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Introduction

Hello fellow mathematicians. I am a programmer (uni student) but I've fallen in love with mathematics especially number theory and although I am quite aware of the difficulty on some of those "unsolved problems" and the fact that solving them is (for somebody like me) nearly impossible. Still I look at it more like a puzzle, something to train the brain on and play with.

However I've been playing around with the Goldbach Conjecturefor several months now and there is this model that I've developed. I simply cant find a mistake in it and nobody wanted to take me seriously enough to even consider checking it.. It would be very kind if somebody would "grade" this and point out possible flaws, it would mean a lot to me :) so here is my model:

The Conjecture

The Goldbach Conjecture states that every positive even integer greater than 2 can be expressed as the sum of two primes. In other words this means that for every even number there has to exist at least one set of primes (px, py) whos sum is equal to it.

My Reasoning

Since some numbers like 10 can be expressed as both (5, 5) and (3, 7) there will be some duplicates however if a continuous pattern could be spotted which when followed has to visit eventually all even numbers then the conjecture would be true.

The Table

This would mean that if we create a prime sum table (like a multiplication table) it would have to contain all even numbers if the conjecture is true.

      3  |  5  |  7  | 11 |
    +----+-----+-----+----+
  3 | 6  | 8   | 10  | 14 |
    +----+-----+-----+----+
  5 | 8  | 10  | 12  | 16 |
    +----+-----+-----+----+
  7 | 10 | 12  | 14  | 18 |
    +----+-----+-----+----+
 11 | 14 | 16  | 18  | 22 |
    +----+-----+-----+----+

So if we had a table with infinitely columns and rows we'd have (with some duplicates) all even numbers. Notice how the main diagonal is constructed with the numbers equal to the n't prime number * 2. Because of this everything below it will be a mirror image of everything above it. So lets just observe the numbers in the triangle above it. All even numbers have to be here eventually for the conjecture to be true. We still have some duplicates however.

If we create a table of fixed dimensions n x n, then the last term on that diagonal will be a unique number which will only exist there. When we expand it to n+1, the new last number on the diagonal will be unique, and the previous one will appear in the triangle above the diagonal. This means that we can exclude the diagonal too (if we're talking about a table of infinite dimensions) and all even numbers MUST explicitly be found above (not including) the diagonal.

Lets replace everything we don't need with zeroes:

       3  |  5  |  7  |  11 |
    +-----+-----+-----+-----+
  3 |  0  |  8  | 10  | 14  |
    +-----+-----+-----+-----+
  5 |  0  |  0  | 12  | 16  |
    +-----+-----+-----+-----+
  7 |  0  |  0  |  0  | 18  |
    +-----+-----+-----+-----+
 11 |  0  |  0  |  0  |  0  |
    +-----+-----+-----+-----+

The Algorithm

If we observe only the first row (ignore all the others). We notice that every next even number is for the prime gap Gn bigger than the previous. After 8 comes 10 because the gap between 5 and 3 is 2. If the gaps were 2 every time, starting from 8, we will have 10, 12, 14 and so on, visit all even numbers.

Since this is not the case (some gaps are 4 6 10.. etc) we will have to represent it differently. Lets look at that table in terms of prime gaps.

       3  |  5  |  7  |  11 |
    +-----+-----+-----+-----+
  3 |  0  |  2  |  2  |  4  |
    +-----+-----+-----+-----+
  5 |  0  |  0  |  2  |  4  |
    +-----+-----+-----+-----+
  7 |  0  |  0  |  0  |  2  |
    +-----+-----+-----+-----+
 11 |  0  |  0  |  0  |  0  |
    +-----+-----+-----+-----+

We can see a pattern unfolding, its the prime gap sequence shifted to the right by 1 position as the row number increases. The sequence 1 2 2 4 2 4.. wont contain 1 because we have excluded number 2 from the table.

The equation to the n'th even number is 2*n. So every time we have a prime gap of 2, the increment of "n" in that equation is 1 (aka visit the next in the sequence). If we have a gap of 4, it will skip one and visit the next for example 6 + 4 = 10 (skip 8, visit 10).

We are going to represent this pattern using binary notation for the gaps. 2 becomes 1 (visit next) 4 becomes 01 (skip one, visit next) 6 becomes 001 (skip two, visit next) and so on

The zeroes already in the table wont have any influence, because they just represent empty space. We get:

0  1  1  0  1
0  0  1  0  1
0  0  0  1  
0  0  0  0  0

Since there is infinitely many gaps, we can add a zero (since it wont change anything) to the missing spot in the 3rd row, 5th column, we get:

0  1  1  0  1
0  0  1  0  1
0  0  0  1  0
0  0  0  0  0

The Compensation Rotation

We must not forget that these numbers were in that triangle, initially and that we ignored the column gaps.

