RE: Goldbach's Conjecture (fwd)

[Jim Choate <ravage@EINSTEIN.ssz.com>]

Forwarded message:

> From: "Blake Buzzini" <bab282@psu.edu>
> Subject: RE: Goldbach's Conjecture (fwd)
> Date: Thu, 19 Nov 1998 23:37:09 -0500

> >From _Elementary Theory of Numbers_ by William J. LeVeque, pg. 6:
> 
> "It was conjectured by Charles Goldbach in 1742 that every even integer
> larger than 4 is the sum of two odd primes.  (All primes except 2 are odd,
> of course, since evenness means divisibility by two.)"

Ok, so this one says it was Goldbach himself and in particular states two
odd primes completely eliminating 4 from the get go ...

> >From _Excursions in Number Theory_ by C. Stanley Ogilvy and John T.
> Anderson, pg. 82:
> 
> "Goldbach's conjecture. Is every even number expressible as the sum of two
> primes?"

This one is the second version...

> >From _Goldbach's Conjecture_ by Eric W. Weisstein
> (http://www.astro.virginia.edu/~eww6n/math/GoldbachConjecture.html):
> 
> "Goldbach's original conjecture, written in a 1742 letter to Euler, states
> that every Integer >5 is the Sum of three Primes. As re-expressed by Euler,
> an equivalent of this Conjecture (called the ``strong'' Goldbach conjecture)
> asserts that all Positive Even Integers >= 4 can be expressed as the Sum of
> two Primes."

And finaly a third completely different slant. They at least get Fermats
contribution right.

> Am I misreading somewhere?

Well I'd say that all three of your references tended to contradict each
other. Which one do you want to stand on?

This is my quote:
 
> 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).

I'll stand by this statement.


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