I've not posted here before but have been an interested reader. I wanted to run an attempted proof of the Twin Prime Conjecture (and perhaps the Polignac Conjecture in general) past the forum members - I know that most attempted proofs are flawed. But I'm optimistic enough about this attempt to try to get some feedback.

It's related to the infinite product fraction Euler inverted to show that the reciprocals of the primes diverge. There's a fairly brief summary at the start, then more detail beneath. If you do have time to take a look I'd be extremely grateful.

http://barkerhugh.blogspot.com/2011/01/twin-primes-and-polignac-conjecture.html

Thanks - I'll try there too. If I get some derision I can live with it - I wouldn't put th the attempt up if I didn't think it had something to it, but it may well be that someone can debunk it for me.

Meanwhile any feedback here will also be appreciated.

Hi, it's always fun to try your hand at maths but please don't be disappointed with more serious responses if you visit some actual maths boards like: http://www.physicsforums.com/ . Mathematicians get a lot of people come to them claiming great proofs when they have lots of mistakes in, and 1 mistake is enough to stop it from being a proof.

Your proof has a lot of erroneous statements in it, try and simplify it to it's actual bare essentials, but I am fairly sure it's false because I think the statement "Any single prime pn can only remove 2/pn of the remaining twin prime pair candidates in the pattern and always removes a symmetrical pattern of pairs." proves your method is insufficient, rather than anything else.

But please feel free to elaborate.

Zurtex,

Firstly, thanks for taking a look. Yes, I'm aware of the danger of being a crank who thinks they've come up with something great. And rather expecting someone to tell me the flaw.

Someone on mersenne thinks the flaw is the jump at the end, saying that there must be a "last prime" that casts out the last eligible twin prime. They could be right, in that it might be a Zeno-style paradox.

I did put up a compressed version here, as a fair amount of the first version wasn't necessary for the specific twin prime problem

http://barkerhugh.blogspot.com/2011/01/twin-prime-proof-compressed-version.html

You say, the statement "Any single prime pn can only remove 2/pn of the remaining twin prime pair candidates in the pattern and always removes a symmetrical pattern of pairs." proves your method is insufficient, rather than anything else."

Not sure what you mean you by saying - the statement above seems essentially correct. But perhaps you're making the same point, that this doesn't imply there can't be a "last prime" that casts out the last pair of TPP candidates?

Your approach is a little if off, it starts with the classical proof by contradiction:

Assume "there are a finite number of twin primes"

But you don't make particular use of this and instead attempt to go on and make a counting argument instead, based on prime density and Sieve of Eratosthenes. The counting argument being that only a finite number of twin prime candidates are removed from the infinite set of twin prime candidates using the Sieve of Eratosthenes.

However if we assume here are a finite number of twin primes then it doesn't matter that there are an infinite set of twin prime candidates.

"...you don't make particular use of this and instead attempt to go on and make a counting argument instead, based on prime density and Sieve of Eratosthenes. The counting argument being that only a finite number of twin prime candidates are removed from the infinite set of twin prime candidates using the Sieve of Eratosthenes."

No, that's not the argument. The number of candidates removed at each stage must obviously also be infinite. pn removes an infinite number of pairs containing multiples of pn. A fraction of infinity is not a finite number.

And it's not a counting argument as the method used clearly couldn't be used to count primes or twin primes.

Also it's not based on twin prime density per se. Again, you couldn't use this technique to calculate that, since the sieve includes twin primes.

"However if we assume here are a finite number of twin primes then it doesn't matter that there are an infinite set of twin prime candidates."

So this isn't really a good objection.

Really my argument is closer to this structure, which is a version of the proof of the infinity of primes:


Assume there is not an infinite number of primes . If so, there must be a number above which there are no more primes.

We use the Sieve of Eratosthenes to cast out potential primes.

There must be a “last prime” that, when added to the pattern, casts out the final remaining candidates. (Meaning that no number larger than this is needed to cast out further candidates).

Any single number p removes a symmetrical repeating pattern of remaining candidates, and can only remove at most 2/p of the remaining candidates in the pattern.

Therefore it is impossible for such a “last prime” to exist and there is an infinite number of primes.


Now, this is a similar structure to Euclid's proof of the infinity of primes, which is valid. In that proof, he shows that if you have an infinite number of primes, then multiply them all together there must be a gap in the pattern and that gap must be a prime or a product of a higher prime than the ones in the original set. What I am aspiring to do is show this can be adapted to twin primes, because they also expand in a symmetrical, fractal pattern.

If there is a flaw I think it must revolve around the question of whether such a "last prime" is indeed necessary to cast out the last twin candidate pair.

The proof does show that given a finite number of primes there must be a gap in the twin prime pattern (indeed, an infinite series of gaps) as well as a gap in the prime pattern. But is it valid to step from that to saying that there is no largest twin prime? That's what I'm pondering now.

I meant 1/p in the prime proof:

Any single number p removes a symmetrical repeating pattern of remaining candidates, and can only remove at most 1/p of the remaining candidates in the pattern.

Sorry, I was interpreting your proof too much, my point is:

"Any single number p removes a symmetrical repeating pattern of remaining candidates, and can only remove at most 2/p of the remaining candidates in the pattern."

If there are a finite number of twin primes, then you need on remove all the twin primes, that will be sufficient. There can be an infinite number of candidates left, that doesn't matter.

Sorry, I was interpreting your proof too much, my point is:

"Any single number p removes a symmetrical repeating pattern of remaining candidates, and can only remove at most 2/p of the remaining candidates in the pattern."

If there are a finite number of twin primes, then you need on remove all the twin primes, that will be sufficient. There can be an infinite number of candidates left, that doesn't matter.