The LAST THEOREM
of FERMAT
by Brian S.
McMillan
According to the
Wikipedia web site for Abelian groups, the Commutative and Associative
Properties dictates that the solutions to the following equations may be
addressed independently of the order in which the product is calculated, and
thereby provide equivalent arguments. In order to illustrate this, the
general form of the equation for all bases as applied to the first two terms
raised to Identical Odd Powers may be written
thus:
WHERE: The
base of the first term is equal to a (4a + c) Integer and the Second Term base
equals a (4b + d) Integer, or the Sums of the bases for both terms (that is less
the Odd Exponent greater than 1) equal a (4n + 0), (4n + 1), (4n + 2) or (4n +
3) Integer.
THEN:
((4a + c)^odd + (4b + d)^odd) / (4(a + b) + (c +
d))
= Integer
OR:
((4a + c)^odd - (4b + d)^odd) / (4 (a - b) + (c - d))
= Integer
___________________________________________________
CLOSURE
PROPERTIES for ODD INTEGERS in MULTIPLICATION and DIVISION
IF:
(4a + 1)(4b + 1) = 4n + 1
AND:
(4a + 3)(4b + 3) = 4n + 1
THEN:
(4a + 1)(4b + 3) = 4n + 3
AND:
(4a + 1)(4b + 1)(4c + 3) = 4n + 3
(Redundant)
SO:
(4a + 1)(4b + 3)(4c + 3) = 4n + 1
(Redundant)
OR:
(4a + 3)(4b + 3)(4c + 3) = 4n + 3 (Redundant)
HENCE:
(4a + 3)(4b + 3)(4c + 3)(4d + 3) = 4n + 1 (Redundant)
So we may see
from the above (Non-Redundant) examples that (4n + 3) Integers are only closed
for the non-matched set and Odd Multiple Terms. For the matched sets and
Even Term Count, it is the (4n + 1) Integers that dominate.
WHERE: odd
= any value Odd Integer exponent
IF:
(4a + 0)^odd =
4n
AND:
(4a + 2)^odd =
4n
OR:
(4a + 1)^odd = 4n +
1
AND:
(4a + 3)^odd = 4n + 3
Because of the
Fundamental Theorem of Integers, we already know that the properties for the
bases of the terms may be exploited by the operations of mathematics to include
or exclude certain types of Integers. It may also be proposed that the new
postulate for Even and Odd Exponents is just an extension of the Fundamental
Theorem. For Example:
For some Even
Integer (a), and Odd Integer (b).
IF:
(a^2 + b^2) = 4n + 1
AND:
2 * Odd = 4n + 2
THEN:
(Even^odd + Odd^odd) = Odd Integer
SO:
(Even^(2*odd) + Odd^(2*odd)) = Odd Integer
THEN:
((Even^2)^odd + (Odd^2)^odd) /(Even^2 + Odd^2) = Odd Integer
OR:
((Even^(even*odd) + Odd^(even*odd)) /(Even^even +
Odd^even)
= Odd Integer
HENCE:
((Even^even)^odd + (Odd^even)^odd) /(Even^even + Odd^even)
= Odd Integer
Since the final
two equations which are listed above, are identical, we may now begin our
polarity argument.
Now comes the
properties for the individual terms as they might be raised to Even and Odd
powers. This particular solution may be applied as a rule of omission for
the problem known as "The Last Theorem of Fermat". This rests upon the
fact that while certain 4n + 1 series numbers may be root-able to the square...
can 4n + 1 + 1 = 4n + 2 series numbers be root-able to the square or the cube,
or to any power for that matter? For example:
WHERE:
Odd + Even = Odd, Odd + Odd = Even, or Even + Even = Even
IF:
(4k + 1)^even = 4n + 1
AND:
(4k + 1)^odd = 4n + 1
AND:
(4k + 3)^even = 4n + 1
THEN:
(4k + 3)^odd = 4n + 3
OR:
(4k + 0)^even = 4n
SO:
(4k + 0)^odd = 4n
AND:
(4k + 2)^even = 4n
THEN:
(4k + 2)^odd = 4n
We have seen
above, that when we attempt to yield a (4n + 2) integer by any and all of the
combinations of possible bases with their corresponding exponents, the reverse
of the operation is also prohibited. So for three out of the four possible
Odd integer combinations shown below, a (4n + 2) integer is the result. We
may now eliminate those three combinations from our mathematical choices.
