Yes, the method uses a sieve, but the problem of endpoints has been overcome.(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Goldbach and Polignac - Proof

Loading...

Similar Threads - Goldbach Polignac Proof | Date |
---|---|

Some ideas concerning the Goldbach conjecture | Dec 25, 2012 |

Has Goldbach's conjecture been solved? | Oct 1, 2012 |

Direction of Goldbach Partitions | Aug 6, 2012 |

Proof of Goldbach,Polignac,Legendre,Sophie Germain conjecture.pdf | Sep 27, 2010 |

Very simple (Dis)proof of Riemann hypothesis, Goldbach, Polignac, Legendre conjecture | Aug 19, 2010 |

**Physics Forums - The Fusion of Science and Community**