Friday, 20 July 2012

A modular problem page 2

I can go some way to generalising the proof I made for 3 factor numbers to numbers with more factors. For instance:

A number has 4 odd prime factors a < b < c < d

Assume
abc = 1 mod d = (0, 0, 0, 1)
abd = 1 mod c = (0, 0, 1, 0)
acd = 1 mod b = (0, 1, 0, 0)
bcd = 1 mod a = (1, 0, 0, 0)

abc + abd + acd +bcd = (1, 1, 1, 1)

Subtract 1

= (0, 0, 0, 0) = 0 or abcd

a >= 3, b >= 5, c >= 7, d >=11

abc =< abcd/11
abd =< abcd/7
acd =< abcd/5
bcd =< abcd/3

1/3 + 1/5 + 1/7 + 1/11 = 886/1185

abcd = 1185

Therefore (1, 1, 1, 1) ≠ abcd + 1

abc + abd + acd +bcd = 1

RAA

Therefore the initial assumption was impossible.

It looks to me as though this works up to 8-factor numbers (and would have worked for 3 factor numbers) but then stops working because 1/3 + 1/5 + .... 1/ 23+ 1/29 > 1 (29 being the 9th odd prime).
A number has 4 odd prime factors a < b < c < d
 Also, if this works at all, then it works for any number with an odd number of factors, for instance for five factors:

abcd + abde + abce +abde + bcde = (1, 1, 1, 1, 1)

Subtract 1

= (0, 0, 0, 0, 0) = 0 or abcde

However abcde+1 is even, and (abcd + abde + abce +abde + bcde) is odd.

so abcde+1 ≠ (abcd + abde + abce +abde + bcde)

(Actually this last bit isn't completely sound - if abcd + abde + abce +abde + bcde = 1 mod abcde but this is a multiple of abcde rather than abcde itself, then the proof doesn't hold - but this at least extends the proof to numbers with a very large number of odd factors since it takes a long time to reach 2 in the series 1/3 + 1/5 + 1/7  + 1/11.... However this series isn't convergent, so you do eventually get to 2)



CRGreathouse also commented that

"You can extend it to 9- or even 10-factor numbers by adding finite checking. Probably too difficult for more than 9 factors if you're doing it by hand, though."
Which is correct - there would be many combinations for 9, 10 and above where the fractions didn't exceed 1, so it does work in theory for a big swathe of highly composite numbers - you just can't make it a general proof without finite checking.

A modular problem

This is something I already put up on My Math Forum.

Let a, b, c be three different prime numbers.

Is it possible that

ab = 1 mod c
ac = 1 mod b
bc = 1 mod a

This is clearly possible if one of the numbers is even as 30 is an example:

6 = 1 mod 5
10 = 1 mod 3
15 = 1 mod 2

But is it possible if they are all odd?

abc has three prime factors, a < b < c
Each number below abc has a unique set of modular residues for a, b, c which we will express as (x, y, z)

Assume
ab = 1 mod c
ac = 1 mod b

Start at (0, 0, 0) = 0
First we add (0, 0, 1) to get ab = 1 mod c = (0, 0, 1)
Then we add (0, 1, -1) to get to ac = 1 mod b = (0, 1, 0)

Now, assume we add (0, 1, 0) + (-1, -1, 0) = (-1, 0, 0)

If (-1, 0, 0)= bc then bc ≠ 1 mod a

We know that (-1, -1, 0) < bc because
(0, 0, 1) = ab
(-1, -1, 0) = (0, 0, 1) – 1 = ab - 1
ab < bc
Therefore (-1, -1, 0) < bc

Is it possible that (-1, 0, 0) ≠ bc?

To get to any other value of (x, 0, 0) we must add or subtract a multiple of bc from (-1, 0, 0)

We can’t subtract because (-1, 0, 0) < bc.

And (-1, 0, 0) + bc > bc.

Therefore (-1, 0, 0) = bc

Therefore bc ≠ 1 mod a

C.R.Greathouse helpfully pointed out that since I hadn't clearly specified that a, b, and c were odd that last bit didn't work - there's a bit more detail on how that can be cleaned up here.

What I really want is a more general version of this proof, that applies to numbers of more than three factors, as that would fit in with the other stuff I've been working on. I've made a start on this here.

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?

Monday, 16 July 2012

Plane substitutions.

OK, to help explain the logic of the attempted twin prime proof, here is an example of how the plane substitutions work. Here's the starting point of the 4n x 9n grid (in other words how it looks in the prime number hypercuboid we started from).



















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

Now I am going to duplicate the 9n +1 and 9n+7 plane. The 9n+2 and 9n+8 plane are coloured black (as well as the 4n+2 column). The 9n+5 is coloured yellow.

