Goldbach's Conjecture: My Semi-Final Answer...

First of all, it's a little misleading to say that the holes on one line are more complex than the frequencies on the same line, when viewed in isolation. After all, they are just negative space of each other. Rather, I would prefer to exploit how the number line doesn't get less complex as you move to the right, but rather less compacted and more serially correlated.


Meaning, if you dropped down onto one spot on a number line of unknown length, and you happened to hit a frequency, it would be a pretty good bet there is another frequency to your right, no matter where you are. If you drop onto a hole, it is slightly less probable, than if you're on a frequency, that there will be a frequency to your right.


Furthermore, I would argue, as you keep moving right from a random spot, and you keep hitting frequencies, the probability that the next slot to your right will contain a frequency rises indefinitely. And so what you have on the high end of a number line is a lot of wasted space. Since all frequencies are products of primes, you're just repeating yourself.


8 7 6 5 4 3 2 1 0

0 1 2 3 4 5 6 7 8


Look at the pair above. We can already see - even at this short length - that if we consider any series of two slots, if that series has at least one frequency, there is much more likely to be at least one frequency in the next two slots to the right. Basically, there is compaction asymmetry in any line long enough to have a frequency to begin with!


So my final argument...


Will have to wait until tomorrow:)


eLROY

So, can I prove there are more likely to be additional frequencies next to series of frequencies? Yes. Each subsequent frequency in an uninterrupted series means zero must be at least some fixed amount further to your left. And that means you have more numbers to be out of phase. But the fact that you already have hit out-of-phase frequencies itself proves you have more numbers to be out of phase and therefore, more numbers.

The Goldbach's Conjecture is NOT something that YOU can prove with your knowledge base. You do believe you have a better mind than thousands of top mathematicians over the world? If you really do, you wont be posting here.


Your posts are like garbage, doing nothing but contaminating the forum.


How much mathematical background do you have? Up to Number Theory? Differential equations? Even if you know those pretty well, your approach will still be like to build spaceship using saw and hammer. While I admire your courage, I also suggest you have better things to do with your time, such as to remember the poker hands you played a little more clearly.


The Goldbach's Conjecture is not something that YOU can do in your life time. Plus, you dont have any idea of what you are doing. While you are trying to attack many advanced problem, I believe you should look at yourself a little more objectively.


I wont respond to your post anymore, so dont bother arguing with me.

This is worse than your stock market posts. Just because your posts are long doesn't make them intelligent.


You haven't come close to PROVING this. A mathematical proof is much different than logically reasoning it out(though you didn't do that either). Everyone knows the Conjecture is true...the hard part is proving it mathematically.

Why don't you tackle some simpler problems first before expecting us to wade through all these long inconclusive posts?


Prove any prime number (greater than 3) squared minus 1 is divisible by 24. For example 5 squared minus 1 equals 24, 7 squared minus 1 equals 2*24, etc.

You're the same one who said your own poker game is a waste of time.


http://www.twoplustwo.com/cgi-bin/newforums/genpok.pl?read=30980


What kind of an attitutude is that?


And how can you know so much about this problem and me? What a paradox, that you know so much, and yet it is all so worthless.


You can be the garbage man with your own life, Soon. But what I am curious is, with all the garbage in this forum, why you have singled out my approach to Goldbach's to throw in the trash?


So, if your saying that my Goldbach posts do not "make your life better" then, well, so what, your life was a lost cause already.


I will not take your defeatist attitude.


Goldbach's Conjecture is like caffeine for me. Hope you find yours someday:)


eLROY

Hey, if I'm spouting hogwash in stocks, please take every liberty to beat me up.


Believe me, if I didn't hope to take punches from someone, I'd be in Yahoo Antiques Group or something.


What a patsy you are.


eLROY

First, a prime square cannot be a frequency of anything, it must emerge from the existing cycle and mark out its own spot on the number line. It cannot be a multiple of 3. This is clear...


