Can Goldbach and Polignac's Conjecture be Proven with a Sieve Method?

[Andy Lee]

Yes, the method uses a sieve, but the problem of endpoints has been overcome.

Let S=(..., -5, -3, -1, 1, 3, 5, ...)

Select any integer N>=6.

Remove from S all multiples of 3 (>=9) and the pairs an equal but opposite distance from N. Then at least 1/3 of the pairs will remain.

Remove from S all multiples of 5 (>=25) and the pairs an equal but opposite distance from N. Then at least 3/5 of the pairs will remain.

Remove from S all multiples of any odd number (not 1) greater than or equal to its square, and the pairs an equal but opposite distance from N. Then at least (x-2)/x of the pairs will remain.

Perform the steps indefinitely as iterations rather than in isolation and at least
(1/3)(3/5)*...*(x-2)/x = 1/x of the pairs will remain.

The expected minimum pairs that remain is thus

E = Lim_x->infinity(1/x)(2x-2) where 2x-2 is the number of elements in S

E = Lim_x->infinity(2-2/x)

E = 2

There are at least 2 pairs of numbers p=(N+2y), q=(N-2y) if N is odd
(or p=((N-1)+2y), q=((N-1)-2y) if N is even) remaining.

Therefore 2N = p+q where p is prime and abs(q) is prime.

So 2N = p+q or 2N = p-q, p, q both prime, for all N>=3.

QED.

Andy Lee

 

[Andy Lee]

One minor correction:

When removing multiples of 3, 5, ..., x,
the multiples will be >=x² OR <=-x²

 

[CRGreathouse]

Perform the steps indefinitely as iterations rather than in isolation and at least
(1/3)(3/5)*...*(x-2)/x = 1/x of the pairs will remain.

The multiplication is not correct.

The expected minimum pairs that remain is thus

E = Lim_x->infinity(1/x)(2x-2) where 2x-2 is the number of elements in S

Not well-defined. The number of elements in S is aleph_0, not 2x-2.

 

[Andy Lee]

The multiplication is not correct.

Why?

Not well-defined. The number of elements in S is aleph_0, not 2x-2.

What is important here is that x approaches infinity. It does not matter which infinity,
so aleph_0 vs lim x_->infinity seems only to be a notational issue.

 

[CRGreathouse]

What is important here is that x approaches infinity. It does not matter which infinity,
so aleph_0 vs lim x_->infinity seems only to be a notational issue.

But at no point during the calculation is 2x-2 the cardinality of S. aleph_0 vs. _{\lim_{x\to \infty}} isn't the issue.

 

[Andy Lee]

But at no point during the calculation is 2x-2 the cardinality of S. aleph_0 vs. _{\lim_{x\to \infty}} isn't the issue.

Yes, you are correct. The cardinality of S is actually >= 2x-2
This seems to strenghten the proof.

 

[CRGreathouse]

Yes, you are correct. The cardinality of S is actually >= 2x-2
This seems to strenghten the proof.

No, it weakens it. A strengthening would be using fewer numbers than the N available to you. A weakening would be using more. A severe weakening would be using infinitely many numbers.

On the other hand, this one is correct: you can find a pair 2N - k, 2N + k with 2N - k and 2N + k relatively prime to 2, 3, ..., sqrt(2N). Unfortunately this doesn't prove GC because the proof allows k to be arbitrarily large, unlike in GC.

 

[Andy Lee]

On the other hand, this one is correct: you can find a pair 2N - k, 2N + k with 2N - k and 2N + k relatively prime to 2, 3, ..., sqrt(2N). Unfortunately this doesn't prove GC because the proof allows k to be arbitrarily large, unlike in GC.

Hence the title of my post!

My proof is of the conjecture:

Every even number greater than 2 can be expressed as either the sum or the difference of exactly two primes.

By the way, I will show in a later post that if a number can be expressed as the difference of two primes then it is true that it must also be expressed as the sum of two primes.

So for now, I have proved "goldbach and polignac". Soon... "goldbach" and "polignac".

