First let’s make a naïve statement about the relationship between
prime distribution and twin prime distribution in the hypercuboid.
In each case the basic topology is the same – we take a “slice”
through the number chain, then do the same at right angles an infinite number
of times. On a very simplistic level, we would expect to end up with a similar
pattern, but with fewer twin primes than primes.
(A potential objection here, is that one could say something similar about the elimination process in the standard Sieve of Eratosthenes, but that Sieve does actually eliminate every number in the end. However, the sieve from the previous page doesn't eliminate every number, it leave gaps for prime numbers, so if the twin prime version could genuinely be shown to be topologically equivalent then it would be valid to conclude that twin primes are infinite  however this step, below, is probably where this proof fails at this point.)
(A potential objection here, is that one could say something similar about the elimination process in the standard Sieve of Eratosthenes, but that Sieve does actually eliminate every number in the end. However, the sieve from the previous page doesn't eliminate every number, it leave gaps for prime numbers, so if the twin prime version could genuinely be shown to be topologically equivalent then it would be valid to conclude that twin primes are infinite  however this step, below, is probably where this proof fails at this point.)
But can we give this naïve idea any greater substance? Two shapes
are topologically equivalent if one can be distorted into the other – does this
apply to the patterns of primes and twin primes in the hypercuboid?
We can simplify
the process by only searching for numbers that are the highest of two twin
primes. We know that these must be on a (p^2)n+p plane, so they must be in one of
the intersections identified as prime numbers in the last version of the
hypercuboid. But we also need to narrow the search to eliminate numbers that
aren't 2 higher than another prime.
So we need to look at numbers that are also on a (p^2)n+p+2 plane, and that are not composite numbers (so aren't on any blue or black planes).
So we need to look at numbers that are also on a (p^2)n+p+2 plane, and that are not composite numbers (so aren't on any blue or black planes).
First, let’s look
at the planes which can be expressed as pn + 2.
For p = 3, we know
that the 9n +5 plane contains the prime number 5, which is the highest of two
twins.
9n + 2 and 9n +8
are 2 higher than 9n and 9n +6 respectively, so can never contain the highest
of a pair of twin primes.
So we colour 9n +
5 red, and we colour 9n + 2 and 9n +8 black.
4n

4n+1

4n+2

4n+3


9n

36

9

18

27

9n+1

28

1

10

19

9n+2

20

29

2

11

9n+3

12

21

30

3

9n+4

4

13

22

31

9n+5

32

5

14

23

9n+6

24

33

6

15

9n+7

16

25

34

7

9n+8

8

17

26

35

We repeat this
process in all other planes. (In the images, I'll stick to the 9 x 4 grid as
the bigger ones are a bit unwieldy).
Next, we want to
try and distort this shape into the same kind of hypercuboid we had for prime
numbers. Having coloured 9n + 2 and 9n +8 black, we know they can’t contain any
numbers that are the higher of two twins – we’ll show this by merging those
planes with the 9n plane and the 9n+6 plane respectively.
(If you don't like the idea of "merging" see Appendix C for an alternative suggestion of removing these planes from the hypercuboid and working with an incomplete number chain).
(If you don't like the idea of "merging" see Appendix C for an alternative suggestion of removing these planes from the hypercuboid and working with an incomplete number chain).
4n

4n+1

4n+2

4n+3


9n(9n+2)

36 (20)

9 (29)

18 (2)

27 (11)

9n+1

28

1

10

19

9n+3

12

21

30

3

9n+4

4

13

22

31

9n+5

32

5

14

23

9n+6 (9n+8)

24 (8)

33 (17)

6 (26)

15 (35)

9n+7

16

25

34

7

We can do the same
for any plane which contains pn +2 numbers other than (p^2)n+p+2. For instance we
colour 25n+7 red and merge 25n+2, 25n+12, 25n+17 and 25n+22 with black planes.
Next we “stretch”
the remaining green planes by simply repeating them:
4n

4n+1

4n+2

4n+3


9n(9n+2)

36 (20)

9 (29)

18 (2)

27 (11)

9n+1

28

1

10

19

9n+1

28

1

10

19

9n+3

12

21

30

3

9n+4

4

13

22

31

9n+4

4

13

22

31

9n+5

32

5

14

23

9n+6 (9n+8)