Okay, take 4,5. To get to the spot to the left of our square, we add our prime times itself minus 1 - because we are starting at its first cycle - and add it to the number to its left. Meaning we add 5 * 4 = 20 to 4, where four is both the number to the left and our prime minus 1.


The number we reach must be a multiple of 3. Well, because our prime is not a multiple of 3, and so the three cylce must be 1 or 2 to its left. If it is 1 to the prime's left - as in the case of 13 - a multiple of 12, plus 12 will be on the three cycle. If the 3 is 2 to the prime's left - as in the case of 5 - then the prime times the number to its left cannot be a multiple of 3.


When a number that is added to the left number is not a multiple of 3, the position of the 3 wave in relationship to the resulting number must change. Since it cannot jump from 1 to the left to 1 to the right - since this hole contains our prime square which cannot be a frequency of 3 - then it must touch down on the number to the left of our prime square, since 2 to the left - or 1 to the left of the sum - would not be a relative move.


Put simply, since 12 is a multiple of 3, 12 plus (13 * 12) - or (13 * 13) minus 1 - will be a multiple of 3. Since 4 is not a multiple of 3, and 5 is prime - and 4 is an even number - then 5 * 4 cannot be a multiple of 3, so our addition must move us relative to the 3 wave.


Which brings us to this number being a multiple of 2^3 as well. An odd number times an even number to the left of it, plus that even number, will always be a multiple of 2^3. Why? Because it's the same as squaring that number, then adding another two times it for good measure. Any even number is 2*x. Any even number squared is 2 squared * x squared.


Any even number squared plus 2 times itself is any number squared plus half of it times 4, meaning half of it times 2 squared. So, we have x*x*2*2 + x*2*2. So we have {x*(x+1)}*(2*2). Since x*(x+1) will always be even - since at least x or x+1 must be even, then there must be a 2 buried inside it, so we get ?*2*2*2!


For anything to be a multiple of 2^n and 3, it must be a multiple of 24.


Done:)


eLROY

But this is a much simpler answer:


http://www.google.com/search?q=cache:tzZOfJske-cC:mathforum.org/dr.math/problems/smythe11.17.97.html+prime+squared+24&hl=en


Instead of saying the number to the left of our square is p*(p-1) + (p-1), as I did, they simpy called that number (p+1)*(p-1).


Then, when you have this p+1 handy, you can dispense with all my gobbledygook, and say that since the 3-wave doesn't touch down at p, it must touch down at either p+1 or p-1 (or cover a space more than 3:). Therefore a multiple of both must have a three frequency in it.


Plus, they used a clever thought that didn't occur to me, that of the even numbers to either side of our prime, in addition to being a multiple of 2, one must also be a multiple of 4, or 2 squared - making their product a product of 2 cubed.


But of course, without thinking in p+/-1 terms, there is no reason that would occur to you, meaning this idea of even numbers two apart. They used three numbers, I only used 2, plus a bunch of odd stuff.


eLROY

And the reason why it didn't occur to me that every other two-wave touchdown must be a 4-wave touchdown is, because in my recent entropy analysis, I don't have any 4-waves - or 6-waves or 8-waves for that matter.


Since I have been thinking only in terms of "filling holes" in "flat" or "binary" resolution, a 2-wave can do it just as well as a 4 or 8. The idea that there are waves of non-prime numbers is extraneous, I think.


But primes^n, now, that is a whole 'nother matter!


eLROY

I don't have time to write 60-120 lines to rebut your crap. All I can do is read it, laugh, and move on.


What about Goldbach's conjecture? Are you still claiming to prove it?

Your exposition is pretty convoluted and you make up terms without definitions, but you seemed to have the right idea. The result works for any number (bigger than 3) that is not a multiple of 2 or 3. For example, 25 squared minus 1 equals 624 equals 26*24, but 25 is not prime.


Here is another problem. Note 2 to power of 4 equals 4 to the power of 2 equals 16. What is the complete set of distinct integer pairs with this property?

I was confused as to why you were singling me out when I never said anything. Sorry again.