From a34a750c84b41763c045bba7f8c7e27df38f6176 Mon Sep 17 00:00:00 2001 From: "R.P. Clark" Date: Sat, 17 Dec 2022 16:08:56 +0000 Subject: [PATCH] now for the finale, can't catch up ^3 and above --- flt.tex | 93 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 93 insertions(+) diff --git a/flt.tex b/flt.tex index e69de29..2cfd6dc 100644 --- a/flt.tex +++ b/flt.tex @@ -0,0 +1,93 @@ +\documentclass[a4paper,10pt]{article} +\usepackage[utf8]{inputenc} + +%opening +\title{Fermat Primes reasoning} +\author{R.P.Clark} + +\begin{document} + +\maketitle + +\begin{abstract} +Fermat reasoning for proving $a^N+b^N = c^N$ does not exist as an integer solution for values of $N \ge 3$. +\end{abstract} + +\section{Definitions} + +\subsection{Numbers as collections or bags of prime numbers} +For the reasoning that follows it is helpful to think of +integers as bags\footnote{Bags are like axiomatic sets except bags can hold duplicate values.} of prime numbers that can be multiplied together, or bag of Prime Numbers (BPN). +For instance the number, say 12 can be factored as $2\times2\times3$ +or $2^2 \times 3$. +\paragraph{Additional as a destructor of co-prime factors} +Consider addition using numbers as bags of prime numbers. +Consider adding 12 and 21 i.e. $12+21=33$. +$$ (2\times2\times3 + 3\times7) $$ +Clearly the common factor of three can be moved outside the brackets +$$ 3 \times (2\times2 + 7) $$ +This means that the co-prime elements within the brackets +will not be preserved, but destroyed / removed from the resultant BPN. +The BPN for 33 is $3\times11$, the 2 and 7 prime numbers +in the brackets have been removed. +\paragraph{Power Balanced Prime Numbers} +A number is prime balanaced for a given integer $N$, if all its prime factors exist $N*k$ +times where $k$ is a positive integer. as the PBPN is relevant for a given N value +PBPN is defined as a boolean function PBPN(N) where it is true or false +according to its BPN content. +For instance, the number 25 i.e. $5^2$ is prime balanced for $N=2$: 100 is also a PBPN(2) +$5^2\times2^2 = 100$. +The number 200 i.e. $5^2\times2^3$ is not a PBPN(2) because the power of 2 ($2^3$) is not cleanly divisible by 2. + +\subsection{Roots: integer or trancendental results} +In order for a root to have an integer result it must have a `Prime~Balanced' +PBPN. If it does not the root will contain a trancedental fractional power. +Consider 200 in the case above. Its square root is $5^1\times2^{3/2}$. +Any number greater than 1 with a fractional root will be transendental +because it cannot resolve back to a BPN. + +\subsection{Relatively Co-Prime} +If two numbers have at least some co-prime factors they are `Relatively co-prime' (RCP). +Thus numbers that can share some prime factors and have co-prime as well +are RCP. + +\section{Back to the Fermat conjecture} +\subsection{Discussion with $N=2$} +$$ a^N + b^N = c^N$$ +For the $c^N$ part to have an integer in $c$ the result of the addition must be +a PBPN. +Also both $a$ and $b$ must be less than $c$. +However they must add up to a PBPN. +This means a and b must be relatively co prime. +Consider the Pythagoras example for a set square 3,4,5. +$$4^2 + 3^2 = 5^2$$ +Viewing this as bags of prime numbers we have $2^3 + 3^2 = 5^2$. +Here the numbers being added are RCP (actually co-prime as well). +The addition has removed the co-prime factors of $2$ and $3$ and resulted +in a PBPN(2) number with 5 as its only prime number. + +Now consider multiplying the numbers in the addition +by a PBPN(2) number. The simplest is $2^2$. So +$$ 2^2\times3^2 + 2^2\times4^2 = 100$$ +or +$$ 2^2(3^2 + 4^2) = 5^2\times2^2 $$ + +The number 100 is true for PBPN(2) and therfore has an integer square root. + +\subsection{Discussion with $N=3$} +For the equation below +$$ a^3 + b^3 = c^3$$ +to be true for some integers a,b and c +the following conditions must be met. +\begin{itemize} + \item a and b must be less than c + \item a and b must be RCP + +\end{itemize} + + + + + + +\end{document}