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added more trivial examples to illustrate the

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presevation of common prime factors under addition.
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Robin P. Clark 2015-07-02 09:57:42 +01:00
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papers/fermat/fermat.tex
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@ -77,7 +77,7 @@ The numbers $a$ and $b$ may have common and uncommon prime factors; these can b
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. $2^5 = 32$.
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}
@ -124,7 +124,7 @@ are the common ones, i.e. cbpf(a,b).
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, $a+b \equiv \prod bfp(a) + \prod bfp(b)$
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.
@ -132,6 +132,66 @@ 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 $a^n+b^n = c^n$.
%
\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.
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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, $7$.
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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.
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%
%
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.
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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:
$11\times 11(\prod\{5,97\} + \prod \{13,43\}) .$
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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 $b$.