327 lines
11 KiB
TeX
327 lines
11 KiB
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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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{proposition}[theorem]{Proposition}
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\newtheorem{corollary}[theorem]{Corollary}
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\newcommand{\pf}{prime~factor}
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\newcommand{\pfs}{prime~factors}
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\def\layersep{1.8cm}
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\linespread{1.0}
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\title{flt primes}
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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.P. Clark}
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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{
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Viewing integers as collections of prime numbers and
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using properties of prime numbers under addition and multiplication
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this paper offers a proof of Fermats last theorem for
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all positive integers $> 2$.
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}
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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 positive integers into constituent products of bags of primes}
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Any positive integer can be represented as a collection (or bag) of prime numbers multiplied together.
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A function $bpf()$ or `bag of {\pfs}' is defined to represent this.
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\begin{equation}
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a^n + b^n = \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 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 to both; $cbpf(a,b)$.
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A `Set' in mathematics is a collection of objects that may have only one of each type of element.
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A `bag' is similar to a Set, except that it may have duplicates.
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The number $32$ is represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$.
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Viewing the addition of $a^n +b^n$ as products of bags of common and uncommon~{\pfs}:
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\begin{equation}
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\label{eqn:primesexpanded0}
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\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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\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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\section{Properties of numbers viewed as products of bags of prime factors}
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\subsection{Conditions where some Primes are guaranteed not preserved in addition}
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%
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% ADDITION DESTROY UNCOMMON PRIME FACTORS
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%
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Adding numbers creates a `dissolving' of prime factors in the result:
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that is addition of numbers causes the uncommon prime factors to become
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lost.
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%
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Consider $43 +21 = 64$. These primes add up to a result with
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a bag of six twos i.e. $bpf(64) = \{2,2,2,2,2,2\}$ or more conventionally $64=2^6$.
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%%
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Prime numbers are unique. Adding to them, or adding other prime numbers to them, takes that unique
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property away.
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%
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If a prime is added to another prime number the result
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cannot be a prime number, simply because all prime numbers above two are odd;
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the result of the addition must even and therefore have at least a prime factor of two.
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%
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Further, if numbers are added, the prime factors of the
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result will not contain any of the uncommon primes.
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That is the only prime factors preserved in the result of addition of $a$ and $b$
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are the common ones, i.e. cbpf(a,b).
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%
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%consider
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%
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Thus only common prime factors in $a$ and $b$ are preserved
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as a result of equation~\ref{eqn:primesexpanded1}.
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This is simply because in addition
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the common prime factors can be extracted, $a+b \equiv \prod bfp(a) + \prod bfp(b)$
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re-writing $\prod cbpf(a,b) \big( \prod ubpf(a) + \prod ubpf(b) \big)$:
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This means the uncommon prime factors of $\big( \prod ubpf(a) + \prod ubpf(b) \big)$
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are lost and the $\prod cbpf(a,b)$ preserved.
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%
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%
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%
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This means for $a+b$ and $a^n+b^n$ the only prime factors preserved (i.e. in $c^n$)
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are those common to $a$ and $b$.
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\subsection{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 representing it as a bag of prime numbers
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$\{2,2,2,2,3,3\}$ or conventionally as $ 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 a whole number $n^{th}$ root all the prime numbers that comprise
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that number must be at the power of n or greater.
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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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%
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% \begin{equation}
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% \label{eqn:primesexpanded2}
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% \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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Extending this concept, taking a number as a bag of prime factors and then taking it to the
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power of $n$, means taking the number of individual primes in the bag of prime factors and multiplying that number by $n$.
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For instance the number 306, as a bag of prime factors is $\{2,3,3,17\}$ i.e. $306=\prod \{2,3,3,17\}$.
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Cubing; $306^3$ gives 28652616: as a bag of prime factors 28652616 is $\{2,2,2,3,3,3,3,3,3,17,17,17\}$.
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Viewing the result of the cubing in terms of bags of primes numbers,
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\begin{itemize}
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\item 306 has 3 twice as a prime factor, $306^3$ has 3 six times as a prime factor:
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\item 306 has 2 once as a prime factor, $306^3$ has 2 3 times as a prime factor:
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\item 306 has 17 once as a prime factor, $306^3$ has 17 3 times as a prime factor.
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\end{itemize}
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% For instance the number
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% \begin{equation}
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% \label{eqn:primesexpanded21}
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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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%
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%
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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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%
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%
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% \begin{equation}
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% \label{eqn:primesexpanded22}
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% \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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\section{Proof by Contradiction.}
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%
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For $a^n + b^n = c^n$ to be true for whole numbers $ > 2$, the highest prime factors on both sides of the equation must be equal.
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%
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That is to say the highest
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prime number in the bag $bpf(a^n + b^n)$
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must be the same as the highest prime factor in the bag $bpf(c^n)$.
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\subsection{Case where the highest prime factor in $pbf(c)$ is a single instance}
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Due to the destruction of non-common prime factors under addition
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both $a$ and $b$ must contain the highest prime in $c$.
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If $a$ and $b$ are whole numbers they either create a result with
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the highest prime more than once, or it is destroyed by addition.
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For $a^n + b^n = c^n$, for the highest prime, this means $a+b=1$.
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This means that where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers.
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This concept can be extended to numbers where there are duplicate highest primes.
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%% Simple case where only one of highest prime factor in c^n
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% describe contradiction for simple case:
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\subsection{Case where the highest prime factor in $pbf(c)$ is a multiple instance}
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%% case where highest prime factor in c^n may be duplicated.
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The highest prime factor in the bag may be duplicated.
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Taking the value $c$ as the product of a bag of prime numbers, it must have
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a largest prime in $c$ (to a power $t$ which is one or more), i.e. $p^t$.
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%
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When $c$ is taken to the power $n$, $c^n$, that
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means this prime factor becomes $p^{tn}$.
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%
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Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{(tn)}$, for that prime in the result.
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%
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Because prime numbers are by definition indivisible by other
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whole numbers, the only way to get a prime number taken to
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a power $p^t$ by addition is to add proportions that add up to one $p^t$.
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%
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This means both a and b must contain this prime factor {\em in some proportion}
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so that $a p^{tn} + b p^{tn} = p^{tn} $ satisfy the highest prime in $c$.
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%
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In order for this to be true $a$ and $b$ must both be fractions of a whole number:
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again this means $a+b$ must equal 1.
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Thus where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers.
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\subsection{trivial case}
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Take the trivial case where $n=2$ and $c$ has the prime number 7 as one of its prime~factors:
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%
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$$ a^n + b^n = 7^n = 49 \; . $$
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%
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In order to get the prime factor 7 in the result both a and b must have the prime number 7 in them.
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That is the numbers $a$ and $b$ must both have the number 7 as a common prime factor
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to get seven as a prime factor in the result.
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Any other number will not give a 7 in the bag of prime numbers representation of the result.
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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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%
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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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%
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%
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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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