Goldbach's conjecture second attempt.

Primenumbers · 2015-06-17 23:17:02

Goldbach's conjecture states; Every even integer greater than 2 can be expressed as the sum of two primes.

How about this;

e= any even no.
c+d= e! factored down with factors of e, but not 2.
i.e. 12! factored down with factors of 12 but not 2 = 12!/3 as many times as possible = 12!/243=1971200.

a+b=(c+d)-e

(1) No.'s <e excluding factors of e, except 2
(2) e factors

f.= factorable by

e is f.(2) not f.(1)
c+d is f.(1) not f.(2)
a+b is f.- not f.(1)(2)

a is f.(2) not f.(1)
b is f.- not f.(1)(2)
c is f.(1) not f.(2)
d is f.(1)(2) not f.-

e= (c-a)+(d-b) = two primes added together. (They are prime because they are not f.(1) or f.(2).)

The above is true because of the following rule:

A=B+C If B is factorable by x but C isn't then A won't be.
i.e. A=B+C
59=21+38 21/7=3 38/7=5 remainder(3)
59/7 = 8 remainder(3). The remainder carries.

Primenumbers · 2015-06-19 00:27:33

I am working on a simpler theory;

e=any even no.
e=(c-a)+(d-b)

1) primes <e excluding 2
2) 2
c and d factorable by 1) not factorable by 2)
a and b factorable by 2) not factorable by 1)
Therefore (c-a) and (d-b) are prime.

I am not however sure whether I need to prove that e will = a certain size....?

Bob · 2015-06-19 20:19:30

hi Primenumbers

I'm not sure I have understood your argument. Let's choose e = 32. So you want two odd primes for c and d. How about c = 23 and d = 13 ? Then you want two even numbers for a and b which have no odd prime factors => a = b = 2.

e = 23 - 2 + 13 - 2

But 23-2 = 21 which is not prime.

Bob

Agnishom · 2015-06-20 03:34:23

I do not see what's going on

Bob · 2015-06-20 03:48:35

Well I don't really, either.

Bob

Primenumbers · 2015-06-22 05:39:32

bob bundy wrote:

So you want two odd primes for c and d.

c and d are not prime. Don't worry, I don't think they work. But take this new idea:

e=any even no.
e= a + b

We want a to be prime but we don't want it to have the same remainders as e, otherwise e-a= non-prime.
Finding the numbers that have a different set of remainders than e will be the same as the number of primes.
Because we are just trying to delete a different set of remainders. (instead of r.=0, r.=x). But hang on a minute r. for 2 will be the same in both sets as e is always even.
Therefore there will always be < no. of primes for the no. of numbers that don't have the same r. for e.
Therefore there will always be an instance for (a) and therefore an instance for (b).
Example: e=84

= 9 ish Therefore primes<e will not be factorable by 2,3,5, and 7.

Finding the no. of primes in e:
2m
6m+3
30m+5 or 25
210m+7 or 7x7 or 7x11 or 7x13 or 7x17 or 7x19 or 7x23 or 7x29

For 84 count the No. of numbers it passes and minus it from 84 then -1 for 1 and + 4 for 2, 3, 5, and 7, I got 23 which is the No. of primes below 84.

Finding the instances of e remainders:
84 has remainder 0 for 2,3, and 7 but remainder 4 for 5. So lets delete the no. of primes that =(5y + 4);
There are 2 no.'s in 30 which have a remainder 4; 19 and 29
30m +19 or 29
84 passes through 5 of them but one of these is 49 so 23-4 but one of these is 2 also so 23-5=18 but don't forget to minus 3 and 7 so 18-2=16......
(5,79)(11,73)(84,71)(17,67)(23,61)(31,53)(37,43)(41,43).

Bob · 2015-06-22 06:27:54

Sorry. I got totally lost at "but we don't want it to have the same remainders as e". After that it just got more and more unintelligible.

Bob

Primenumbers · 2015-06-22 08:00:01

bob bundy wrote:

Sorry. I got totally lost at "but we don't want it to have the same remainders as e".

We have to delete remainders for e otherwise if a had r.=x, then e-a=b: e(with r.x)- a(with r.x)=r.0 for b, so b would not be prime.
Does that help Bob?

Think of it this way;
Way of finding primes=
2m+1
6m+1 or 5
30m+1 or 7 or 11 or 13 or 17 or 19 or 23 or 29
210m + 1 or 11 or 13.....(no.'s not factorable by 2,3,5 or 7)...........209

e might have a remainder if you try to factor 3, 5 or 7.
e will never have a remainder for 2.

1) Let's pretend e does have a remainder for 3. I.e. remainder for 3=x
There will only be 1 no. with r.x in 6.

2) Let's pretend e does have a remainder for 5 also. I.e. remainder for 5=y
There will only be 2 no.'s with r.y in 30.

3) Let's pretend e does have a remainder for 7 also. I.e. remainder for 7=z
There will only be 8 no.'s with r.z in 210.

You can now see that minusing these no.'s from no. of primes there will still be enough left to make a and b.

Bob · 2015-06-22 19:57:40

hi Primenumbers

Thanks. You have the advantage over me as (1) you've been thinking about this for a while; and (2) you've got lots of this inside your own head anyway. I'll try to follow what you have posted but it will take me a while.

Sorry to be negative but the conjecture has been around since 1742 and no mathematician has resolved it yet, so you are being very optimistic that any solution will fit into a short post. I'm not saying you shouldn't try and I wish you luck with it but, if I had to bet on it, I'd put my money on there being a flaw somewhere.

LATER EDIT:

there will still be enough left to make a and b

As mathematicians have worked on the conjecture for some time, and with super computers available, I'm sure it has been tested for all even numbers up to a very big value. You have to show an algorithm that will definitely work for all evens. Saying there will be enough left is going to be too vague to constitute a proof.

Bob

Last edited by Bob (2015-06-23 00:19:58)

Primenumbers · 2015-06-23 02:19:03

Hi Bob,
Thanks for looking at my posts. I will try to recreate it into simpler style and put it in an algorithm. Might take me a little while. Get back to you later................Primenumbers.

Primenumbers · 2015-06-23 04:45:54

You were right............it doesn't work.

I made a big mistake:

Primes minus No.'s with remainder e would = (not >1) using my method..........which is not true.

Math Is Fun Forum

#1 2015-06-17 23:17:02

Goldbach's conjecture second attempt.

#2 2015-06-19 00:27:33

Re: Goldbach's conjecture second attempt.

#3 2015-06-19 20:19:30

Re: Goldbach's conjecture second attempt.

#4 2015-06-20 03:34:23

Re: Goldbach's conjecture second attempt.

#5 2015-06-20 03:48:35

Re: Goldbach's conjecture second attempt.

#6 2015-06-22 05:39:32

Re: Goldbach's conjecture second attempt.

#7 2015-06-22 06:27:54

Re: Goldbach's conjecture second attempt.

#8 2015-06-22 08:00:01

Re: Goldbach's conjecture second attempt.

#9 2015-06-22 19:57:40

Re: Goldbach's conjecture second attempt.

#10 2015-06-23 02:19:03

Re: Goldbach's conjecture second attempt.

#11 2015-06-23 04:45:54

Re: Goldbach's conjecture second attempt.

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