24 (8)

33 (17)

6 (26)

15 (35)

9n+7

16

25

34

7

9n+7

16

25

34

7

This hypercuboid
will contain repeats of many numbers.
(If we return to
the infinite hotel metaphor, this hypercuboid would contain a lot of rooms with two
numbers on the door and a lot of rooms that share the same number, which would
be useless for a real hotel, but luckily it is only imaginary).
Finally we put the
9n+3 and 9n+5 planes together.
4n

4n+1

4n+2

4n+3


9n(9n+2)

36 (20)

9 (29)

18 (2)

27 (11)

9n+1

28

1

10

19

9n+1

28

1

10

19

9n+3

12

21

30

3

9n+5

32

5

14

23

9n+4

4

13

22

31

9n+4

4

13

22

31

9n+6 (9n+8)

24 (8)

33 (17)

6 (26)

15 (35)

9n+7

16

25

34

7

9n+7

16

25

34

7

At this point, we
have a shape that is very similar to the pattern for prime numbers (image
repeated below):
4n

4n+1

4n+2

4n+3


9n(9n+2)

36 (20)

9 (29)

18 (2)

27 (11)

9n+1

28

1

10

19

9n+2

20

29

2

11

9n+3

12

21

30

3

9n+4

4

13

22

31

9n+5

32

5

14

23

9n+6 (9n+8)

24 (8)

33 (17)

6 (26)

15 (35)

9n+7

16

25

34

7

9n+8

8

17

26

35

We needn’t worry
too much about the 4n+2 plane being all black in the twin prime version as we
knew from the start it contains no twin primes. The important thing is that in
all other dimensions the pattern of black planes (definitely composite) and
green planes (coprime to p, in this case coprime to 2 and 3) is equivalent.
We can therefore conclude that the pattern of intersections between the green
and black planes is also topologically equivalent.
The slight
complication is that each number will now lie on two red planes, one for (p^2)n +
p and one for (p^2)n + p +2, and that we don’t yet know much about the pattern of
intersections in between these two sets of planes. We can't at this stage rule out the possibility that above a certain stage every number that is on a red pn plane falls against a number that is on two red pn+2 planes (so would have been coloured blue).
If it is possible that we reach a point where all future numbers lie on more than two red planes or are coloured black or blue, there isn't an infinitude of twin primes.
If it is possible that we reach a point where all future numbers lie on more than two red planes or are coloured black or blue, there isn't an infinitude of twin primes.
To make this easier to visualise, I’m
going to colour the pn+5 planes yellow
4n

4n+1

4n+2

4n+3


9n(9n+2)

36 (20)

9 (29)

18 (2)

27 (11)

9n+1

28

1

10

19

9n+1

28

1

10

19

9n+3

12

21

30

3

(9n+5)

32

(5)

14

(23)

9n+4

4

13

22

31

9n+4

4

13

22

31

9n+6 (9n+8)

24 (8)

33 (17)

6 (26)

15 (35)

9n+7

16

25

34

7

9n+7

16

25

34

7

Every number that
is intersected by at least one red plane must also be intersected by at least
one yellow plane. (Because it must be two higher than either a prime or a
multiple of a prime).
And every number
that is intersected by at least one yellow plane must also be intersected by at
least one red plane, for the same reason.
We can see that
the pattern created by the intersections of the red planes will be similar to the pattern created by the intersections of the yellow planes.
The problem at
this point is to rule out numbers in this
pattern that are intersected by only one red plane and more than one yellow
plane, or by only one yellow plane and more than one red plane.
For instance, the
number 23 appears on two pn+2 planes (7n+2 and 3n+2). If these were both yellow then this would rule it out from being the
higher of two twin primes (they aren't  49n+23 is black, while 9n+5 is yellow). It is only on one red plane (529n+23), so it is a
prime number. If a number of this sort can remain in the pattern and appear to
be a twin prime, then this method would fail to pick out the real twin primes.
However, if a number of this sort will always be on a black plane (as 23 is), then we
don’t have a problem – the pattern of intersections between black planes will
remove all nontwins and leave only numbers that actually are the higher of two
twin primes.
In the hypercuboid
for prime numbers, we ended up with some blue numbers. For instance the number
30 lies on the 25n+5 plane, the 9n+3 plane and the 4n+2 plane. It isn’t on any
black planes, but we marked it as blue in the pattern – it is composite because
it appears on more than one red plane.
The good news is
that when we are looking for twin primes, it seems likely that no numbers will end up coloured blue
– all even numbers end up coloured black, and it is impossible for an odd
number to lie on more than one red or yellow plane without also lying on a
black plane.
(I had a flawed bit of algebra here before  this is actually quite a complex modular problem, which I have given only a partial proof for here and here.
So the two flaws in this attempt at a proof are this modular problem (partially solved) and whether any real sense can be made of the idea of topological equivalence (below).
So the two flaws in this attempt at a proof are this modular problem (partially solved) and whether any real sense can be made of the idea of topological equivalence (below).
8. Topological Equivalence
We’ve now shown
(if the modular problem is proved, which it isn't yet) that the only primes that will be left in this pattern (where only one red
plane and only one yellow intersects a number) must be the highest of two twin
primes.
All primes that
are not the higher of two twin primes have been sieved out of the pattern, as
they will be intersected by a black plane.
(The only
exception to this is 3 because it is 2 higher than 1 – thus it isn’t removed by
being in a black plane or intersected by a yellow plane).
This means we can
legitimately treat the red and yellow plane in the grid above as a single plane
– the pattern of black, green and red/yellow planes is topologically the same
as the pattern for prime numbers. This means that for every prime in the
original pattern, there must be a prime in this pattern.
(Note, this doesn't mean that yellow planes and red planes always intersect the same numbers, only that where a number isn't intersected by a black plane, it can only be intersected by one red and one yellow plane).
(Note, this doesn't mean that yellow planes and red planes always intersect the same numbers, only that where a number isn't intersected by a black plane, it can only be intersected by one red and one yellow plane).
Many numbers will
be repeated within the pattern. For instance in the grid above we can already
see repeats of 7, 13, 19 and 31. (This makes sense as we know that primes are
denser than twin primes).
For each prime
number in the original hypercuboid, there is an equivalent prime in this
hypercuboid – this means that there are an infinite number of positions in
which there is a prime – and each of these must contain the highest of a
pair of twin primes.
This can only be
because either
1) there is an infinitude of twin primes
1) there is an infinitude of twin primes
or
2) at least one number is being repeatedly infinitely in the pattern.
2) at least one number is being repeatedly infinitely in the pattern.
We can however
rule out the latter possibility. When we “stretch” the pattern, let’s always
replace a black plane with a green plane with the highest modulo value
possible. In the example above we replaced 9n+2 with 9n+1, 9n+5 with 9n +4, and
9n+8 with 9n +7.
So for higher
primes, let’s use the highest modulo values possible – for instance
Replace 25n+ 2
with 25n+ 19,
Replace 25n+ 12
with 25n +21
Replace 25n+17
with 25n+23
Replace 25n+22
with 25n+24.
Replace 49n+2 with
49n+43
Replace 49n+16
with 49n+45
And so on…
By adopting this
strategy, we gradually increase the modulo residues we are using in the
stretching process. This set of substitutions won’t, for instance, lead to any
more repeats of the number 13 in the hypercuboid as it has a modulo residue of
13 for all numbers higher than 13.
As the modulo
residues used for substitutions increase, so do the numbers that can’t be
repeated again, so it is not possible for any number to repeated infinitely in
the pattern.
This leaves us
with only one other explanation of the topological equivalence – we must have
an infinitude of numbers that are the highest of two twin primes each being
repeated a finite number of times.
Therefore there is an infinitude of twin primes.
Extending this to Ktuples.
The prime ktuple
conjecture states that every admissible prime constellation occurs infinitely
often.
An admissible
constellation is one that is defined by a set of patterns of modulo residues,
which “do not include the complete modulo set of residue classes (i.e. the values
0 through p − 1) of any prime p less than or equal to k”.
If the strategy
above, of replacing and “stretching” modulo residue planes, works, then we might be ableprove the infinitude of any admissible prime constellation.
Not worth worrying about that unless the observed flaws in the proof above can be fixed though.
Go to Appendix
Go to Appendix
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