4n
4n+1
4n+2
4n+3
9n
36
9
18
27
9n+1
28
1
10
19
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+7
16
25
34
7
9n+8
8
17
26
35



 If we omit the black planes at 9n+2 and 9n+8 we have two horizontal black planes as in the original diagram (in the proof I talk about merging planes but I think it makes more sense to simply discard them.) The grid contains some repetitions and some numbers in the number chain are missing (but we know they aren't twin primes).

9n
36
9
18
27
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+5
32
5
14
23
9n+6
24
33
6
15
9n+7
16
25
34
7
9n+7
16
25
34
7

 Next we need to extend this up to the 25n plane. The first few slices look like this (before they are sorted by their residue mod 25):


36
9
18
27
28
1
10
19
28
1
10
19
12
21
30
3
4
13
22
31
32
5
14
23
24
33
6
15
16
25
34
7
16
25
34
7

72
45
54
63
64
37
46
55
64
37
46
55
48
57
66
39
40
49
58
67
68
41
50
59
60
69
42
51
52
61
70
43
52
61
70
43

108
81
90
99
100
73
82
91
100
73
82
91
84
93
102
75
76
85
94
103
104
77
86
95
96
105
78
87
88
97
106
79
88
97
106
79

To make it easier to see what the next stage looks like, I'm going to leave out the black planes and the red plane (in which there are no primes other than 3, which is an exception - it isn't a twin because it is only on one red plane. )

So we start with this pattern:



4n+1
4n+3
9n+1
1
19
9n+1
1
19
9n+4
13
31
9n+4
13
31
9n+5
5
23
9n+7
25
7
9n+7
25
7

Here, the numbers in the green and yellow planes up to 900 have been rearranged by their residue mod 25 (picture this as 25 slices of a cuboid):


25n+2, 25n+12, 25n+17 and 25n+22 are black planes as they can't contain the higher of two twin primes. 25n+5 is red, 25n+7 is yellow.

Next, we are going to substitute:

25n+3 for 25n+2
25n+13 for 25n+12
25n+18 for 25n+17
25n+ 23 for 25n+22


I've highlighted 5 - this is the only number on this grid that won't appear on any more red or yellow planes. There are two instances of 7 (one of which is highlighted) - these will both be on the 49+7 red plane in the next dimension and then no other red planes.

The first prime that is a non-twin is 11 - this has already been removed from the pattern when we coloured the 9n+2 plane black - we chose to repeat the 9n+1 plane, so you could see the second instance of 19 as a replacement for 11.

Every number that is only in one red plane when this process is reiterated in infinite dimensions will also be on only one yellow plane, so we can treat the yellow plane as being the same as a green plane.*

Given this, the intersection of green, black and red planes in the grid above is the same topological pattern as we would find if we were looking only for primes, but we have multiple repetitions of individual numbers within the pattern.

Also, I talk about rearranging the numbers according to modulo residue above - it would maybe be better to visualise this as a hypercuboid that from the start contains duplications of some planes that aren't either pn or pn+2  and doesn't include any pn+2 planes other than the (p^2)n+p+2 planes.

Hope that all makes sense?

In theory we need to make these plane substitutions in every dimension - I do have one or two thoughts on why that might be problematic, I'll come back to talk about that soon.

*See the algebra in part 7 of the original proof.

Sunday, 15 July 2012

Twin Prime Proof Appendix


Here’s a bit more detail on how the first hypercuboid was constructed – this is the simple version before we added the complication of residues of squares. This is a version that is looking for pn and pn+2 numbers (eg twin primes) starting from the point where we have narrowed the possible numbers down to the 2n+1 planes and the 3n+1 planes:


1
7
13
19
25
31
37
43
49
55
61
67
73
79
85
91
97
103
109
115
121
127
133
139
145
151
157
163
169
175
181
187
193
199
205

Numbers coloured grey are either 5N or 5N +2.
Numbers coloured blue are either 7N or 7N+2.

We turn this into a box by adding the following five slices starting from each of the five columns:


1
31
61
91
121
151
181
211
241
271
301
331
361
391
421
451
481
511
541
571
601
631
661
691
721
751
781
811
841
871
901
931
961
991
1021
1051
1081
1111
1141
1171
1201
1231
1261
1291
1321
1351
1381
1411
1441
1471
1501
1531
1561
1591
1621
1651
1681
1711
1741
1771
1801
1831
1861
1891
1921
1951
1981
2011
2041
2071
2101
2131
2161
2191
2221
2251
2281

7
37
67
97
127
157
187
217
247
277
307
337
367
397
427
457
487
517
547
577
607
637
667
697
727
757
787
817
847
877
907
937
967
997
1027
1057
1087
1117
1147
1177
1207
1237
1267
1297
1327
1357
1387
1417
1447
1477
1507
1537
1567
1597
1627
1657
1687
1717
1747
1777
1807
1837
1867
1897
1927
1957
1987
2017
2047
2077
2107
2137
2167
2197
2227
2257
2287

