Tuesday, 17 July 2012

A simpler thought on twin primes

C.R.Greathouse from My Math Forum has succeeded in convincing me I was either wrong or overcomplicating things (or just being a crank...) in a few previous posts, which I've taken down for now. The "stretching and merging" created some problems, at the very least it made it pretty hard to conceive of how individual substitutions of numbers actually affect the pattern.

This is a simplified approach, which doesn’t depend on any stretching or merging - I'm not offering it as a proof, but to try and communicate the bit I find interesting and see if there is any way to improve this.

Again, this is based on the idea of numbers laid out in a hypercuboid.

In the diagrams above I am just colouring any pn plane other than (p^2)n+p black and any pn+2 plane other than (p^2)n+p+2 black. Every number starts out white. Once a number is black it stays black.

(I’ve started from the point where even numbers are already coloured black) - the 49n grid is incomplete but I hope there is enough for clarity.)

We leave the (p^2)n+p plane and the (p^2)n+p+2 plane white (except where they are intersected by a black plane).

I would like to argue that:

Every number will end up white or black. Any number >3 that ends up white is a twin prime. If there is an infinitude of white squares in this pattern (once it has been continued infinitely) then there is an infinitude of twin primes.

This seems counter-intuitive because we are doing nothing to mark how the (p^2)n+p numbers intersect with the (p^2)n+p+2 numbers. I believe we don’t need to because any number >3 that doesn’t end up black in this grid must be on both a (p^2)n+p plane and  a (p^2)n+p+2 plane, and can’t be on more than one of each.

However, for this to hold, I do have to make an assumption :

Let a, b, c be three different odd prime numbers.
Assume it is not possible that
ab = 1 mod c
ac = 1 mod b
bc = 1 mod a

(I've proved this bit here.)

(This is possible if one of the numbers is even as 2, 3, 5 is an example: 6 = 1 mod 5, 10 = 1 mod 3, 15 = 1 mod 2).

We also need to assume the same principle for numbers with more factors, for instance for 4 factors, assume that it is not possible that
abc = 1 mod d
abd = 1 mod c
acd = 1 mod b
bcd = 1 mod a

I've proved this for up to 8 factors and for any number with an odd number of factors here. But it is still a big assumption to assume it holds for all numbers.

That rather big assumption has an interesting consequence.

If true, I think it means we don’t have to worry about how the (p^2)n+p plane and (p^2)n+p+2 planes intersect in the grids above. Which means that this is a simpler and more effective sieving process than it might look on first sight. Here's why:

For an odd composite number X to appear on a (p^2)n + p plane, it must be expressible as p(pn+1).

This means that the product of the prime factors of X other than p must be expressible as pn+1.In other words if the factors are a and b, a = 1 mod b.

For a number with two odd factors, for instance 3 x 13, it is clearly impossible that
a = 1 mod b
b = 1 mod a

So any composite number with two odd factors is on at least one black plane. (For instance 39 is on the 169n+39 black plane, 9 is on the 9n black plane.)

And if the assumption above about 3+ factor numbers is true then any number that is on more than one (p^2)n + p plane is on at least one black plane.

The same also holds for any number on more than one (p^2)n + p + 2 plane (for the same reason, as this is also dependent on the factors of (p^2)n + p.)

So any number which isn't on a black plane is on only one (p^2)n + p plane and only one (p^2)n + p + 2 plane, and is thus a twin prime.

I find this interesting because rather than looking at the interaction between two patterns (pn and pn+2) we only need to look at the one pattern.

Now, what I was previously trying to do is to find a tricksy way to show that the pattern of intersections between the black planes is topologically equivalent in this grid and in the prime one where we don’t eliminate any pn+2 planes. But doing it with “duplicating and/or merging” creates other problems as Mr Greathouse patiently prodded me towards seeing.

However, it still seems to me that the situation in the grid above has some kind of topological equivalence to the situation in the prime grid. In each dimension we simply take a black slice through part of the hypercuboid –  the black slices are just wider in the twin version.

(For instance, on a really basic topological level - imagine we rearranged each dimension with the black planes in the middle and the white ones on either side, like this:

There is no logical reason we can't do that as we aren't losing any numbers and can rearrange the planes in the same order in each version. Then we could physically widen (without duplicating) the white planes and narrow (without deleting) the black planes in the twin version of the image (top) and end up with the same area of the cross section being coloured black or white as on the prime version (bottom) - this wouldn't give us any more or less numbers in the white and black areas, but it would be one way of bringing out the topological similarity (- it's a damn sight closer than a doughnut and coffee mug, at least).

I know it wouldn't be fair to assume that a process of this sort necessarily leads to an infinitude of gaps, but we do already know that in the prime version it does leave an infinitude of gaps. 

So 1) is the assumption above correct for 8+factor numbers and 2) is it fair therefore to assume that there is an infinitude of gaps in this pattern? If not, there's probably not a way forward... I wonder if there is a more precise, formal way to look at the topological equivalence question though?

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