is it possible to fake the amsart class format?

Question

I often use the amsart documentclass. I like it because it is very compact (it also seems to be an industry standard in my area). What I don't like about it is that I'm not able to use kpfonts or ntheorem with it.

So my question is: does anyone have a solution to this problem?

One solution might be that I'm incompetent with latex and therefore it is actually possible to use kpfonts with amsart.

The other solution I thought is by using some package like titlesec to get all the formatting (including toc?) the same as amsart. So then the question becomes: have you already figured out how to do that and would you mind telling me? (I agree this solution is a bit dumb)

EDIT: an example of what goes wrong with ntheorem is as follows

\documentclass{amsart}
\usepackage{ntheorem}
\begin{document}
    MWE
\end{document}

it brings up an error (Package ntheorem Error: theoremstyle plain already defined)

Answer 1

Why cannot you use kpfonts with amsart?

This works for me:

\documentclass{amsart}
\usepackage{kpfonts,lipsum}
\begin{document}
\title{Title}
\author{A. U. Thor}
\date{\today}
\maketitle
\lipsum[1]
\end{document}

Update OP asked to put theorem number on margins using amsart.

Actually amsart provides full access to formatting theorem headings with the last argument of \newtheoremstyle command. See the commented code in amsclass.dtx. Here is how to produce margin number:

\documentclass{amsart}
\thispagestyle{empty}
\newtheoremstyle{marginnum}%
   {\topskip}%            Space above
   {\topskip}%            Space below
   {\itshape}%            Body font
   {}%                    Indent amount
   {\bfseries}%           Theorem head font
   {.}%                   Punctuation after theorem head
   {\newline}%            Space after theorem head
   {\makebox[0pt]{\thmnumber{#2}\hspace{2em}}\thmname{#1}\thmnote{ (#3)}}
                     %    Theorem head spec 
\theoremstyle{marginnum}
\newtheorem{thm}{Theorem}
\begin{document}
\begin{thm}[Fermat]
  For $n>2$ the equation $a^n+b^n=c^n$ cannot be satisfied with
  positive integer $a$, $b$ and $c$.
\end{thm}
\end{document}