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+
+
+%%% OUTLINE
+
+
+ 
+
+%\documentclass[twocolumn]{article}
+\documentclass{article}
+%\documentclass[twocolumn,10pt]{report}
+\usepackage{graphicx}
+\usepackage{fancyhdr}
+%\usepackage{wassysym}
+\usepackage{tikz}
+\usepackage{amsfonts,amsmath,amsthm}
+\usetikzlibrary{shapes.gates.logic.US,trees,positioning,arrows}
+%\input{../style}
+\usepackage{ifthen}
+\usepackage{lastpage}
+
+\def\layersep{1.8cm}
+
+\linespread{1.0}
+
+\begin{document}
+                                   % numbers at outer edges
+\pagenumbering{arabic}                        % Arabic page numbers hereafter
+\author{R.Clark$^\star$,  \\     
+         $^\star${\em Energy Technology Control, UK. r.clark@energytechnologycontrol.com} \and $^\dagger${\em University of Brighton, UK} 
+}
+
+%\title{Developing a rigorous bottom-up modular static failure mode modelling methodology}
+\title{fermat}
+%\nodate
+\maketitle 
+ 
+\today 
+
+\paragraph{Keywords:} fermat; prime;
+%\small
+
+\abstract{ % \em 
+} % abstract
+
+
+
+\section{Introduction}
+Fermat's Last Theorem
+states that no three positive integers a, b, and c can satisfy the 
+equation $a^n + b^n = c^n$ for any integer value of n greater than two.
+
+
+
+\section{Breaking these positive integers into constituent primes}
+
+Any positive integer can be represented as a collection (or bag) of prime numbers multiple together.
+A function $bpf()$ or `bag of prime factors' is defined to represent this.
+\begin{equation}
+	\prod{bpf(a)}^n + \prod{bpf(b)}^n = c^n
+\end{equation}
+
+The function $bpf()$ will always contain 1.
+
+The numbers $a$ and $b$ may have common and will have uncommon prime factors; these can be collected into
+three bags, those only in a $ubpf(a)$, those only in b, $ubpf(b)$ and those common, $cbpf(a,b)$.
+
+\begin{equation}
+\label{eqn:primesexpanded0}
+    2 \prod{cbpf(a,b)}^n  \prod{ubpf(a)^n}  +  \prod{cbpf(a,b)}^n \prod{ubpf(b)^n} = c^n
+\end{equation}
+
+this can be re-written as
+
+\begin{equation}
+\label{eqn:primesexpanded1}
+    2 \prod{cbpf(a,b)}^n   \big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}   \big)  = c^n
+\end{equation}
+
+These are all prime numbers, and although some may be repeated within their bags
+a prime number can only exist in one of the bags.
+
+Also all these prime numbers are greater than two and therefore odd.
+
+So this becomes a product of a list of prime numbers in ${cbpf(a,b)}$.
+The common prime factors between a and b multiplied
+by the uncommon prime numbers.
+Let $\prod{ubpf(a)^n}  +    \prod{ubpf(b)^n = k$.
+
+\begin{equation}
+\label{eqn:primesexpanded2}
+    2 \prod{cbpf(a,b)}^n   k  = c^n
+\end{equation}
+
+
+
+
+Adding two prime numbers at any power greater than 1
+and then taking a root means getting an irrational number.
+
+
+
+% 
+% \begin{equation}
+% \label{eqn:primesexpanded}
+%      \prod{cbpf(a,b)}^n  \big( \prod{ubpf(a)^n}  +   \prod{ubpf(b)^n} \big) = c^n
+% \end{equation}
+% 
+% \begin{equation}
+% 	a^n + b^n = \prod{bpf(c)^n}
+% \end{equation}
+% 
+% 
+% %assuming its true $c^n$ must be $ 2 \prod{cpf(a,b)}^n \prod{upf(a)^n} \prod{upf(b)^n} $
+% 
+% 
+% % 
+% % \begin{equation}
+% % 	2 \prod{cpf(a,b)}^n  =  \frac{\prod{bpf(c)^n}}{\prod{upf(a)^n} \prod{upf(b)^n}}
+% % \end{equation}
+
+\section{conditions for having a integer root}
+
+To have an integer root $n$ all prime numbers that comprise the number to be rooted must be at least
+to the power of $n$.
+Consider the square root of 144.
+This can be written as
+$12 \times 12$ or breaking it down into prime numbers
+$2  \times 2  \times 3 \times 2  \times \times 2 \times 3$ or $ 2^4 \times 3^2 $.
+Taking the square root means halving the powers $ \sqrt{2^4 \times 3^2} = 2^2 \times 3$.
+
+To get an nth root you need all the prime numbers that comprise
+that number to be at the power of n or greater.
+
+That means that ${ubpf(a)^n}$ and ${ubpf(b)^n}$ multiply the prime numbers in $cbpf(a,b)$
+$n$ times each.
+These are a component of $c^n$.
+
+
+\begin{equation}
+\label{eqn:primesexpanded1}
+    2 \prod{cbpf(a,b)}^n      = \frac{c^n}{\big( \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}  \big) }
+\end{equation}
+
+% 
+% \begin{equation}
+% \label{eqn:primesexpanded1}
+%         \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}    = \frac{c^n}{ \prod{cbpf(a,b)}^n }
+% \end{equation}
+% 
+% 
+% $c^n$ must contain $ \prod{cbpf(a,b)}^n $
+% %Try to find a and b such that a^2 + b^2 = 144;
+% 
+% 
+% \begin{equation}
+% \label{eqn:primesexpanded1}
+%         \prod{ubpf(a)^n}  +    \prod{ubpf(b)^n}    = \frac{c^n}{ \prod{cbpf(a,b)}^n }
+% \end{equation}
+% 
+
+%It should be even because its multiplied by 2.
+% It must have all the common factors of $a$ and $b$ twice but the uncommon factors only once.
+% This seems to be an apparent contradiction.
+% It means the $2 \prod{cpf(a,b)}^n $ term is  multiplied by at least one other prime number. % and therefore cannot have an nth root.
+% A number must consist of n times of all its prime number can give an integer nth root.
+% Because a and b are different they must consist of at least one difference in prime numbers.
+% 
+% Taking equation~\ref{eqn:primesexpanded}
+% and re-writing:
+% \begin{equation}
+% \label{eqn:primesexpanded2}
+% 	\sqrt[n]{2}^n \prod{cbpf(a,b)}^n  \prod{ubpf(a)^n}  \prod{ubpf(b)^n} = c^n
+% \end{equation}
+% 
+% 
+% Taking the nth root of both sides of equation~\ref{qn:primesexpanded2} gives
+% 
+% \begin{equation}
+% \label{eqn:primesexpanded2}
+% 	\sqrt[n]{2}  \prod{cbpf(a,b)} (\prod{ubpf(a)}   \prod{ubpf(b)}) = c 
+% \end{equation}
+% 
+
+
+
+Which means that a product of $c$ is a root of 2, it is therefore irrational
+and not a whole number.
+
+
+If $c$ is even 2 can be divided from each side until only
+both $c$ and $ \prod{cbpf(a,b)}  \prod{ubpf(a)}   \prod{ubpf(b)} $
+are odd. The $\sqrt[n]{2}$ term remains. The result $c$ is therefore irrational.
+
+%Adding $a^n$ and $b^n$ where a and b are different means adding primes to th power of N
+%which means they have no integer nth root.
+
+{
+\footnotesize
+\bibliographystyle{plain}
+\bibliography{../../vmgbibliography,../../mybib}
+}
+
+\today
+%\today
+\end{document}
+