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Topic: Stanford Univ professors endorsing Goldbach proof to arxiv
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plutonium.archimedes@gmail.com

Posts: 9,926
Registered: 3/31/08
Stanford Univ professors endorsing Goldbach proof to arxiv
Posted: May 13, 2014 12:34 AM
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Proof of Goldbach conjecture using successive perfect squares and x/Ln(x)

_____
Proof
_____

Now, what I have done here is try to list all the two-prime-composites in order of their product. Here are all possible combinations of the odd primes listed according to their product. And I separated them according to Perfect Squares

2 2 = 4 (4)

4


3 3 = 9 (6)

9

3 5 = 15 (8)

16


3 7 = 21 (10)
5 5 = 25 (10)

25

3 11 = 33 (14)
5 7 = 35 (12)


36

3 13 = 39 (16)
7 7 = 49 (14)


49

3 17 = 51 (20)
5 11 = 55 (16)
3 19 = 57 (22)


64

5 13 = 65 (18)
3 23 = 69 (26)
7 11 = 77 (18)


81

5 17 = 85 (22)
3 29 = 87 (32)
7 13 = 91 (20)
3 31 = 93 (34)
5 19 = 95 (24)

100

3 37 = 111 (40)
5 23 = 115 (28)
7 17 = 119 (24)
11 11 = 121 (22)


121


3 41 = 123 (44)
3 43 = 129 (46)
7 19 = 133 (26)
3 47 = 141 (50)
11 13 = 143 (24)


144



5 29 = 145 (34)
3 51 = 153 (54)
5 31 = 155 (36)
7 23 = 161 (30)
13 13 = 169 (26)


169

5 37 = 185 (42)
11 17 = 187 (28)


196

7 29 = 203 (36)
5 41 = 205 (46)
11 19 = 209 (30)
5 43 = 215 (48)
7 31 = 217 (38)
13 17 = 221 (30)


225

.
.
.

Now to prove Goldbach, I need to use the AP-postulate that between successive perfect squares always exists two distinct primes starting with the perfect squares of between 4 and 9 and we have primes 5,7 and the AP-Postulate proved to be true.


Here is the proof of the AP-postulate which is a stronger version of the Bertrand's postulate.

_____________________
Proving the AP postulate
_____________________

The AP-Postulate is making the intervals between primes smaller than in the doubling of intervals in the Bertrand's, but, also by increasing the number of primes to be two rather than a solo one prime:

The intervals are successive perfect-squares as this:

2^2 to 3^2 to 4^2 to 5^2 to 6^2 . . .

And calling a pair of generalized primes as p,q, I need to show this is true:

2^2 p,q 3^2 p,q 4^2 p,q 5^2 . . .

So, where Bertrand's postulate is one prime interspliced between successive N to 2N to 4N etc, the AP postulate is a pair of primes interspliced between successive perfect squares. The AP postulate is far more demanding for it involves smaller intervals and involves the existence of a pair of primes not a singular solo prime.

Now in the proof of AP-postulate the two primes are distinct.

For the generalized pair of primes p,q, I must show that we can do this:

2^2 p,q 3^2 p,q 4^2 p,q 5^2 p,q 6^2 p,q 7^2 . . .

Now because the Prime Counting Function (PCF) produces a lot more primes than needed in Bertrand's it is obvious PCF produces a lot more primes than needed for AP-postulate because, again, for the simple explanation that N/Ln(N) is monotonic increasing function. And even though the intervals are shortened compared to Bertrand and we require two distinct primes, the function N/Ln(N) is still capable of a lot of room to satisfy those requirements.


4/Ln(4) = 4/1.38 = approx 2.8
9/Ln(9) = 9/2.19 = approx 4.1
subtracting we have 1.3 prime between 4 and 9, yet actually we have two


9/Ln(9) = 9/2.19 = approx 4.1
16/Ln(16) = 16/2.77 = approx 5.7
subtracting we have 1.6 primes between 9 and 16, yet we actually have two


16/Ln(16) = 16/2.77 = approx 5.7
25/Ln(25) = 25/3.21 = approx 7.7
subtracting we have 2 primes between 16 and 25, yet we actually have three


25/Ln(25) = 25/3.21 = approx 7.7
36/Ln(36) = 36/3.58 = approx 10.0
subtracting we have 2.7 primes between 25 and 36, and we actually have two

But our computations stop there because clearly from 16 to 25 and 25 to 36 because N/Ln(N) is monotonic increasing the number of primes between successive perfect squares is going to be always larger than 2.

