diff --git a/papers/fermat/fermat.tex b/papers/fermat/fermat.tex index 49c936f..c91094f 100644 --- a/papers/fermat/fermat.tex +++ b/papers/fermat/fermat.tex @@ -23,12 +23,22 @@ \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} +\newcommand{\pf}{prime~factor} +\newcommand{\pfs}{prime~factors} + \def\layersep{1.8cm} \linespread{1.0} \title{flt primes} + + + + + + + \begin{document} - % numbers at outer edges + % numbers at outer edges \pagenumbering{arabic} % Arabic page numbers hereafter \author{R.P. Clark} @@ -39,10 +49,13 @@ \paragraph{Keywords:} fermat; prime; %\small -\abstract{ % \em -} % abstract - +\abstract{ +Viewing integers as collections of prime numbers and +using properties of prime numbers under addition and multiplication +this paper offers a proof of Fermats last theorem for +all positive integers $> 2$. +} \section{Introduction} Fermat's Last Theorem @@ -53,9 +66,9 @@ equation $a^n + b^n = c^n$ for any integer value of n greater than two. \section{Breaking positive integers into constituent products of bags of primes} Any positive integer can be represented as a collection (or bag) of prime numbers multiplied together. -A function $bpf()$ or `bag of prime factors' is defined to represent this. +A function $bpf()$ or `bag of {\pfs}' is defined to represent this. \begin{equation} - \prod{bpf(a)}^n + \prod{bpf(b)}^n = c^n + a^n + b^n = \prod{bpf(a)}^n + \prod{bpf(b)}^n = c^n \end{equation} %The function $bpf()$ will always contain 1. @@ -66,7 +79,7 @@ A `Set' in mathematics is a collection of objects that may have only one of each A `bag' is similar to a Set, except that it may have duplicates. The number $32$ is represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$. -Viewing the addition of $a^n +b^n$ as products of bags of common and uncommon~prime~factors: +Viewing the addition of $a^n +b^n$ as products of bags of common and uncommon~{\pfs}: \begin{equation} \label{eqn:primesexpanded0} \prod{cbpf(a,b)}^n \prod{ubpf(a)^n} + \prod{cbpf(a,b)}^n \prod{ubpf(b)^n} = c^n \; , @@ -83,8 +96,9 @@ this can be re-written as: \section{Properties of numbers viewed as products of bags of prime factors} -\subsection{ Primes are guaranteed not preserved in addition } +\subsection{Conditions where some Primes are guaranteed not preserved in addition} % +% ADDITION DESTROY UNCOMMON PRIME FACTORS % This means for $a^n+b^n$ the only prime factors guaranteed to be in $c^n$ are are those common in $a$ and $b$. @@ -100,11 +114,14 @@ If a prime is added to another prime number the result cannot be a prime number, simply because all prime numbers above two are odd; the result of the addition must even and therefore have at least a prime factor of two. % +Prime numbers are unique. Adding to them, or adding other prime numbers to them, takes that unique +property away. +% Thus only common prime factors in $a$ and $b$ are preserved as a result of equation~\ref{eqn:primesexpanded1}. % -\subsection{conditions for having a integer root} +\subsection{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 $n$. @@ -130,39 +147,48 @@ that number must be at the power of n or greater. % Adding two prime numbers at any power greater than 1 % and then taking a root means getting an irrational number. -\subsection{} - -\begin{equation} -\label{eqn:primesexpanded21} - \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} $ - - +Extending this concept, taking a number as a bag of prime factors and then taking it to the +power of $n$, means taking the number of individual primes in the bag of prime factors and multiplying that number by $n$. +For instance the number 306, as a bag of prime factors is $\{2,3,3,17\}$ i.e. $306=\prod \{2,3,3,17\}$. +Cubing; $306^3$ gives 28652616: as a bag of prime factors 28652616 is $\{2,2,2,3,3,3,3,3,3,17,17,17\}$. +Viewing the result of the cubing in terms of bags of primes numbers, +\begin{itemize} +\item 306 has 3 twice as a prime factor, $306^3$ has 3 six times as a prime factor: +\item 306 has 2 once as a prime factor, $306^3$ has 2 3 times as a prime factor: +\item 306 has 17 once as a prime factor, $306^3$ has 17 3 times as a prime factor. +\end{itemize} +% For instance the number +% \begin{equation} +% \label{eqn:primesexpanded21} +% \prod{cbpf(a,b)}^n \big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) = c^n +% \end{equation} % % \begin{equation} -% 2 \prod{cpf(a,b)}^n = \frac{\prod{bpf(c)^n}}{\prod{upf(a)^n} \prod{upf(b)^n}} +% 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} +% +% +% 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:primesexpanded22} +% \prod{cbpf(a,b)}^n = \frac{c^n}{\big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) } % \end{equation} -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:primesexpanded22} - \prod{cbpf(a,b)}^n = \frac{c^n}{\big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) } -\end{equation} - - -\section{contradiction} +\section{Proof by Contradiction.} % 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. % @@ -177,6 +203,10 @@ both $a$ and $b$ must contain the highest prime in $c$. If $a$ and $b$ are whole numbers they either create a result with the highest prime more than once, or it is destroyed by addition. +For $a^n + b^n = c^n$, for the highest prime, this means $a+b=1$. +This means that where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers. +This concept can be extended to numbers where there are duplicate highest primes. + %% Simple case where only one of highest prime factor in c^n % describe contradiction for simple case: @@ -188,25 +218,26 @@ Taking the value $c$ as the product of a bag of prime numbers, it must have a largest prime in $c$ (to a power $t$ which is one or more), i.e. $p^t$. % When $c$ is taken to the power $n$, $c^n$, that -means this prime factor becomes $p^{t+n}$. +means this prime factor becomes $p^{tn}$. % -Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{(t+n)}$, for that prime in the result. +Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{(tn)}$, for that prime in the result. % Because prime numbers are by definition indivisible by other whole numbers, the only way to get a prime number taken to a power $p^t$ by addition is to add proportions that add up to one $p^t$. % This means both a and b must contain this prime factor {\em in some proportion} -so that $a p^{t+n} + b p^{t+n} = p^{t+n} $ satisfy the highest prime in $c$. - -In order for this to be true $a$ and $b$ must both be fractions of a whole number. +so that $a p^{tn} + b p^{tn} = p^{tn} $ satisfy the highest prime in $c$. +% +In order for this to be true $a$ and $b$ must both be fractions of a whole number: +again this means $a+b$ must equal 1. +Thus where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers. \subsection{trivial case} - -Take the trivial case where $c$ has the prime number 7: +Take the trivial case where $n=2$ and $c$ has the prime number 7 as one of its prime~factors: % -$$ a^n + b^n = 7^n = 49 $$ +$$ a^n + b^n = 7^n = 49 \; . $$ % In order to get the prime factor 7 in the result both a and b must have the prime number 7 in them. That is the numbers $a$ and $b$ must both have the number 7 as a common prime factor