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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{  
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, $a+b \equiv \prod bfp(a) + \prod bfp(b)$
re-writing $\prod cbpf(a,b) \big( \prod ubpf(a) + \prod ubpf(b) \big)$:
This means the uncommon prime factors of $\big( \prod ubpf(a) + \prod ubpf(b) \big)$
are lost and the $\prod cbpf(a,b)$ preserved.
%
%
%
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 $ > 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:
\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 $ > 2$; $a^n + b^n \neq c^n$ 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
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}

\today
%\today
\end{document}