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From: Archimedes Plutonium on 13 Aug 2010 04:50 Alright, I have fiddled around with Goldbach, nay, since 1991, almost 20 years and it is time I finally conquer it. Although, I must say, the idea of a Galois Algebra interchange is very tempting as a proof. But in 1991 I sought for a conventional proof and tonight continues that saga. What I did in 1991 was set up "partitions" such as for the 8 partition is this: 4 4 5 3 6 2 7 1 8 0 And for the 10 partition is this: 5 5 6 4 7 3 8 2 9 1 10 0 And in the 1990s the arguement I was proferring was that by the Chebychev theorem a prime in the left column had to line up with a prime in the right column such as (5,3) for the 8-partition. And the reason they had to line up was because multiplication would be missing the number 15 = 5x3. So that argument, remotely resembles my recent argument of the Galois Algebra interchange between multiplication and addition. So here is a Conjecture that is the equivalent of the Goldbach Conjecture: Given any Even Natural >2, that within the partition of that Even Natural a pair of primes will line up forming a Goldbach summand. If I can prove that, I have proven Goldbach. Now this is rather refreshing to look at this rather than the Goldbach conjecture itself. So what in mathematics could possibly exist to force every partition of an Even Natural >2 to have a pair of primes? And the only thing I can think of is multiplication. In the 10-partition we notice two Goldbach summands of (5,5) and (7,3). And in every one of these partitions the largest multiplication is the square numbers at the beginning of (4,4) for the 8 partition and (5,5) for the 10-partition. And the Goldbach prime pair is going to be either a smaller product than the square product or actually equal to the square product such as (5,5). So that 4x4 is larger than the 3x5. So then, this leads me to consider that the proof of Goldbach hinges on a simple idea that the all possible summations of Naturals to equal an Even Natural >2, must possess at least one prime pair in the partition of that Even Natural. This is a Goldbach Equivalent statement. Now why must it possess that prime pair which is a Goldbach summand? And the answer is not sweet but ugly in complexity. It has something to do with multiplication and the spacing between all the numbers that have only two prime factors in their prime decomposition. That list goes like this: 2x2 = 4 2x3 = 6 2x5 = 10 2x7 = 14 3x5 = 15 3x7 = 21 And further complicated by those products with three or more prime factors: 2x2x2 = 8 2x3x2 = 12 But the Goldbach is specific to two primes as summand pairs. So,clearly, the above is a equivalent statement to the Goldbach conjecture. And it draws me to consider that it is no less easier to make progress. Why in the world should those partitions have a Goldbach prime pair? Is there something in geometry about squares and rectangles having the same *semiperimeter* but varying in area? And where we have prime number distance. So, I am really beginning to think that Goldbach is like Fermat's Last Theorem where there are boatloads of counterexamples in the Infinite Integers and where the proof of Goldbach is to go through all the numbers from 0 to 10^500 to see if they obey, and if they do, then Goldbach would be true. Has anyone tried a Hensel p-adic modification as to whether the p- adics provide a counterexample to the Goldbach? In AP-adics, this number is surely Even: .. . . 14131211109876543210 but it probably has no Goldbach prime summands. So, more and more, it is looking to be the case that either Goldbach is true from a Galois Algebra interchange or true because it obeys all the numbers from 0 to 10^500, but that no conventional proof can be had. Archimedes Plutonium http://www.iw.net/~a_plutonium/ whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies |