Robin_PHD/papers/fermat/fermat.tex

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\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 $32$ for instance, can be 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}

That is to say the inner bracket on the left hand side is an addition of co-prime numbers,
\begin{equation}
\label{eqn:primesexpanded2}
    \prod{ubpf(a)^n}  \perp   \prod{ubpf(b)^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, $a+b \equiv \prod bpf(a) + \prod bpf(b)$
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 $a^n+b^n = c^n$.
%
%%% NIEL HARDINGS BIT
%
For the classic right angle triangle $3^2+4^2=5^2$ the prime numbers in the 
addition are not preserved. As products of bags of prime numbers this is 
$\prod \{3,3\} + \prod\{2,2,2,2\} = \prod \{5,5\}$.
%
\subsubsection{Trivial example, single prime factor preserved}
%
Consider $bpf(182)=\{2,7,13\}$ and $bpf(2365)=\{5,11,43\}$ these have no common prime factors
so adding them should result in a number which when factored contains none of the primes in $182$ and $2365$.
Adding $182+2365=2574$; so taking prime factors $bpf(2574)=\{3,3,283\}$.
The loss of uncommon prime factors for this case of addition is shown to be true.

%
Now consider two numbers with one common factor $bpf(49665)=\{3,7,5,11,43\}$ and 
$bpf(322) = \{2,7,23\}$. These have one common factor, $7$.
%
The addition of these numbers is: $ 322 + 49665 \equiv \prod \{2,7,23\} + \prod \{3,7,5,11,43\} $; as the 7 is added twice
it can be taken outside of the addition and multiplied by what remains of it
$7(\prod \{2,23\} + \prod \{3,5,11,43\})$. 
%
This means the unique properties
of the uncommon prime factors (those within the bracket) will be destroyed,
and thus they will not appear in the result
of the addition. The number 7 will be multiplied by the number in brackets, 
and thus that prime factor will survive in the result.
%
So, $ 322 + 49665 = 49987$: $bpf(49987) = \{7,37,193\}$.
As expected the common prime factor,7, exists in the result of the addition
but the uncommon prime factors have disappeared.

\subsubsection{Trivial example, multiple prime factor preserved}

Consider a repeated prime factor (i.e. a prime $p$ to the power $t$ $p^t$). The same rules apply.
%
For this prime factor to be preserved in the result
of an addition, it must be in both summed quantities at an equal or greater power (or
number of duplicates of that prime in the bag).
%
Consider two numbers with $11^2$ in one number and $11$ in the other: $bpf(110)=\{11,2,5\}$
and $bpf(67639)=\{13,11,11,43\}$. The only common prime factor is 11 once.
%
%
%

Adding $67639+110 = 67749$, $bpf(67749) = \{3,11,2053\}$. As expected the prime factor
11 only appears once in the result, because only one 11 can be taken as a common factor; $11(\prod\{3,2053\}+\prod \{13,11,43\})$
 addition has been to the lone 11 within the brackets and thus `dissolved' it from the result.

To get $11^2$ preserved as a prime factor is must appear twice, i.e. on both sides of the 
addition. 
%
%
Taking a new number with two prime factors of 11, say 58685,
$bpf(58685)=\{5,11,11,97\}$ and adding this 
to $bpf(67639)=\{13,11,11,43\}$:
$58685+67739 = 126324 \equiv \prod\{5,11,11,97\}+\prod\{13,11,11,43\} $;
because $11^2$ can be taken out of the bracket it can be re-written thus:
$11\times 11(\prod\{5,97\} + \prod \{13,43\}) .$
%
%
This means the result $58685+67739 = 126324$, should contain $11$ twice as a prime factor
but all the uncommon prime factors should not be 
present in the result, i.e. $bpf(126324)=\{2,2,3,3,11,11,29\}$.
%
%
%
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 $b$.
Another way to look at this is the number of common prime factors is multiplied by
the addition in the brackets.
%% BONFIRE OF THE PRIMES:wq

\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 $bpf(c)$ 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, a and b must contain a proportion
of the highest prime factor in $c$; this means a and b {\em for that prime factor} would have to add up to one.
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 $bpf(c)$ 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 for the highest prime factor, $p^t$, $a+b$ must equal 1. In other words a proportion of the 
highest prime factor in $c$ must exist in $a$ and $b$ such that they add up to one.
Thus where $a$ and $b$ are $ > 1$; $a^n + b^n \neq c^n$ for whole numbers.

\subsection{trivial case: single prime number in $c^n$}

Take the trivial case where $n=3$ and  $c$ has the prime number 7 as one of its prime~factors:
%
$$ a^n + b^n = 7^n = 343 \; . $$
%
In order to get the prime factor $7^3$ in the result both $a$ and $b$ must have some proportion of $7^3$ in them.
That is the numbers $a$ and $b$ must both have the number $7^3$ as a common prime factor
to get $7^3$ as a prime factor in the result.
Any other number will not give a $7^3$ in the bag of prime numbers representation of the result.
Thus to make $ a^n + b^n = 343 $ both a and b could contain fractional quantities of  $7^3$
but not both.
Thus for whole numbers, where $\prod bpf(c^n)$ contains a single prime  $ a^n + b^n \neq c^n  \; where \; n < 2$.

\subsection{trivial case: multiple prime numbers in the bags}

Take the trivial case where $n=3$ and  $c$ has the prime numbers 13 and 11 as its prime~factors:
%
say $ c = \prod \{ 13,13,13,11 \} = 24167$ cubing this gives $ \prod \{13,13,13,13,13,13,13,13,13,11,11,11\} $ or $ \prod \{13^9,11^3\} $ .
A strategy of trying to preserve the prime factors under addition can now be attempted.
To preserve the primes both $13^3$ and $11$ must be present in both a and b.
So trying $a = \{13^3,11\}$ and  $b = \{13^3,11\}$ taking cubes gives $a^3 + b^3 = \prod \{13^9,11^3,2\}$
Here both primes $13^3$ and $11$ have been preserved in the addition but there is an extra factor in the result, i.e. the $2$.
Adding any other prime factors to either $a$ or $b$ makes the result too
large. Adding the minimum quantity to both in order to preserve the prime factors
gives a result with the prime factor $2$ in it as well.
%
% 
\paragraph{Looking at just the highest prime factor}
For numbers to be equal their highest prime factors must have the same index (or power).
To get the result $c^n$ from the addition $13^3$ must be present in both $a$ and $b$, 
if it is present singularly in $a$ and $b$, it will 
be present twice in the result (i.e. adding the prime $2$) to the result product.
\paragraph{thinking about preserving $13^3$ in the result $c^n$}
So to preserve $13^3$ in the result; consider $a = \prod \{ 13,13,13\}$ and $b = \{ 13,13,13\}$.
Adding them cubed; $a^3+b^3 = \prod \{ 13^9\} + \prod \{ 13^9\}$ which can be re-written as
 $ \prod \{13^9\} (1 + 1) \}$ which gives $\prod \{ 13^9,2 \}$
; the extra prime factor of 2 means that while $13^3$ was preserved a new prime factor popped up in the result.
%; as $c$ is to the power of n
%it should be $2^3$.
In general this means $a$ and $b$ being whole numbers
cannot  make the equation $a^n+b^n=c^n$ true.
%
Or in  other words it comes back to the addition $ a^n + b^n = c^n $ preserving the common prime factors in the result $c^n$,
but not the uncommon factors. 
%
\paragraph{Case where $a^n$ and $b^n$ may have a large number of uncommon factors}
A prime number may be produced by the addition
that is larger than any found in $a$ or $b$.
If this occurs, the new larger prime will not be present in $c^n$.
Thus for whole numbers, where $\prod bpf(c^n)$ contains multiple primes  $ a^n + b^n \neq c^n  \; where \; n < 2$.