13
43
73
103
133
163
193
223
253
283
313
343
373
403
433
463
493
523
553
583
613
643
673
703
733
763
793
823
853
883
913
943
973
1003
1033
1063
1093
1123
1153
1183
1213
1243
1273
1303
1333
1363
1393
1423
1453
1483
1513
1543
1573
1603
1633
1663
1693
1723
1753
1783
1813
1843
1873
1903
1933
1963
1993
2023
2053
2083
2113
2143
2173
2203
2233
2263
2293

19
49
79
109
139
169
199
229
259
289
319
349
379
409
439
469
499
529
559
589
619
649
679
709
739
769
799
829
859
889
919
949
979
1009
1039
1069
1099
1129
1159
1189
1219
1249
1279
1309
1339
1369
1399
1429
1459
1489
1519
1549
1579
1609
1639
1669
1699
1729
1759
1789
1819
1849
1879
1909
1939
1969
1999
2029
2059
2089
2119
2149
2179
2209
2239
2269
2299

25
55
85
115
145
175
205
235
265
295
325
355
385
415
445
475
505
535
565
595
625
655
685
715
745
775
805
835
865
895
925
955
985
1015
1045
1075
1105
1135
1165
1195
1225
1255
1285
1315
1345
1375
1405
1435
1465
1495
1525
1555
1585
1615
1645
1675
1705
1735
1765
1795
1825
1855
1885
1915
1945
1975
2005
2035
2065
2095
2125
2155
2185
2215
2245
2275
2305

And as before, note that we can rearrange these to make the planes clearer, for instance:









5n+1
5n+2
5n+3
5n+4
5n+5
7n+1
1
127
43
169
85
7n+2
121
37
163
79
205
7n+3
31
157
73
199
115
7n+4
151
67
193
109
25
7n+5
61
187
103
19
145
7n+6
181
97
13
139
55
7n+7
91
7
133
49
175


7n+1
7n+2
7n+3
7n+4
7n+5
7n+6
7n+7
11n+1
1
331
661
991
1321
1651
1981
11n+2
211
541
871
1201
1531
1861
2191
11n+3
421
751
1081
1411
1741
2071
91
11n+4
631
961
1291
1621
1951
2281
301
11n+5
841
1171
1501
1831
2161
181
511
11n+6
1051
1381
1711
2041
61
391
721
11n+7
1261
1591
1921
2251
271
601
931
11n+8
1471
1701
2131
151
481
811
1141
11n+9
1681
2011
31
361
691
1021
1351
11n+10
1891
2211
241
571
901
1231
1561
11n+11
2101
121
451
781
1111
1441
1771


7n+1
7n+2
7n+3
7n+4
7n+5
7n+6
7n+7
11n+1
463
793
1123
1453
1783
2113
133
11n+2
673
1003
1333
1663
1993
13
343
11n+3
883
1213
1543
1873
2203
223
553
11n+4
1093
1423
1753
2083
103
433
763
11n+5
1303
1633
1963
2293
313
643
973
11n+6
1513
1843
2173
193
523
853
1183
11n+7
1723
2053
73
403
733
1063
1393
11n+8
1933
2263
283
613
943
1273
1603
11n+9
2143
163
493
823
1153
1483
1813
11n+10
43
373
703
1033
1363
1693
2023
11n+11
253
583
913
1243
1573
1903
2233

For the proof above, these diagrams would need to be extended to 25n x 49n grids and 49n x 121n grids, but it is easier just to imagine those…

Appendix B

Just a quick thought on modulo residues:


I find it easiest to think of modulo residues this way – any prime number p has a modulo residue of p for every other prime. For instance, 19 is defined by the infinite set of pn + 19 residues. However, for numbers below 19, you need to convert this into numbers in the appropriate modulo range, so this is equivalent to 2n+1, 3n+1, 5n+4, 7n+5, 11n+8, 13n+6, 17n+2, 19n (or 19n+19) and every pn+19 thereafter. 

Appendix C

Some people might not like the "merging" stage of the attempted proof. (it's certainly a bit dodgy as things stand!) An alternative method would be to simply remove the pn+2 planes from the grid at the same time as repeating other green planes. This would leave us with an incomplete number chain, with repetitions. It would still be topologically equivalent to the prime number hypercuboid. And when it comes to the p^2n+p and p^2n+p+2 planes, we could simply ignore one of them since both will contain the same numbers (>3) in locations where they are not intersected by black planes.

(See also this page for an attempt to simplify the patterns of coloured planes)

Appendix D

Sometimes in the proof there may be some confusion between planes and hyperplanes - I've not been too pedantic about this because the only truly important thing about the multi-dimensional geometry of the hypercuboid is that each number lies on the intersection of an infinite number of planes and that an (admissible) infinite range of modulo residues uniquely defines any number.