Take a look at the first vector 0 1 1 0 1 This means the following:

6  8  10  12  14
0  1   1   0   1

For row1 this would mean: skip 6, visit 8, visit 10, skip 12 visit 14, so the numbers 8 10 and 14 can be found in the first row.

Lets take a look at row2

8  10  12  14  16
0  1   1   0   1

We start with the number 8 as offset, because of the diagonal shift that was created by the vertical prime gap that we "ignored" initially. Lets create a new table, with the binary pattern for each row but take the offset into account. We get:

6   8   10  12  14
0   1    1   0   1

8  10   12  14  16
0   0    1   0   1

10  12  14  16  18
 0   0   0   1   0

16  18  20  22  24
 0   0   0   0   0

We have apples and oranges here really. To bring it all into the same "unit of measure" if you want to call it like that, we have to shift the 2nd row to the right by 1 position, the 3rd row again by 1 position, the 4th row by 3 positions etc.

How do we know how many times we have to shift to the right for the nth row? 1, 1, 3, 1, 3, 1, 3, 5.. I've gotten this in excel with a bigger table and it is always the n'th prime gap - 1 times. We do this by adding zeroes (which are neutral) to the left side and moving everything n positions to the right. Lets perform this final step to get:

6  8  10  12  14
0  1   1   0   1
0  0   0   1   0
0  0   0   0   0
0  0   0   0   0

We've gotten so many zeroes because the table dimensions I've chosen were really small (the frame we've observed) so a lot information had gone "out of sight" from the shiftings.

The Final Step

Basically we've had different information about the same thing just from "different angles". Now that we've aligned it we can perform a bitwise OR operation on all rows to get: {0, 1, 1, 1, 1}

The meaning of this array is that in the observed triangle (above the diagonal, excluding the diagonal) in a table of fixed dimensions (4 x 4) we haven't visited 6, we visited 8, 10, 12, 14.

The reason 6 wasn't visited because of the exclusion rule of the diagonal. We can say that the numbers 4 and 6 are special cases but they can be constructed as 4 = 2 + 2, 6 = 3 + 3.

Because 1 initially meant "from current position move by 2" (as in visit the next even number) and since we have no zeroes left (or holes if you want to put it) after performing this algorithm, my statement is that the pattern is "complete"

Conclusion

Since we can go to any place on the grid (does't have to be the beginning), create a table onwards from that position of (n x n) dimensions (aka take a slice of infinity), perform the algorithm and get an array {1, 1, n times, 1} and since we can do this onwards from that position for a table with the dimensions (n + 1) x (n + 1) and get an array of {1, 1, (n + 1) times, 1} we can say that the pattern would be complete for (n + 2) x (n + 2); (n + 3) x (n + 3) and eventually (infinity) x (infinity).

This would mean that in the final step we'd get the array {1, 1, inf. times, 1} where when we start from 6 for every 1 we increment by 2 so we visit 8, 10, 12, 14 .. etc all even numbers up to infinity

So since (I think) I've proven that this table contains eventually all even numbers it can be said that every even number has to have at least one pair of primes (px, py) whos sum is equal to itself or that every even number greater than 6 can be written as the sum of two primes (with the special cases 4 = 2 + 2 and 6 = 3 + 3) hence the conjecture is true.

I am sorry for the long post, hope I've explained it clear enough. Thanks for reading :)!

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In section "The Table" you claim that the number in position $(n,n)$ in the $n \times n$ table will be found above the diagonal in the $n+1 \times n+1$ table. The $n$-th row corresponds to the $n$-th prime $p_n \ge 3$. Your claim is equivalent to saying that $2p_n$ can be written as a sum of two primes of which one is $p_{n+1}$. This does not work for precisely where your above table stops, i.e. at $n=4$, because $2 p_4 = 2 \cdot 11 = 22 = 13 + 9$ and $9$ is not prime.

If you don't understand my explanation, just add one more row and column to your above table (corresponding to $13$), and then check whether $22$ can be found above the diagonal. (It can't.)

I haven't read further than that.

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  • $\begingroup$ I understand what you say, the number 22 can be found as 17+5 tho and it will be above the diagonal. You are right, that claim of mine was not correct. I will have to take a look and try to figure out when it has to appear in the triangle. It'd be awesome if you found more mistakes as I think there will be more than one now! Thank you! $\endgroup$ – Ilhan Sep 10 '15 at 15:45
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    $\begingroup$ @Ilhan: In general, that $2p$ can be written in at least two ways as a sum of primes (one way being $p+p$) seems a conjecture in itself. Prove this one first, if you want to use it later. $\endgroup$ – Alex M. Sep 10 '15 at 15:55

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