Goldbach's Conjecture

[science_man_88]

Originally Posted by CRGreathouse

Why test larger numbers? Just prove the (very minor) result.




Since that will have (in all likelihood) over 4 x 10^23 Goldbach partitions, I doubt there's time to list them all...

yeah i tried for the other number and i ran out of characters by over 5000 characters trying to post it lol.

the hard part is to find a way other than more test (example a well known formula) that can prove the result i can't prove already with examples.

I'm supposedly a crank NOTE: conjecture is great for cranks once they get logic added as they love to conjecture we can get within the mind of someone who proposes something unproven (our own kind)

funny how i put my semi proof together in my mind during a night when i couldn't sleep from overactive mind caused by excitement and caffeine lol.

[3.14159]

You observed that 0.5a ± b is a prime number in certain cases.

So a has to be an even number, and b has to be a number coprime to a.

I think what you're observing is an example of a classic arithmetic progression, an ± b = p, where b is coprime to an, which makes an infinite amount of examples in the case that it is.

[science_man_88]

Originally Posted by 3.14159

You observed that 0.5a ± b is a prime number in certain cases.

So a has to be an even number, and b has to be a number coprime to a.

I think what you're observing is an example of a classic arithmetic progression, an ± b = p, where b is coprime to an, which makes an infinite amount of examples in the case that it is.

well if you look at my earlier code but with the a=a+1 in the part told later on you can turn my formula into the formula you gave:

c = (.5*c+n)+(.5*c-n)

c=2n as in yours

the others = p1 and p2 so the above becomes:

2n = p1+p2 which is what you stated.

[3.14159]

c = (.5*c+n)+(.5*c-1)

Don't you mean, -n? You were making the proposition that 0.5a ± b = p.

If c = 2n; Both of the integers in parentheses are prime:

So, if you're proposing that (a + b) + (a - b) = 2n, where a - b and a + b are both primes, you will have to prove that this is the case for every even number.

Ex: 20

20 ± 7: 13, 27
27 = 3^3. Fail.

20 ± 9: 11, 29
You're in luck here. I thought this would be a counterexample.

Well, as I have stated many times before: Run some tests. Then, develop a proof. (The latter is way easier said than done.)

Okay: One last attempt at looking for a counterexample:

600 ± 7: 593, 607.
DAMMIT.

Hey: This could make a nice factoring algorithm, too.

1. Find intsqrt(a).
2. Intsqrt(a) ± b should give you a factor if its composite
3. If nothing is found down to.. 1, it's a pseudoprime. (It only works with primes that share an identical difference with the sqrt. of the integer to be factored. I find that degrading.)

Sounds a bit like reverse trial division.

[science_man_88]

Originally Posted by 3.14159

Don't you mean, -n? You were making the proposition that 0.5a ± b = p.

If c = 2n; Both of the integers in parentheses are prime:

So, if you're proposing that (a + b) + (a - b) = 2n, where a - b and a + b are both primes, you will have to prove that this is the case for every even number.

Ex: 20

20 ± 7: 13, 27
27 = 3^3. Fail.

20 ± 9: 11, 29
You're in luck here. I thought this would be a counterexample.

Well, as I have stated many times before: Run some tests. Then, develop a proof. (The latter is way easier said than done.)

Okay: One last attempt at looking for a counterexample:

600 ± 7: 593, 607.
DAMMIT.

Hey: This could make a nice factoring algorithm, too.

1. Find intsqrt(a).
2. Intsqrt(a) ± b should give you a factor if its composite
3. If nothing is found down to.. 1, it's a pseudoprime. (It only works with primes that share an identical difference with the sqrt. of the integer to be factored. I find that degrading.)

Sounds a bit like reverse trial division.

well actually i think a in your case is .5*c but in thinking of it could we not pick a large even number and according to this we can create a tree(I realized my statement about 2 and 3 working with n = 0 isn't true i think):

c
┌┴┐
p+p
etc.

these p can follow the same process getting at least 2 new primes found(can be repeated)

[science_man_88]

I know you won't see it until you post again but CRGreathouse I see this as a way to find new primes given certain rules i haven't quite found (wonder if we can apply this to Mersennes specifically) if this is all true and the equal spacing seems to so far as my testing says then with a fuel rules to figure the non primes out could we not use prime(.5*c) prime(.5*c+n) prime(.5*c-n) to prove one prime is equally distant from 2 other primes the hard part is that 9 is odd but not prime but it fits the primes surrounding for odd numbers we'd need rules to figure out which ones to eliminate(and more than 6x+1 or 6x-1 as it still wouldn't eliminate 25).

[CRGreathouse]

Originally Posted by science_man_88

I know you won't see it until you post again but CRGreathouse I see this as a way to find new primes given certain rules i haven't quite found

I haven't seen anything suggesting that. If you find anything, please post.

[science_man_88]

5 is odd and prime it is equidistant from 2 primes(3,7), same with 7(3,11),same with 11(3,19, and 5,17), the bad part is I've showed so can odd numbers like 9,15,25 etc. but they aren't prime could we check this method and apply a few rules to speed it up ?

[3.14159]

5 is odd and prime it is equidistant from 2 primes(3,7), same with 7(3,11),same with 11(3,19, and 5,17), the bad part is I've showed so can odd numbers like 9,15,25 etc. but they aren't prime could we check this method and apply a few rules to speed it up ?

In the case of a ± b, you should only use numbers coprime to a. Using numbers that have a common divisor with a leads to the numbers resulting from that being divisible by the common divisor.

well actually i think a in your case is .5*c but in thinking of it could we not pick a large even number and according to this we can create a tree(I realized my statement about 2 and 3 working with n = 0 isn't true i think):

A tree showing the decomposition of an even number as the sum of two prime numbers? That seems like a decent idea.

these p can follow the same process getting at least 2 new primes found(can be repeated)

Decomposing prime numbers into sums of prime numbers: They'll eventually decompose into a sum of 2s and 3s, if you allow repetition. If not, it'll decompose into the primes that are not the sums of other primes.

Ex: 127 eventually might decompose to: 2(59) + 3(3)

[CRGreathouse]

Originally Posted by science_man_88

5 is odd and prime it is equidistant from 2 primes(3,7), same with 7(3,11),same with 11(3,19, and 5,17), the bad part is I've showed so can odd numbers like 9,15,25 etc.

Can you prove that this always happens -- that for large enough n, there is a prime p < n such that 2n - p is composite?

[science_man_88]

Originally Posted by 3.14159

In the case of a ± b, you should only use numbers coprime to a. Using numbers that have a common divisor with a leads to the numbers resulting from that being divisible by the common divisor.



A tree showing the decomposition of an even number as the sum of two prime numbers? That seems like a decent idea.



Decomposing prime numbers into sums of prime numbers: They'll eventually decompose into a sum of 2s and 3s, if you allow repetition. If not, it'll decompose into the primes that are not the sums of other primes.

Ex: 127 eventually might decompose to: 2(59) + 3(3)

well if it shows all primes under it could we then find all primes under a given even number ? try it with 164200 for me.

never mind it probably won't get all primes. also try for power of 2 maybe.