329 lines
11 KiB
TeX
329 lines
11 KiB
TeX
|
|
|
|
%%% 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}
|
|
|
|
\newtheorem{theorem}{Theorem}[section]
|
|
\newtheorem{lemma}[theorem]{Lemma}
|
|
\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
|
|
\pagenumbering{arabic} % Arabic page numbers hereafter
|
|
\author{R.P. Clark}
|
|
|
|
\maketitle
|
|
|
|
\today
|
|
|
|
\paragraph{Keywords:} fermat; prime;
|
|
%\small
|
|
|
|
|
|
\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
|
|
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 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 {\pfs}' is defined to represent this.
|
|
\begin{equation}
|
|
a^n + b^n = \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 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 to both; $cbpf(a,b)$.
|
|
A `Set' in mathematics is a collection of objects that may have only one of each type of element.
|
|
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~{\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 \; ,
|
|
\end{equation}
|
|
|
|
this can be re-written as:
|
|
|
|
\begin{equation}
|
|
\label{eqn:primesexpanded1}
|
|
\prod{cbpf(a,b)}^n \big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) = c^n \; .
|
|
\end{equation}
|
|
|
|
|
|
|
|
\section{Properties of numbers viewed as products of bags of prime factors}
|
|
|
|
\subsection{Conditions where some Primes are guaranteed not preserved in addition}
|
|
%
|
|
% ADDITION DESTROY UNCOMMON PRIME FACTORS
|
|
%
|
|
Adding numbers creates a `dissolving' of prime factors in the result:
|
|
that is addition of numbers causes the uncommon prime factors to become
|
|
lost.
|
|
%
|
|
Consider $43 +21 = 64$. These primes add up to a result with
|
|
a bag of six twos i.e. $bpf(64) = \{2,2,2,2,2,2\}$ or more conventionally $64=2^6$.
|
|
%%
|
|
Prime numbers are unique. Adding to them, or adding other prime numbers to them, takes that unique
|
|
property away.
|
|
%
|
|
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.
|
|
%
|
|
Further, if numbers are added, the prime factors of the
|
|
result will not contain any of the uncommon primes.
|
|
That is the only prime factors preserved in the result of addition of $a$ and $b$
|
|
are the common ones, i.e. cbpf(a,b).
|
|
%
|
|
%consider
|
|
%
|
|
Thus only common prime factors in $a$ and $b$ are preserved
|
|
as a result of equation~\ref{eqn:primesexpanded1}.
|
|
This is simply because in addition
|
|
the common prime factors can be extracted, $a+b \equiv \prod bfp(a) + \prod bfp(b)$
|
|
extracting the common prime factors this becomes $\prod cbpf(a,b) \big( \prod ubpf(a) + \prod ubpf(b) \big)$:
|
|
this means the uncommon prime factors of $\big( \prod ubpf(a) + \prod ubpf(b) \big)$
|
|
are lost and the $\prod cbpf(a,b)$ preserved.
|
|
%
|
|
Because of this property of addition of numbers in relation to preserved
|
|
prime factors, it can be used to make inferences on the equation $a^n+b^n = c^n$.
|
|
%
|
|
%
|
|
This means for $a+b$ and $a^n+b^n$ the only prime factors preserved (i.e. in $c^n$)
|
|
are those common to $a$ and $b$.
|
|
|
|
|
|
\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 $n$.
|
|
Consider the square root of 144.
|
|
This can be written as
|
|
$12 \times 12$ or representing it as a bag of prime numbers
|
|
$\{2,2,2,2,3,3\}$ or conventionally as $ 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 a whole number $n^{th}$ root all the prime numbers that comprise
|
|
that number must be at the power of n or greater.
|
|
|
|
% 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}
|
|
% \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.
|
|
|
|
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}
|
|
% 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}
|
|
|
|
|
|
\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.
|
|
%
|
|
That is to say the highest
|
|
prime number in the bag $bpf(a^n + b^n)$
|
|
must be the same as the highest prime factor in the bag $bpf(c^n)$.
|
|
|
|
\subsection{Case where the highest prime factor in $pbf(c)$ is a single instance}
|
|
|
|
Due to the destruction of non-common prime factors under addition
|
|
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 $ > 1$; $a^n + b^n \neq c^n$ 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:
|
|
\subsection{Case where the highest prime factor in $pbf(c)$ is a multiple instance}
|
|
%% case where highest prime factor in c^n may be duplicated.
|
|
|
|
The highest prime factor in the bag may be duplicated.
|
|
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 $p^{tn}$.
|
|
%
|
|
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^{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 $ > 1$; $a^n + b^n \neq c^n$ for whole numbers.
|
|
|
|
\subsection{trivial case}
|
|
|
|
Take the trivial case where $n=2$ and $c$ has the prime number 7 as one of its prime~factors:
|
|
%
|
|
$$ a^n + b^n = 7^n = 49 \; . $$
|
|
%
|
|
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
|
|
to get seven as a prime factor in the result.
|
|
Any other number will not give a 7 in the bag of prime numbers representation of the result.
|
|
|
|
|
|
|
|
|
|
%
|
|
% \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}
|
|
|