addition of primes disolving effect described.
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@ -18,20 +18,20 @@
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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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\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.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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\author{R.P. Clark}
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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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@ -50,86 +50,104 @@ 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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\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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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 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 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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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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Thus 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 bag of prime factors:
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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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\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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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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\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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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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\subsection{ Primes are guaranteed not preserved in addition }
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Only common prime factors are guaranteed preserved as a result of equation~\ref{eqn:primesexpanded1}.
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%
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This means for $a^n+b^n$ the only prime factors guaranteed to be in $c^n$
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are are those common in $a$ and $b$.
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%
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Adding numbers creates a dissolving of the prime factors in the result.
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Addition of primes causes the highest prime factors to become
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lost, but increases the number of smaller prime factors.
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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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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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% \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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\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 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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$12 \times 12$ 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 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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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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\subsection{}
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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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\begin{equation}
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a^n + b^n = \prod{bpf(c)^n}
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\end{equation}
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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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% \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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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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@ -137,10 +155,65 @@ 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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\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{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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%% 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 $p^{t+n}$.
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%
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Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{(t+n)}$, 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^{t+n} + b p^{t+n} = p^{t+n} $ satisfy the highest prime in $c$.
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In order for this to be true $a$ and $b$ must both fractions of a whole number.
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\subsection{trivial case}
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Take the trivial case where $c$ has the prime number 7:
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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 result.
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%
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% \begin{equation}
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% \label{eqn:primesexpanded1}
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@ -181,15 +254,15 @@ These are a component of $c^n$.
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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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%
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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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25
papers/fermat/to_primes.c
Normal file
25
papers/fermat/to_primes.c
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#include <stdio.h>
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int main()
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{
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int number, div;
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printf("Enter a number to know its prime factor: ");
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scanf("%d", &number);
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printf("\nThe prime factors of %d are: \n\n", number);
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div = 2;
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while (number != 0) {
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if (number % div != 0)
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div = div + 1;
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else {
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number = number / div;
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printf("%d ", div);
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if (number == 1)
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break;
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}
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}
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return 0;
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}
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