% \subsection{Another way to look at it}
% 
% Take the highest prime factor in the result $c^n$.
% Try to reproduce it using integers with $a^n$ and $b^n$.
% 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% IS it true that if there are any common factors a^n +b^n = c^n cannot exist
% even for n >= 2 ????
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Hole in the theory..... pythagorean squares.... and bollocks}

Pythagorean squares all contain uncommon prime factors in their addition and create
a large prime to the power of 2 as a result.
Fermat's last theorem is for powers, the $n$ factor, of greater than 2.
Consider where all the cubed prime factors in $a$ and $b$ are uncommon (the is $a$ and $b$ are co-prime).
The result of the addition will create a number that is the product of prime
numbers; why could some combination not produce a number with $n$ times the power of
all its primes.
Consider that although the numbers $a$ and $b$ are co-prime, there will
be prime numbers `missing' from both $a$ and $b$. If both numbers
added are odd, the prime factor 2 has to pop-up.
If 2 is a common prime factor but 3 is not that will pop-up and so on.

But I am stumped here. If all the numbers in $a$ and $b$ are co-prime, i.e $a \perp b$,  
$a^n + b^n$ could, magically, produce a result where the result $n$ number of the same prime, 
or even several prime factors, but all to the power of $n$, i.e. $c^n$ where $c$ 
is the prime addition of $a^n + b^n$. I just cannot see why not, so all this has proved nothing. Arrrrghhhhh.
Probably good for the soul to look at prime numbers a-lot.....

Maybe you can take prime triplets say, and try to work back in some general way and show no $a^3 + b^3$ integer solution exists?
Or a general way of showing $a^n + b^n \; where \; n > 2$ cannot produce a simple $n$ number of all prime factors shown in the result....
That is the last thing........ that is the only way flt  can have an exception, where $a \perp b$, AND 
it produces a $c^n$ where c is an integer.....

BUT IS IT  A HOLE

IT JUST MEANS THERE IS ONLY one special case for an flt 

Common primes are disproved in this way,
only where $a^n + b^n = \forall p \in c^n | p \perp a \wedge p \perp b$ can actually work and this
means $a \perp b$.


$ a \perp b \perp c $ this is actually how it works for Pythagorean triangles.
Is there anyway to go back from $\forall p \in c^n | p \perp a \wedge p \perp b $

\section{Further work}

Using the rule of adding uncommon factors, an iterative `fishing' method for finding prime numbers can be applied.
An addition is made,  and the lower prime factors of the result are added as uncommon factors in a subsequent addition
until finally a larger single prime number is found.
This can be done with single primes, or for finding larger primes in less iterations, with multiple powers of primes.
For example consider
%
starting with the first two uncommon primes squared,
$$\prod \{2,2\} + \prod \{3,3\} = \{13\};$$
this gives the prime number 13.
%octave:26> factor(2^2+3^2)
%ans =  13
%
%
%octave:28> factor(2^2+3^2*13^2)
%ans =
By placing 13 squared as an uncommon factor, more primes are shaken out:
$$\prod \{2,2\} + \prod \{3,3,13,13\} = \prod \{5,5,61\}.$$
% 
Making the prime number 5 uncommon in the addition gives:
$$\prod \{2,2\} + \prod \{3,3,5,5,13,13\} = \prod \{17,2237\};$$
%octave:29> factor(2^2+3^2*13^2*5^2)
%ans =
%
removing the 17 by making it an uncommon factor in the addition:
   %  17   2237
$$\prod \{2,2\} + \prod \{3,3,5,5,13,13,17,17\} = \{10989229\};$$
%octave:30> factor(2^2+3^2*13^2*5^2*17^2)
%$ans =  10989229
gives a larger prime number 10989229 and so on.

\subsection{Fishing with cubic primes}
Fishing with primes cubed reveals larger primes quicker: take
$$ \prod \{2,2,2,5,5,5,11,11,11\} + \prod \{3,3,3,7,7,7,13,13,13\} = \prod \{ 383 ,  56599 \} \; ,$$
Adding 56599 as a single uncommon factor;
$$ \prod \{2,2,2,5,5,5,11,11,11\} + \prod \{3,3,3,7,7,7,13,13,13,56599\} = \prod \{ 151, 359, 1553, 13679 \} \; , $$
Removing 359 reveals a large prime:
$$ \prod \{2,2,2,5,5,5,11,11,11\} + \prod \{3,3,3,7,7,7,13,13,13,56599,359\} = \{413419682557097\} $$
 
% 
% \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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\bibliography{../../vmgbibliography,../../mybib}
}

\today
%\today
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