now for the finale, can't catch up ^3 and above

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+\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}