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From: Archimedes Plutonium on 13 Aug 2010 04:50
Alright, I have fiddled around with Goldbach, nay, since 1991, almost
20 years and
it is time I finally conquer it. Although, I must say, the idea of a
Galois Algebra interchange
is very tempting as a proof. But in 1991 I sought for a conventional
proof and tonight
continues that saga.

What I did in 1991 was set up "partitions" such as for the 8 partition
is this:

4 4
5 3
6 2
7 1
8 0

And for the 10 partition is this:

5 5
6 4
7 3
8 2
9 1
10 0

And in the 1990s the arguement I was proferring was that by the
Chebychev theorem
a prime in the left column had to line up with a prime in the right
column such as (5,3)
for the 8-partition. And the reason they had to line up was because
multiplication would
be missing the number 15 = 5x3. So that argument, remotely resembles
my recent argument of the Galois Algebra interchange between
multiplication and addition.

So here is a Conjecture that is the equivalent of the Goldbach
Conjecture:
Given any Even Natural >2, that within the partition of that Even
Natural a pair of
primes will line up forming a Goldbach summand.

If I can prove that, I have proven Goldbach. Now this is rather
refreshing to look at this
rather than the Goldbach conjecture itself.

So what in mathematics could possibly exist to force every partition
of an Even Natural >2 to
have a pair of primes? And the only thing I can think of is
multiplication. In the 10-partition
we notice two Goldbach summands of (5,5) and (7,3).

And in every one of these partitions the largest multiplication is the
square numbers at the beginning of (4,4) for the 8 partition and (5,5)
for the 10-partition. And the Goldbach prime pair
is going to be either a smaller product than the square product or
actually equal to the square product such as (5,5). So that 4x4 is
larger than the 3x5.

So then, this leads me to consider that the proof of Goldbach hinges
on a simple idea that the all possible summations of Naturals to equal
an Even Natural >2, must possess at least one prime pair in the
partition of that Even Natural. This is a Goldbach Equivalent
statement.

Now why must it possess that prime pair which is a Goldbach summand?
And the answer
is not sweet but ugly in complexity. It has something to do with
multiplication and the spacing
between all the numbers that have only two prime factors in their
prime decomposition. That list goes like this:

2x2 = 4
2x3 = 6
2x5 = 10
2x7 = 14
3x5 = 15
3x7 = 21

And further complicated by those products with three or more prime
factors:

2x2x2 = 8
2x3x2 = 12

But the Goldbach is specific to two primes as summand pairs.

So,clearly, the above is a equivalent statement to the Goldbach
conjecture. And it draws me to
consider that it is no less easier to make progress. Why in the world
should those partitions have a Goldbach prime pair?

Is there something in geometry about squares and rectangles having the
same *semiperimeter* but varying in area? And where we have prime
number distance.

So, I am really beginning to think that Goldbach is like Fermat's Last
Theorem where there are
boatloads of counterexamples in the Infinite Integers and where the
proof of Goldbach is to
go through all the numbers from 0 to 10^500 to see if they obey, and
if they do, then Goldbach
would be true.

Has anyone tried a Hensel p-adic modification as to whether the p-
adics provide a counterexample to the Goldbach? In AP-adics, this
number is surely Even:

.. . . 14131211109876543210
but it probably has no Goldbach prime summands.

So, more and more, it is looking to be the case that either Goldbach
is true from a
Galois Algebra interchange or true because it obeys all the numbers
from 0 to 10^500,
but that no conventional proof can be had.

Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
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