IF:
(4a + 1)^x + (4b + 1)^x = 4n + 2
AND:
(4a + 3)^x + (4b + 3)^x = 4n + 2
THEN:
(4a + 1)^even + (4b + 3)^even = 4n + 2
SO:
(4a + 1)^odd + (4b + 3)^odd = 4n
The final example
of interest, where (4a + 1)^odd + (4b + 3)^odd = 4n, is a bit tricker, but it
may also be shown that a 4n + 2 integer will always be embedded within the
remainder. If we think outside of the box and simply reverse the sine and,
therefore the order for one of the terms in the equation, and using the
Commutative and Associative Properties as applied to the Postulate of Exponents
with regard to the validity for Subtraction as well as Addition of the terms,
then our Polarity Solution is easily revealed along with the first six possible
combinations of our Proof. For example:
If (a) is less
than or equal to (b) in any of the pertinent examples that follow, then we will
get a negative number and the appropriate integer for that series may be added
instead of subtracted from the yield to divide by 4 and therefore verify that
Integers series designation. However, for example, simply subtracting 1
from 3 to get 2 will obviously achieve the same result, thus the usefulness of
the Induction approach.
WHERE: odd
= same value Odd Integer exponent
HENCE:
(4a + 1)^odd - (4b + 3)^odd = 4n + 2
OR:
(4a + 3)^odd - (4b + 1)^odd = 4n + 2
The Transmutation
of Properties within The Fundamental Theorem of Integers dictates that the last
two examples listed immediately above may be considered identical with regard to
the non-rootability of the yield. If we transpose the properties in the
third equation above so that (1 + 3) = (2 + 2) then we will have a yield that is
the Sum of two (4n + 2) type Integers each raised to an even power. Since
according to the Distributive Property, this yield will always be divisible by
2^(even+1) with an Odd Integer as a remainder, then we will always have a (4n +
2) type Integer as part of that yield, and therefore, no root will be
possible. For example:
IF:
(4a + 1)^even + (4b + 3)^even = 4n + 2
AND:
(4a + 2)^even + (4b + 2)^even = 4n
THEN:
((4a + 2)^even + (4b + 2)^even) /2^(even+1) = Odd
SO:
((4a + 2)^even + (4b + 2)^even) /2^even = 4n + 2
This next set of
equations utilizes the same argument, only in this case it is a (4n + 3) series
Integer which is guaranteed in the yield for which it has no Even Integral Root
and we have easily disposed of the following combinations in our Fermat
Argument. For example:
IF:
(4a + 0)^even - (4b + 1)^even = 4n + 3
AND:
(4a + 0)^even - (4b + 3)^even = 4n + 3
THEN:
(4a + 2)^even - (4b + 1)^even = 4n + 3
SO:
(4a + 2)^even - (4b + 3)^even = 4n + 3
Finally, it has
already been shown when two of the terms are either of the same Odd Integer,
that no root is possible. Since the three pairs of expressions represented
here are equivalent wether the sine for the terms is positive or negative, then
our argument is now complete.
IF:
(4a + 0)^odd + (4b + 1)^odd = 4n + 1
AND:
(4a + 0)^odd + (4b + 3)^odd = 4n + 3
THEN:
(4a + 1)^odd - (4b + 1)^odd = 4n + 0
SO:
(4a + 3)^odd - (4b + 3)^odd = 4n + 0
OR:
(4a + 1)^odd + (4b + 1)^odd = 4n + 2
HENCE:
(4a + 3)^odd + (4b + 3)^odd = 4n + 2
PS. If I
have left out anything, please bring it to my attention. Thank you.
