- #1
Andy Lee
- 37
- 0
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
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