1

I tried to use package paracol because I want to put similar theorems near one another, so reader can compare them. But I put attention that original text grow up from 290 to 320. Looking what happens I find small example

\documentclass{amsart}
\scrollmode
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{paracol}
\begin{document}
\title{Test File}

\begin{abstract}
In this paper I test shaded environment.
\end{abstract}
\maketitle

\newenvironment{Shaded}{%
\def\FrameCommand{\fboxsep3pt \colorbox{shadecolor}}%
 \MakeFramed {\FrameRestore}}%
{\endMakeFramed}

\newenvironment{framedPage}[1][\hsize]
{\MakeFramed{\hsize#1\advance\hsize-\width \FrameRestore}}%
{\endMakeFramed}

\newenvironment{shadedPage}[1][\hsize]
{
\def\FrameCommand{\colorbox{shadecolor}}%
\MakeFramed{ \FrameRestore}}%
{\endMakeFramed}

\setlength{\columnseprule}{0.5pt}

\begin{paracol}{2}
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left 

It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation
satisfies to equation
\[
v^i=v^ja_i^j
\]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
\[
\delta^i_k=a^i_k
\]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.
\switchcolumn%
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left 
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left 

It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation
satisfies to equation
\[
v_i=v_ja^j_i
\]
Choosing values of coordinates as
$v_i=\delta^k_i$
where we selected $k$ we get
\begin{equation}
\label{identity row}
\delta^k_i=\delta^k_ja^j_i
\end{equation}
From \eqref{identity row} it follows
\[
\delta^k_i=a^k_i
\]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.
\end{paracol}


\end{document}

You can see that on second page there few empty lines.

2

There are exactly equal if you use the same content for both columns.

z

\documentclass{amsart}
\scrollmode
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{paracol}
\begin{document}
\title{Test File}

\begin{abstract}
In this paper I test shaded environment.
\end{abstract}
\maketitle

\newenvironment{Shaded}{%
\def\FrameCommand{\fboxsep3pt \colorbox{shadecolor}}%
 \MakeFramed {\FrameRestore}}%
{\endMakeFramed}

\newenvironment{framedPage}[1][\hsize]
{\MakeFramed{\hsize#1\advance\hsize-\width \FrameRestore}}%
{\endMakeFramed}

\newenvironment{shadedPage}[1][\hsize]
{
\def\FrameCommand{\colorbox{shadecolor}}%
\MakeFramed{ \FrameRestore}}%
{\endMakeFramed}

\setlength{\columnseprule}{0.5pt}

\begin{paracol}{2}
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
\[v^i=v^ja_i^j\]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
\[\delta^i_k=a^i_k\]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.

\switchcolumn%
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
\[v^i=v^ja_i^j\]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
\[\delta^i_k=a^i_k\]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.
\end{paracol}

In order not to be confused with the text of two similar theorems one after the other, I suggest defining new commands, for example \firsttheorem and \secondtheorem each one holding the content of its theorem.

Then use

\begin{paracol}{2}
  \firsttheorem 
\switchcolumn%
  \secondtheorem
\end{paracol}

As in this code.

\documentclass{amsart}
\scrollmode
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{paracol}
\begin{document}
\title{Test File}

\begin{abstract}
In this paper I test shaded environment.
\end{abstract}
\maketitle

\newenvironment{Shaded}{%
\def\FrameCommand{\fboxsep3pt \colorbox{shadecolor}}%
\MakeFramed {\FrameRestore}}%
{\endMakeFramed}

\newenvironment{framedPage}[1][\hsize]
{\MakeFramed{\hsize#1\advance\hsize-\width \FrameRestore}}%
{\endMakeFramed}

\newenvironment{shadedPage}[1][\hsize]
{
\def\FrameCommand{\colorbox{shadecolor}}%
\MakeFramed{ \FrameRestore}}%
{\endMakeFramed}

\setlength{\columnseprule}{0.5pt}

\newcommand{\firsttheorem}{% first theorem
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
\[v^i=v^ja_i^j\]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
\[\delta^i_k=a^i_k\]
    Since $k$ is arbitrary, we get the conclusion $a=\delta$.       
    }


\newcommand{\secondtheorem}{% a similar theorem
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
\[v^i=v^ja_i^j\]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
\[\delta^i_k=a^i_k\]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.       
}


\begin{paracol}{2}
\firsttheorem   
\switchcolumn%
\secondtheorem
\end{paracol}

\end{document}

UPDATE

The gap in the second page is originated by the combination of paracol and amsart.

The gap disappears using a standard class as article and add \usepackage{amsmath, amsthm} to make the amslatex commands available.

See also Article vs amsart

d

3
  • The theorem is like it is. I cannot put Leibnitz theorem instead of Pifagor theorem. I also do not like this empty space in the beginning of the second page. Mar 25 at 21:39
  • 1
    @Aleks Kleyn Please see the updated answer. Mar 25 at 22:30
  • Yes. This works better and may be will give some key to find answer. Thank you Mar 25 at 22:57

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