Signed, Brian S.
McMillan
On the Web:
http://www.pumpraser.com/fermatwilsongoldbach.html
http://www.pumpraser.com/fermatmersenne.html
http://www.pumpraser.com/fermatgoldbach.html
For Mathematical
Physics:
http://www.pumpraser.com/physics.htm
Brian S.
McMillan, Copyright 2007-2008
___________________________________________________
THE
by Brian S.
McMillan
For any Integer N
greater than 4,
THEN:
(N - 1)! + 1 = 4n + 1
For any integer N
greater than 5,
THEN:
(N - 2)! - 1 = 4n + 3
IF:
((P - 1)! + 1) /P = Integer
AND:
((P - 2)! - 1) /P = Integer
Let us try a
reducto-ad-absurdum strategy via the Wilson Theorem or it's variant, for
example: If we assume that after a point there are no more 4n + 1
Primes.
WHERE:
P = (4k + 3) only...
IF:
(((4k
+ 1) - 1)! + 1) = 4n + 1
AND:
(((4k + 1) - 1)! + 1) /(4a + 3) /(4b + 3) = Integer
We may see that
we do NOT have to divide the 4n + 1 yield in the very first example by the (4k +
1) factor to see that it must be a 4n + 1 Integer. Since we have assumed
that (4n + 1) Primes are NOT infinite, and we have used a (4k + 1) number in our
example it must mean that the (4n + 1) yield must be composite and factorable
only by Pairs of (4n + 3) Primes. Because (4k + 1) must be a compound of
at least one pair of (4n + 3) Primes and that same composite must evenly divide
the yield. We now have a contradiction.
THEN:
(((4k + 3) - 1)! + 1) = 4n + 1
SO:
(((4k + 3) - 1)! + 1) /(4k + 3) = (4n + 3)
AND:
(((4k + 3) - 1)! + 1) /(4k + 3) = (4a + 3)(4b + 3)(4c + 3)
In the last
example above, it has been shown that if there are no more (4n + 1) Primes, then
once we divide our (4k + 3) Prime into the (4n + 1) yield, the remainder (4n +
3) in the second to the last equation above must either be Prime or a product
composed of 3 more (4n + 3) Primes multiplied together. We now have
another contradiction.
OR:
(((4k + 3) - 2)! - 1) = (4k + 3)
THEN:
(((4k + 3) - 2)! - 1) /(4k + 3) = Integer
HENCE:
(((4k + 3) - 2)! - 1) /(4k + 3) = (4a + 3)(4b + 3)
__________________________________________________
EUCLID-FERMAT
PROOF for the INFINITUDE of (4k + 1) PRIMES
Now the following
strategy that I have devised is not quite finished and in fact it may be reduced
to an absurd argument concerning Pseudo-Primes or Composite Integers of any form
eventually only being reducible to a closed (4n + 3) Prime Factorization.
I will send out the full argument later... it is very short, but
effective. It has to do with all (4n + 3) numbers being separated by 2^2
or 4, and therefore the product of these Primes cannot account for the (4n + 1)
climb in the Natural Integers. Furthermore, if at some point all (4n + 1)
Integers become composite, then according to
contradiction.
CLOSURE
PROPERTIES for ODD INTEGERS in MULTIPLICATION and DIVISION
IF:
(4a + 1)(4b + 1) = 4n + 1
AND:
(4a + 3)(4b + 3) = 4n + 1
THEN:
(4a + 1)(4b + 3) = 4n + 3
OR:
(4a + 3)(4b + 3)(4c + 3) = 4n + 3
HENCE:
(4a + 3)(4b + 3)(4c + 3)(4d + 3) = 4n + 1
So we may see
from the above table that (4n + 3) Integers are only closed for the non-matched
set and Odd Multiple Terms. For the matched sets and Even Term Count, it
is the (4n + 1) Integers that dominate.
For the product
of all possible ascending Prime members of a given set + or - 1.
