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\title{flt primes}
\begin{document}
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\author{R.P. Clark}
 
\maketitle 
 
\today 

\paragraph{Keywords:} fermat; prime;
%\small

\abstract{ % \em 
} % abstract



\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 prime factors' is defined to represent this.
\begin{equation}
	\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.
Thus the number $32$ is represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$.

Viewing the addition of $a^n +b^n$ as products of bag of  prime factors:
\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{ Primes are guaranteed not preserved in addition }

Only common prime factors are guaranteed preserved as a result of equation~\ref{eqn:primesexpanded1}.
%
This means for $a^n+b^n$ the only prime factors guaranteed to be in $c^n$
are are those common in $a$ and $b$.
%
Adding numbers creates a dissolving of the prime factors in the result.
Addition of primes causes the highest prime factors to become
lost, but increases the number of smaller prime factors.
%
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 $64=2^6$.
%
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.


\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.

\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} $


% 
% \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{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.


%% 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 $p^{t+n}$.
%
Therefore, for that highest prime in $c$,   $a^n + b^n$ must add up to $p^{(t+n)}$, 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 fractions of a whole number.

\subsection{trivial case}


Take the trivial case where $c$ has the prime number 7:
%
$$ a^n + b^n = 7^n = 49 $$
%
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 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.

{
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\today
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