# Goldbach and Polignac - Proof

1. Jan 24, 2010

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

E = Limx->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

2. Jan 25, 2010

### Andy Lee

One minor correction:

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

3. Jan 25, 2010

### CRGreathouse

The multiplication is not correct.

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

4. Jan 25, 2010

### Andy Lee

Why?

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.

5. Jan 25, 2010

### CRGreathouse

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.

6. Jan 25, 2010

### Andy Lee

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

7. Jan 25, 2010

### CRGreathouse

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.

8. Jan 25, 2010

### Andy Lee

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".

9. Jan 25, 2010

### CRGreathouse

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

10. Jan 25, 2010

### Andy Lee

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?

11. Jan 25, 2010

### Andy Lee

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

12. Jan 25, 2010

### CRGreathouse

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

13. Jan 26, 2010

### Andy Lee

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

14. Jan 26, 2010

### CRGreathouse

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.

Last edited: Jan 26, 2010
15. Jan 26, 2010

### ramsey2879

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.

16. Jan 26, 2010

### CRGreathouse

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).

17. Jan 26, 2010

### Andy Lee

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.

18. Jan 26, 2010

### 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

Last edited: Jan 26, 2010
19. Jan 26, 2010

### CRGreathouse

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

20. Jan 26, 2010

### Andy Lee

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.