WHERE:
(2 * 3 * 5.. * P + 1) = (4n + 3) Odd
THEN:
(2 * 3 * 5... * P - 1) = (4n + 1) Odd
The Euclid
Strategy, represented immediately above, demonstrates that none of the members
of the set may be factors of the yield, and that if (4n + 1) Primes are not
infinite, then they must eventually be compounds of only (4n + 3) Prime
Numbers. Since composites of (4n + 3) Primes may then only be reducible by
an Odd multiple of (4n + 3) Primes, then all (4n + 3) Integers must be Prime,
hence the contradiction. Since this is not possible, our Argument and
therefore our Proof is now complete. Thank you.
IF:
2 * Odd + 1 = 4n + 3
AND:
2 * Odd - 1 = 4n + 1
THEN:
2 (4n + 3) + 1 = 4 (2n + 1) + 3
SO:
2 (4n + 1) + 1 = 4 (2n) + 3
OR:
2 (4n + 3) - 1 = 4 (2n + 1) + 1
THEN:
2 (4n +
1) - 1 = 4 (2n) + 1
HENCE:
2P + 1 = 4n + 3
OR:
2P - 1 = 4n + 1
THEN:
2 * Odd = 4n + 2
HENCE:
2P = 4n + 2
New Postulate
(brief) these are just examples:
WHERE: a or
b = any Integer greater than 1
WHERE: x =
Odd Prime or Odd Composite, y = Any Integer Greater Than Zero
IF:
(a^(x,y) + b^(x,y)) /(a^y + b^y) = Integer
OR:
(a^(x,y) + b^x) /(a^(x,y/x) + b^(x/x)) = Integer
THEN:
((a^y)^x + (b^y)^x) /(a^y + b^y) = Integer
HENCE:
((a^y)^x + (b^z)^x) /(a^y + b^z) = Integer
___________________________________________________
Polarity Argument
(first variation)
WHERE: n =
Any Integer
IF:
(a^(4n+2) + b^(4n+2)) /(a^2 + b^2) = Integer
THEN:
(a^((4n+2)/2) + b^((4n+2)/2)) /(a + b) = Integer
Note: The
last equation above is the standard for Odd Powers which is already known (All
be it modified) as for the first example, this has not yet been fully explored
by main stream mathematics.
Note: In
the first example immediately above, where the a and b terms are any combination
of Odd Integers or an Even with Odd Integer combination as raised to a 4n + 2
exponent, while the yield is always going to be divisible by the Sum of the two
bases squared, a 4n + 1 Integer will always be part of the yield. This
default rule follows from the result as outlined in the Fundamental Theorem of
Integers when any Odd Number is raised to an Even Power. It is in the
tailoring of this example which allows the exploration of very large Prime
Numbers to be reckoned from these remainders, and applied within the Science of
Public Key Cryptography.
POLARITY-INDUCTION
OMISSION STRATEGY
A SPECIAL CASE
SERIES for n^2 + 1 = NP
by Brian S.
McMillan
NEW POSTULATE
(Brief)
Where any Odd
Integer is a Common Factor of the Exponents for at least two of the terms in a
Three Term Equation, then the Sum of those two terms when raised to their
respective Exponents will always equal a Composite Integer, and therefore, may
never be Prime. Since this is easily demonstrated by dividing the Sum of
the bases for those two terms as raised to the remaining factors for their
respective Exponents, that is less the Odd Common Factor, then we now have the
basis for our argument.
Since the
following examples may contain any combination of integers to represent the
bases within the terms, and it has been known for some time that this type of
argument may be applied using Odd Exponents only, and with the exception of the
specific case for Fermat Numbers and theoretically any Integral base raised to
the higher powers of 2, this stands as the more generalized proposition in the
case of both Even and Odd Exponents. A proof for The Last Theorem of
Fermat is therefore implied, and may be reduced to a single three term
equation. Furthermore, a strategy for both the Goldbach and Opperman
Conjectures is likewise revealed. Thank you.
The New Postulate
for Even as well as Odd Exponents, completely removes the necessity for
recognition of the sine or properties of the Integers which comprise the bases
of the terms and will work equally well for any combination of Even or Odd bases
through Subtraction as well as Addition between the terms. Any equation
which may incorporate the more Modular Argument for the Fermat Little Theorem as
well as the Mersenne Series may also be addressed, and therefore establishes a
Proposition which is totally independent of the considerations of Primality or
Co-Primality within the bases of those same terms, and should ultimately reveal
the very underpinnings for the operation of those mathematical strategies.
However, as with any theoretical approach, there may be instances where there is
a convenient overlap in our ability to represent a proof. Happily, there
is overlap-a-plenty. This will become clearer a little farther down this
page while using a combined Polarity-Property Strategy. While this may
ultimately be unnecessary, it will provide a basis for further proof of the
arguments.
Note: Euler
and Lucas are both credited with a proof for the denumeration of Fermat Numbers
greater than F4, however these proofs have not convinced everyone. This
is, theoretically, the only remaining impediment to a truly brief and simple
proof for the (so called) Last Theorem of Fermat.
Because n^2 + 1
may only be Odd when n = Even Integer
A Special Case
for the Exponent when equal to a (4k + 2) series Integer.
WHERE:
Odd is Greater Than 1 for all Examples
IF:
(4k + 2) = 2 * Odd
AND:
(a^(2*odd) + b^(2*odd)) /(a^2 + b^2) = Integer
THEN:
((Even^odd)^2 + 1) /(Even^2 + 1) = Integer
WHERE:
(Even^odd) = n
HENCE:
n^2 + 1 = NP_For Some n = Even^odd
A Special Case
for the Exponent when equal to a (4k + 0) series Integer, where k = Odd, a
general example may be represented thus.
IF:
(4k + 0) = 4 * Odd = 2 * (4k + 2)
AND:
(a^(4*odd) + b^(4*odd)) /(a^4 + b^4) = Integer
THEN:
((Even^(2*odd))^2 + 1) /(Even^4 + 1) = Integer
OR:
((Even^(4k+2))^2 + 1) /((Even^2)^2 + 1) = Integer
HENCE:
((Even^(2^n*Odd))^2 + 1) /((Even^(2^n))^2 + 1) = Integer
To be continued.
Brian S.
McMillan, Copyright 2008
New Postulate
continued (brief) these are just a limited set of examples:
WHERE: x =
Prime or Odd Composite, y = Even Integer, z = Any
Integer
AND: a or b are
greater than 1
IF:
(a^(x*y*z) + b^x) /(a^(x*y*z/x) + b^(x/x)) = Integer
THEN:
(a^(x*y*z) + b^(x*y)) /(a^(x*y*z/x) + b^(x*y/x)) = Integer
WHERE: x =
3 or Odd Composite containing 3 as a factor, y = Even Integer, z = Any
Integer
AND: a or b are
greater than 1
IF:
((a^(x*y*z) +
b^x) /(a^(x*y*z/x) + b^(x/x)) - a^(x*y*z/x) * b^(x/x))^(1/2) = Integer
THEN:
((a^(x*y*z) +
b^(x*y)) /(a^(x*y*z/x) + b^(x*y/x)) - a^(x*y*z/x) * b^(x*y/x))^(1/2) = Integer
For some Even
Integer (a), and Odd Integer (b).
IF:
(a^2 + b^2) = 4n + 1
AND:
2 * Odd = 4n + 2
THEN:
(Even^odd + Odd^odd) = Odd Integer
SO:
(Even^(2*odd) + Odd^(2*odd)) = Odd Integer
THEN:
((Even^2)^odd + (Odd^2)^odd) /(Even^2 + Odd^2) = Odd Integer
OR:
((Even^(even*odd) + Odd^(even*odd) /(Even^even +
Odd^even)
=
Odd Integer
HENCE:
((Even^even)^odd + (Odd^even)^odd) /(Even^even + Odd^even)
= Odd Integer
Since the final
two equations which are listed above, are identical, we may now begin our
polarity argument.
