Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

Topic: FLT is a corollary of Beal and more difficult; Proof of Weak Goldbach
#1475 Correcting Math

Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  
plutonium.archimedes@gmail.com

Posts: 8,724
Registered: 3/31/08
FLT is a corollary of Beal and more difficult; Proof of Weak Goldbach
#1475 Correcting Math

Posted: Jan 28, 2014 4:03 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply


FLT is a corollary of Beal and more difficult; Proof of Weak Goldbach #1475 Correcting Math

Alright I learned something today about the difference between a theorem proof and a corollary proof (or Lemma). I was under a mistaken impression that the more general of a statement to prove means the harder it is. But in fact, most corollaries and lemmas are proven after the general statement is proven. I was searching around on the Web and then I remembered my High School and College days of proving theorems and remembered _finally_ that once the theorem was proven the corollaries or lemmas fell out naturally. So I was forgetful of that experience. If we were to prove a corollary or lemma in isolation of the general statement, we would not get far. In the same manner, FLT is a corollary or lemma of Beal, and we must prove Beal first and then the FLT falls into place with an easy proof. Now this has drastic implications for Wiles's alleged FLT, for if the order of a proof is Theorem and then Corollary, means that Wiles's offering is likely a fake proof.

So now I have to go back and have the Beal proof come first and the FLT fall out as a corollary of Beal. No use in trying to prove FLT in isolation of Beal, because it is fraught with cases, numerous cases and almost unprovable, but with Beal proved, it eliminates all those cases and we only have to concern ourselves with the prime divisor of A, B, C for A+B=C.

But in that search I ran across the Weak Goldbach conjecture. I sadly was thinking it had already been proven, but it turns out it is not yet proven true. So if my methods are good enough to prove the Theorem of Strong Goldbach, those same methods should prove Weak Goldbach.

______________________________
Weak Goldbach proof getting started
______________________________



Weak Goldbach Conjecture:
Every odd number beyond 7 can be written as the sum of three odd primes.

Alright, I am going to do the same that I did on Strong Goldbach. First I need to set up Addition Columns, but even numbers were a breeze, and here I have to improvise to some degree.

We start with 11

0 11

0  11 
1  10
2   9
3   8
4   7
5  6

and we adjoin to 3 8 the 8 addition column:

0  8 
1  7 
2  6 
3  5 
4  4 

And we see that we have 3 + 3 + 5

Now in the Strong Goldbach we find the Odd Composite Numbers composed of just two primes such as the 3x5=15 in the 8 Column and then we include the 11 column by replacing the 3 8 with the 8 replaced by 3 5.

We do the same for 13, then 15, then 17, etc etc.

Now we have to improvise a little bit because we have Composites composed of 3 primes such as 3 +3+5 = 11 and then 3 + 5 + 5 = 13.

We focus on three prime composites as multiplication 3*3*3 = 27, 3*3*5= 45, 3*5*5= 75, 5*5*5=125, etc etc.

In Strong Goldbach we used the Prime Counting Function x/Ln(x) on perfect squares which guaranteed every even number is the sum of two primes. Here we have three primes to worry about, so I need to adapt the Prime Counting Function to that of perfect cubes.

3^3
3*3*3
3*3*5
4^3
3*5*5
5*5*5

5^3

Fine so far, but have to see if this readaption will work.

Too late tonight. Will resume tomorrow.

What I expect is that the Three Prime Composites will follow the Prime Counting Function x/Ln(x) and then that proves every odd number beyond 7 is the sum of three odd primes.

And here is a case where the Corollary is harder to prove than the theorem of Strong Goldbach. So in a sense, the Weak Goldbach is analogous to the FLT while the Beal is analogous to the Strong Goldbach.

--       

Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.        

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe        

Archimedes Plutonium



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.