Guess I will have to also show (since I posted on my facebook wall that I solved both the strong and the weak Conjecture) the solution for the weak Conjecture as well:
Christian of the Goldbach family said also that he regarded it as evident that every odd number greater than seven was the sum of three primes, something which, according to him, should be evident if every even number was the sum of two primes.
So, we now find out the truth, or rather, you do, since I would be lying if I said I did not know the answer already, right?
It must be boring being a teacher, but some appear to make a living out of it, so, each to his own, I suppose.
Anyway, to the matter at hand, or rather, on the screen... this is also available on mobile, although it is not formatted differently from the computer/laptop version, and so you have to zoom in a lot, but, hey that is modern technology for you.
?!
Christian of the Goldbach family said also that he regarded it as evident that every odd number greater than seven was the sum of three primes, something which, according to him, should be evident if every even number was the sum of two primes.
So, we now find out the truth, or rather, you do, since I would be lying if I said I did not know the answer already, right?
It must be boring being a teacher, but some appear to make a living out of it, so, each to his own, I suppose.
Anyway, to the matter at hand, or rather, on the screen... this is also available on mobile, although it is not formatted differently from the computer/laptop version, and so you have to zoom in a lot, but, hey that is modern technology for you.
?!
Suppose one had an odd number θ and this odd number was so large that primes (p1;p2;p3;...;pn) existed where pn is the greatest prime just less than the root of θ ;-
Then 1<n<θ could be written fully in terms of factor sets by:
θ = [Fp1] +[Fp2] +[Fp3] +...+[Fpn] +[Rpn]
(where the factor sets and remainder set are as previously defined in older posts, all up to the value of θ )
Now, since an odd number is of the form (2d-1), where d={1;2;3;...}, then it follows that there is likely to be, sometimes, an even number involved in the summation, and the ONLY even number that is prime is '2' and this means that, actually, even for the STRONG conjecture, the factor set CAN have ONE member involved in the pairing of the primes;- the factor prime.
However, that is not the point.
Now, since an odd number is of the form (2d-1), where d={1;2;3;...}, then it follows that there is likely to be, sometimes, an even number involved in the summation, and the ONLY even number that is prime is '2' and this means that, actually, even for the STRONG conjecture, the factor set CAN have ONE member involved in the pairing of the primes;- the factor prime.
However, that is not the point.
If the factor set primes are not factors of the odd number in question, then they will most likely (meaning DEFINITELY, since these are numbers, not rocket science or probability..whatever "probability" really means) going to pair up with other members of of other factor sets and then each pair will pair with a remainder element according to a definite ratio, an INDISCRIMINATE, definite ratio.
In fact, one can casually say that a third of Rpn will be taken by the sharing with the factor set Z3, with a fifth of the remainder being claimed by Z5, and a seventh of whatever remains after that by Z7, an eleventh of whatever remains after Z7 by Z11, and whatever is left over after that by Z13, so that, up to the value of pn, the remainder after summation, φrpn would be:
φrpn= [Rpn] - ([Fp1] +[Fp2]+[Fp3]+...+[Fpn])[Rpn],
= [Rpn] {1- ([Fp1] +[Fp2]+[Fp3]+...+[Fpn])
This means that, this remainder will then have to be split three ways so that one gets the number of triple primes that give the sum of the odd number, θ , such that, in effect,
p(triples)= (1/3)φrpn
Actually, if there are NO primes less than the odd number θ ( and this applies for even Σ as well in the strong conjecture) then there is no need for pn or such and the prime that is considered is just'1', such that Rpn ={1} only and therefore
p(triples)= (1/3).3 (since the odd number would be 3)
=> p(triples)=1
There would be 1 triple that gives a sum of the odd number.
So,one can consider the solution to the Goldbach Conjectures available, and fully solved. I can now post THIS on my facebook wall and dust my hands of it!
Enough already, LET US LOOK at
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