To Be Continued.
Brian S.
McMillan, Copyright 2008
The Fundamental
Theorem of Integers (brief)
The Natural
Integers possess fundamental, skeletal properties, which may be altered or
changed by the various operations of arithmetic, i.e. Addition, Subtraction,
Multiplication, Division, Roots, Exponents... Commutative, Associative,
Distributive, and by default the (Primary) Identity for some of the Integers
remain non-mutable within the bounds of the operations. These Integers
which possess non-mutable (Primary) identities, are responsible for the dominate
characteristics observed within the Even and Odd Integers. One of the many
uses for this approach is embedded in the exclusion principles that these
properties obey in relation to the Natural Integers, allowing or disallowing
certain Integral Operations. It was Pierre de Fermat, around the
historical period of 1650 A.D., which first categorized these properties.
This work would not have been possible without him.
I have discovered
a previously non-explored branch of Simplified Integral Algebra, or Geometry as
your preference, since it is easily and economically applicable to the
cardinality arguments of the Euclidian Plane, as well as interchangeable with
all Gaussian Induction Formats. I believe this to be as fundamentally
important as anything that has been handed down to us by the Ancient
Greeks. If any of you wish to look at it, the click-on is below, and the
primary table appears as a small series of equations about three paragraphs of
equations from the top of the page. I've listed one of them below. I
am honored to bring this to you all. Thanks Again.
Yours,
Brian S. McMillan
http://www.pumpraser.com/fermatmersenne.html
Now comes the
properties for the individual terms as they might be raised to an Odd
power. This particular solution may be applied as a rule of omission for
the problem known as "The Last Theorem of Fermat". This rests upon the
fact that while certain 4n + 1 series numbers may be root-able to the square...
can 4n + 1 + 1 = 4n + 2 series numbers be root-able to the square or the cube,
or to any power for that matter? For example:
WHERE:
Odd + Even = Odd, Odd + Odd = Even, or Even + Even = Even
IF:
(4k + 1)^even = 4n + 1
AND:
(4k + 1)^odd = 4n + 1
AND:
(4k + 3)^even = 4n + 1
THEN:
(4k + 3)^odd = 4n + 3
OR:
(4k + 0)^even = 4n
SO:
(4k + 0)^odd = 4n
AND:
(4k + 2)^even = 4n
THEN:
(4k + 2)^odd = 4n
___________________________________________________
We have seen
above, that when we attempt to yield a (4n + 2) integer by any and all of the
combinations of possible bases with their corresponding exponents, the reverse
of the operation is also prohibited. So for three out of the four possible
Odd integer combinations shown below, a (4n + 2) integer is the result. We
may now eliminate those three combinations from our mathematical choices.
