1998-11-20 - RE: Goldbach’s Conjecture (fwd)

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From: Jim Choate <ravage@EINSTEIN.ssz.com>
To: cypherpunks@EINSTEIN.ssz.com (Cypherpunks Distributed Remailer)
Message Hash: 8780151bccf5810343fc55a2dca812140a73a8cb75e161b37a5b18f242d57568
Message ID: <199811200339.VAA08272@einstein.ssz.com>
Reply To: N/A
UTC Datetime: 1998-11-20 03:56:28 UTC
Raw Date: Fri, 20 Nov 1998 11:56:28 +0800

Raw message

From: Jim Choate <ravage@EINSTEIN.ssz.com>
Date: Fri, 20 Nov 1998 11:56:28 +0800
To: cypherpunks@EINSTEIN.ssz.com (Cypherpunks Distributed Remailer)
Subject: RE: Goldbach's Conjecture (fwd)
Message-ID: <199811200339.VAA08272@einstein.ssz.com>
MIME-Version: 1.0
Content-Type: text



Forwarded message:

> From: "Blake Buzzini" <bab282@psu.edu>
> Subject: RE: Goldbach's Conjecture
> Date: Thu, 19 Nov 1998 22:17:54 -0500

> I could be wrong, but I thought Goldbach's conjecture was that every even
> number could be expressed as the sum of *two* primes. This doesn't prohibit

No, that was Fermat, Goldbach just says every even number greater than two
can be represented as a sum of primes. Basicaly Fermat says that if we have
n primes we can reduce them to 2 primes only, in all cases. Which happens to
exclude using equilateral triangles as a test bed since you can't tile a
equilateral with only two other equilaterals, you could use rectangles though.
So basicaly from a geometric perspective Fermat says that given a rectangle of
even area it is possible to divide it with a bisector into two rectangles of
prime area.

It's interesting that Fermat doesn't mention that the only prime that can
use two as a factor is 4. And you can't factor 2 at all since we eliminate
1 as a potential candidate (another issue of symmetry breaking simply so we
don't have to write '....works for every prime but 1' on all our theorems).

> repetition. Therefore, under Goldbach's conjecture:
> 
> 4 -> 2 + 2
> 6 -> 3 + 3 but NOT 2 + 2 + 2
> 8 -> 5 + 3 but NOT 2 + 2 + 2

The real issue for me is the interaction of primes (ie n * 1 = n) and the
identity theorem (ie n * 1 = n). They're opposite sides of the same coin.

It doesn't really matter now since it doesn't look like I'm going to get a
copy of EURISKO in this lifetime to play with.


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