 

[CRGreathouse]

My proof is of the conjecture:

Every even number greater than 2 can be expressed as either the sum or the difference of exactly two primes.

It doesn't prove that. Compare your statement to mine.

 

[Andy Lee]

It doesn't prove that. Compare your statement to mine.

Your concern seems to be that I allow 2N-k could be negative.

But if -a and b are relatively prime to 2, 3, ... , sqrt(2N) where a, b >0
then a and b are also relatively prime to 2, 3, ... , sqrt(2N).

So why the concern?

 

[Andy Lee]

On the other hand, this one is correct: you can find a pair 2N - k, 2N + k with 2N - k and 2N + k relatively prime to 2, 3, ..., sqrt(2N). Unfortunately this doesn't prove GC because the proof allows k to be arbitrarily large, unlike in GC.

Why do you suggest the upper limit is sqrt(2N)?
This is not in the proof.

 

[CRGreathouse]

Why do you suggest the upper limit is sqrt(2N)?
This is not in the proof.

Do you really think you've proved that the numbers are relatively prime to all integers?

 

[Andy Lee]

Do you really think you've proved that the numbers are relatively prime to all integers?

No. Just that the upper limit in your earlier statement should be sqrt(2N+k).

 

[CRGreathouse]

No. Just that the upper limit in your earlier statement should be sqrt(2N+k).

That doesn't appear anywhere in your proof. Try rewriting your proof using that, and you should be able to see why it doesn't work. In particular, for any bound B you can show that there are infinitely many numbers summing to a given number and relatively prime to 2, 3, ..., B, but you can't show that the smallest (in absolute value) is less than B^2.

 

[ramsey2879]

Why?

The multiplication is not correct because since 9 is not prime then multiples if nine were already removed by the time you reached x = 7. So you have 1/7 = 1/9 which is not so.

 

[CRGreathouse]

The multiplication is not correct because since 9 is not prime then multiples if nine were already removed by the time you reached x = 7. So you have 1/7 = 1/9 which is not so.

That's why I said that it was wrong. I see now that Andy is using only the odds, not the odd primes, so the multiplication is correct (though weaker than it would be if he used just the primes).

 

[Andy Lee]

The multiplication is not correct because since 9 is not prime then multiples if nine were already removed by the time you reached x = 7. So you have 1/7 = 1/9 which is not so.

Recall, the multiplication is to determine the least fraction of numbers remaining.

Remove multiples of 3 from S, then at least 1/3 of the numbers remain.
Remove multiples of 3 and 5 from S, then at least 1/3*3/5 of the numbers remain.
Remove multiples of 3, 5, and 7 from S, then at least 1/3*3/5*5/7 of the numbers remain.
Remove multiples of 3, 5, 7, and 9 from S, then AT LEAST 1/3*1/5*5/7*7/9 of the numbers remain.

I realize removing multiples of 9 is not necessary, but it does not make the last statement incorrect.

 

[Andy Lee]

I see now that the cardinality of the set S is really x+2, so we have

E = Limx->infinity(1/x)(x+2) where x+2 is the number of elements in S

E = Limx->infinity(1+2/x)

E = 1

 

[CRGreathouse]

I see now that the cardinality of the set S is really x+2, so we have

E = Limx->infinity(1/x)(x+2) where x+2 is the number of elements in S

E = Limx->infinity(1+2/x)

E = 1

So does this prove that there is exactly one pair (p, q) of primes for each k with p - q = k?

 

[Andy Lee]

So does this prove that there is exactly one pair (p, q) of primes for each k with p - q = k?

I'm not sure what your k is, but I believe I have proven that every even number
greater than 4 can be expressed as the sum or the difference of at least one pair of primes.

Recall, E gives the minimum.

 

[CRGreathouse]

I just wanted to point out that Underwood's book Mathematical Cranks (pp. 175-177) discusses your proof. Of course not your exposition, since the book was published 18 years ago, but the core of the proof is the same. If you're still interested in this, Andy, you should look up the book and see what Underwood has to say. His exposition is much better than mine, to be sure.