IF:
(4a
+ 1)^x + (4b + 1)^x = 4n + 2
AND:
(4a + 3)^x + (4b + 3)^x = 4n + 2
THEN:
(4a + 1)^even + (4b + 3)^even = 4n + 2
SO:
(4a + 1)^odd + (4b + 3)^odd = 4n
TRANSPOSITION OF
BASE PROPERTIES WITHIN THE TERMS
It may be noted
that the base Integers for all three terms may be altered within certain
guidelines when two of the three terms are raised to an Odd Exponent. The
usefulness of this approach is embedded within the number of different ways that
these base Integers may be represented as well as to provide conclusive proof
that The Generalized Proposition for Exponents has no dependency what-so-ever
regarding the Properties for the base of the terms except with regard to the
individual properties of those terms in relation to the Divisor Sum as both
being raised to Even Powers, and therefore may lead to a more cogent theory of
Numerical Groups. However, because this Transposition of Properties does
not work for all Even Powers, this is solid testimony to the correct
interpretation for The Fundamental Theorem of Integers. For Example:
For the Variable
Integers (a, b) contained within the bases of the terms, we may take the Sum of
(a + b) = c and subtract any set of Integers from and up to (c - 1) to yield c -
d = e, so that now we have a new a and b which is still divisible by our
original base Sum. For Example:
IF:
(a + b) = (d + e)
THEN:
((4d + 1)^odd + (4e + 3)^odd) /((4a + 1) + (4b + 3)) = Odd
SO:
(4a + 1 + 4b + 3) = (4a + 4b + 1 + 3) = (4 * (a + b) + 4)
To further
explore this we may even transpose the variables themselves within the main
terms. For Example:
IF:
(a + b) = (e + d)
THEN:
((4e + 1)^odd + (4d + 3)^odd) /((4a + 1) + (4b + 3)) = Odd
SO:
(4e + 1 + 4d + 3) = (4e + 4d + 1 + 3) = (4 * (e + d) + 4)
Finally we may
now fully transpose the bases of the terms through Property Mutation. For
example, if the Properties for the Base Integers is one of an Odd Sine, lets say
a (+ 1 or + 3), as long as the end result equals the Key Property Sum for our
original two base terms, then we may now convert these two base values to
reflect the Sum of two Even Integer Types. Even though the Exponents are
involved in moulding the final yield, while the total may be different, it is
still divisible by our original Summand. For Example:
IF:
(a + b) = (e + d)
THEN:
((4e + 2)^odd + (4d + 2)^odd) /((4a + 1) + (4b + 3)) = Even
SO:
(4e + 1 + 4d + 3) = (4d + 4e + 1 + 3) = ( 4e + 2 + 4d + 2) = 4a + 4b + 4 etc.
A NUMERICAL
EXAMPLE:
IF:
((4 * 31 + 1)^3 + (4 * 09 + 3)^3) = 2012444
THEN:
((4 * 31 + 1)^3 + (4 * 09 + 3)^3) /((4 * 31 + 1) + (4 * 09 + 3) = 12271
Odd Integer
For the Sum of
Variables (31 + 09) = 40, we may now change any combination of those variables
as long as they add up to 40, such as (20 + 20) = 40, and our original Divisor
Sum will still factor our altered Yield. For example:
WHERE:
((4 * 20 + 1)^3 + (4 * 20 + 3)^3) = 1103228
THEN:
((4 * 20 + 1)^3 + (4 * 20 + 3)^3) /((4 * 31 + 1) + (4 * 09 + 3) = 6727 Odd
Integer
If we now take
our argument to include the altered properties for the original Base Integers,
as long as they also add up to the same Summands then we will be able to
duplicate the same results. Such as, (4a + 1) + (4b + 3) = (4 * (a + b) +
4), where the + 4 may be converted to (2 + 2), thus our property mutation within
the Fermat Categories. For example:
IF:
((4 * 20 + 1)^3 + (4 * 20 + 3)^3) /((4 * 31 + 1) + (4 * 09 + 3)) = 6727
Odd Integer
THEN:
((4 * 20 + 2)^3 + (4 * 20 + 2)^3) /((4 * 31 + 1) + (4 * 09 + 3)) = 6724
Even Integer
With this type of
weapon in our mathematical arsenal, we may now address the Goldbach Conjecture
with renewed vigor.
I appreciate all
your attention in this matter,
Brian S.
McMillan
___________________________________________________
P.S. Also,
when 4n + 2 numbers are multiplied by any Odd Integer at all, they remain a 4n +
2 number along with its non root-able property. Thank you.
P.P.S.
Furthermore, the sum of any two terms when raised to Any power greater than 2,
will never yield a Prime Number, when the exponents of the two terms both share
at least (one) Odd Prime as a common factor. It matters not that the
exponents of the two terms ARE or ARE NOT equal. The only exception to the
above is the trivial result where both terms equal 1, then obviously the result
will always equal 2. Thank you.
On the Web:
http://www.pumpraser.com/fermatgoldbach.html
For Mathematical
Physics:
http://www.pumpraser.com/physics.htm
Brian S.
McMillan, Copyright 2007-2008