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