ satisfy the Diophantine equation
$$
�2X^{3}-2MY^{3}=AX-BY.
$$
By the proof of Theorem 2, we have
$$
�P_{k-1}\le 10^{8}M^{4}-1.
$$
Since $k\ge n$, we have a contradiction. The theorem is proved. $\bullet$
\bigskip

By direct calculation, we see that $n(M)=1$ for
$$
�M=2,9,11,12,13,14,16,19,20,28,29,30,32,33,35,36,38,39,40,43,44,45,46,47,50,...
$$



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{5. A hypothesis about remainders}

Let $x_{1},x_{2},\ldots$ be a sequence on $[0,1]$. Assume that
$$
F(t)=\lim_{n\to\infty}\frac{m_{n}(t)}{n}
$$
for all $0\le t\le 1$, where $m_{n}(t)$ is a number of $x_{1},\ldots, x_{n}$ such that
$0\le x_{k}\le t$. The function
$$
f(t)=\frac{d}{dt}F(t)
$$
is called the density of the sequence $x_{1},x_{2},\ldots\;$. If $f(t)=1$, then
$x_{1},x_{2},\ldots$ has the uniform distribution. For example, let $x$ be irrational,
that the sequence
$$
x_{n}=nx-[nx]
$$
has the uniform distribution.

Consider the remainders
$
�r_{0},r_{1},r_{2},\ldots
$
of the continued fraction
$$
\sqrt[3]{M}\;=\;a_{0}+r_{0}\;=\;a_{0}+\frac{1}{\displaystyle a_{1}+r_{1}}\;=\;
  a_{0}+\frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}+r_{2}}}
\;=\;
 a_{0}+\frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}
 +\frac{1}{\displaystyle a_{3}+r_{3}}}}\;=\;\cdots\;
$$
\bigskip

\noindent{\bf Hypothesis.} {\it The sequence
$$
�\xi_{n}=\log_{2}(1+r_{n})
$$
has the uniform distribution.}
\bigskip

\noindent {\bf Example.} If we calculate 1001 elements
$\xi_{0},\xi_{1},\ldots,\xi_{1000}$ for $\sqrt[3]{2}$, then
$$
�\frac{1}{1001}\sum_{k=0}^{1000}\xi_{k}=0.5015...\approx\frac{1}{2},
$$
$$
�\frac{1}{1001}\sum_{k=0}^{1000}\xi_{k}^{2}=0.3321...\approx\frac{1}{3},
$$
$$
�\frac{1}{1001}\sum_{k=0}^{1000}\xi_{k}^{3}=0.2470...\approx\frac{1}{4},
$$
$$
�\frac{1}{1001}\sum_{k=0}^{1000}\xi_{k}^{4}=0.1959...\approx\frac{1}{5}.
$$
Analogous results we have for other $\sqrt[3]{M}$.
\bigskip


\noindent{\bf Remark.} The hypothesis is naturally connected with the known Gauss
theorem~\cite{Hi1}. Let
$$
x\;=\;r_{0}\;=\;\frac{1}{\displaystyle a_{1}+r_{1}}\;=\;
  \frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}+r_{2}}}
\;=\;
 \frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}
 +\frac{1}{\displaystyle a_{3}+r_{3}}}}\;=\;\cdots\;
$$
where $0\le x\le 1$. Let  $x$ be a random variable with the uniform distribution on
$[0,1]$. Then $r_{n}$ is a random variable on $[0,1]$. Let $f_{n}(t)$ be the probability
density of $r_{n}$. By the Gauss theorem
$$
\lim_{n\to\infty}f_{n}(t)=\frac{1}{\ln2}\,\frac{1}{1+t}.
$$
This means that the random variables
$$
�\xi_{n}=\log_{2}(1+r_{n})
$$
converge to the uniform distribution.










\begin{thebibliography}{9}

\bibitem{Hi1} A.J.Khinchin. Continued fractions. Nauka, Moscow, 1978.



\bibitem{Pol1}
G.Polya, G.Szeg$\ddot{o}$. Aufgaben und Lehrs$\ddot{a}$tze aus der Analysis.
Sprnger-Verlag, Berlin, 1964


\end{thebibliography}

Vladimir Malyshev

wmal@ryb.adm.yar.ru







\end{document}
