# paracol creates empty lines

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

\MakeFramed {\FrameRestore}}%
{\endMakeFramed}

\newenvironment{framedPage}[1][\hsize]
{\endMakeFramed}

{
\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
$$\label{identity col} \delta^i_k=\delta^j_ka^i_j$$
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
$$\label{identity row} \delta^k_i=\delta^k_ja^j_i$$
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.

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

\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

\MakeFramed {\FrameRestore}}%
{\endMakeFramed}

\newenvironment{framedPage}[1][\hsize]
{\endMakeFramed}

{
\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
$$\label{identity col} \delta^i_k=\delta^j_ka^i_j$$
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
$$\label{identity col} \delta^i_k=\delta^j_ka^i_j$$
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

\MakeFramed {\FrameRestore}}%
{\endMakeFramed}

\newenvironment{framedPage}[1][\hsize]
{\endMakeFramed}

{
\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
$$\label{identity col} \delta^i_k=\delta^j_ka^i_j$$
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
$$\label{identity col} \delta^i_k=\delta^j_ka^i_j$$
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.