QED


And I need the AP-Postulate applied to two-prime-composites. So I have to adapt the AP-Postulate to that of the multiplication of two primes, distinct or the same two primes as seen in the table above.

Now in the proof of the AP-postulate, I used the Prime Counting Function N/Ln(N) to see how many primes exist between successive perfect squares.

To prove Goldbach, all I need is prove that 1 set of primes (distinct or the same) forms a two-prime-composite between each successive perfect square.

Now I am going to make a rather dazzling twist of logic. We know the amount of Primes follows this function of N/Ln(N). We know every Two-Prime-Composite is a composite consisting of only two primes, and since they are composed of only two primes, we can say that their Product follows the function of N/Ln(N).

So let us count up the number of Two-Prime-Composites above from 4 to 100
and we have a total of 20 such Two-Prime-Composites starting with 2x2 and ending with 5x19.

Now the function N/Ln(N) for 100 is 100/4.60 = approx 21.7 and the regular primes are 25 in that interval and the Two-Prime-Composites are 20 in that interval, so we see the Two-Prime-Composites coming closer to the function than the regular primes.

Now look at how many two-prime-composites from 2x2 to 225 perfect-square and we have 42 of them. Now let us see what N/Ln(N) gives and we have 225/5.41 = approx 41.5. So that our replacement of regular primes by that of two-prime-composites product is a good choice of replacement for both are monotonic increase as a function.

So, now, what is needed to prove Goldbach is simply show that at least one pair of two-prime-composites exists in between two successive perfect squares.

Between 4 and 9 is 3x3 or 3+3 and between 9 and 16 is 3x5 or 3+5.

4


3 3 = 9 (6)

9

3 5 = 15 (8)

16

Now let me compute with the N/Ln(N) for successive perfect squares:

25/Ln(25) = 25/3.21 = 7.7 approx
16/Ln(16) = 16/2.77 = 5.7 approx
subtracting leaves 2

36/Ln(36) = 36/3.58 = 10.0 approx
25/Ln(25) = 25/3.21 = 7.7 approx
subtracting leaves 2.3

Successive counts delivers 2 or greater than 2 primes, either distinct or the same primes for a two-prime-composite residing between those two successive perfect squares. And since the function N/Ln(N) is monotonic increasing we have our adaption of the AP-Postulate, that between every successive perfect-square starting with 4, there exists at least one of two-prime-composites.

Now we are not quite finished with the proof for we need one last thing to do. To relate the two-prime-composites with the even numbers. And that relationship is very easy to see in the Addition Columns below of all the even numbers. For every even number ends its column as a perfect square. So the proof of Goldbach is that there exists, always at least one two-prime-composite inside a even number column, matched and paired up, hence Goldbach.

0 4
1 3
2 2

0 6
1 5
2 4
3 3

0 8
1 7
2 6
3 5
4 4


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


0 12
1 11
2 10
3 9
4 8
5 7
6 6


0 14
1 13
2 12
3 11
4 10
5 9
6 8
7 7

0 16
1 15
2 14
3 13
4 12
5 11
6 10
7 9
8 8
.
.
.


QED

--

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



Date Subject Author
5/13/14
Read Stanford Univ professors endorsing Goldbach proof to arxiv
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5/14/14
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5/15/14
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5/19/14
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5/24/14
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5/26/14
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5/30/14
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6/5/14
Read Re: Stanford Univ professors endorsing Goldbach proof to arxiv
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6/5/14
Read Re: Stanford Univ professors endorsing Goldbach proof to arxiv
Richard Tobin
6/5/14
Read Re: Stanford Univ professors endorsing Goldbach proof to arxiv
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6/5/14
Read Re: Stanford Univ professors endorsing Goldbach proof to arxiv
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6/11/14
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6/26/14
Read symmetry in even & odd numbers Re: Stanford Univ professors endorsing
Goldbach proof to arxiv #1858 Correcting Math
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6/29/14
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7/26/14
Read time to do the Geology text new edition Re: Stanford Univ professors
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7/16/14
Read Re: Stanford Univ professors endorsing Goldbach proof to arxiv
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7/28/14
Read why no-one used a stiffened Bertrand Postulate Re: Stanford Univ
professors endorsing Goldbach proof to arxiv
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8/3/14
Read Re: Stanford Univ professors endorsing Goldbach proof to arxiv
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8/4/14
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