27.9.19

Goldbach Conjecture

While reading about the Goldbach conjecture recently, I thought of Z35.

According to the Goldbach Conjecture every even integer greater than 2 can be represented as the sum of two prime integers. But, I wondered, how does this conjecture affect odd integers? I also wondered if this conjecture has some application to some of the conjectures I made related to the product of two distinct prime integers greater than 2.

By definition of even and odd integers, if x is an odd integer, then x+1 is an even integer and x-1 is also an even integer.

A few years ago I conjectured that there are 2 non-trivial integers a and b such that a2 = 1 mod n and b2 = 1 mod n where a + b = n whenever n is the product of two distinct prime integers greater than 2. I called a and b "lonely squares". There are also 2 trivial lonely squares, namely 1 and n-1.

Z35 is very special because:
35 = 5*7, which is not an even integer and yet:
292 = 1 mod 35
and 29+7 = 36 = 1 mod 35

Furthermore,
29+5 = 34 mod 35
342 = 1 mod 35

Note: 29 and 6 are the only non-trivial "lonely square" in Z35.

1 is a trivial "lonely square" since 1^2 = 1 for all n where n = p*q where p and q are distinct odd primes.

34 is also a trivial "lonely square" because, as I proved previously, given any n = p*q where p and q are distinct odd primes, then (n-1)^2 = 1 mod n

29 + 34 is clearly not equal to 35, which is okay since 34 is not prime. The interesting part in this case is the fact that there is a direct relation between the odd integer x=35, its surrounding even integers 34 and 36, the prime integers 5 and 7 of which 35 is a product, and the so-called "lonely squares" of Z35

Another finite field with these interesting properties is Z15 because:
15 = 3*5
11^2 = 1 mod 15
11+ 5 = 16 = 1 mod 15

Furthermore,
11+3 = 14
14^2 = 1 mod 15

Note: 11 and 4 are the only non-trivial "lonely squares" in Z15. And 14 is a trivial one.

Unfortunately, this is not applicable to all products of primes but it is interesting to see where it pops up and why.

To understand why this happens, one must explore the commonalities between the two examples. Both 35 and 15 have a common factor, namely 5. However, Z55 and Z85 do not exhibit the same properties as Z15 and Z35. So this is most likely a dead end.

The other thing that both examples have in common is that q = p+2, also known as twin primes. Indeed, it appears that this has something to do with why Z15 and Z35 have such interesting properties.

For example, 11*13 = 143 and the field Z143 exhibits the same interesting properties as Z15 and Z35 since:
143 = 11*13
131^2 = 1 mod 143
131+13 = 144 = 1 mod 143

Furthermore,
131+11 = 142 mod 143
142^2 = 1 mod 143

Is this true for all products of twin primes?

19.3.19

The skewy revisited

A few years ago I boldly introduced a new term called "skewy", which I defined as the smallest square root of (n+1/2)2 mod n where n is the product of two distinct odd primes.



The skewy has an interesting relation to the Collatz conjecture and to the prime factorization of n.

Here are a few claims to consider:

Claim #1:   4*((n+1)/2)2 = 1 mod n

Claim #2:   The row at 4 of the exponentiation table of Zn consists of all squares co-prime with n.

Claim #4:   If s is the skewy, then 4*(s3 mod n) mod n = s.

Claim #5:   4*(((n+1)/2)3 mod n) mod n = (n+1)/2.

26.1.19

The Collatz Connection

There is one integer sequence, which is keeping me up at night sometimes if I start thinking about it. It is the sequence of odd integers, which do not have 5 as a Collatz exit point. Why don't they? What is so special about them? But most importantly, how can we predict the next term of this sequence? What is the formula? This sequence is not in OEIS. The first few terms that I found are:

21,75,85,113,151,201

Not found in OEIS
The Collatz Conjecture is important for integer factorization. I recently read a very nice explanation of why this problem is important. I came up with a couple of interesting sequences in this vicinity: A276260 and A276290 but the sequence described above is something that troubles me in a mathematical sense.

On a personal note, I just learned today that the mathematician who first thought of the Collatz problem died while attending a Mathematics Conference in the city I was born in.

19.1.19

Dreaming of Collatz Exit Points

A few years ago I talked about Collatz exit points. Some Collatz exit points are reached a lot more frequently than others.  In fact, it appears that most integers reach only one such point.

Here are the first few Collatz exit points: 5, 21, 85, 341, 1365, 5461, 21845, 87381, 349525, 1398101, 5592405, 22369621, 89478485, 357913941, 1431655765, 5726623061, 22906492245, 91625968981, 366503875925, 1466015503701, 5864062014805, 23456248059221, 93824992236885, 375299968947541

Note: I call them Collatz exit points because for each ai in the set above 3*ai + 1 = 2i+1

As I already mentioned here there are only 3 odd integers below 100, for which 5 is not a Collatz exit. Below 200 there are also only 3, below 300 there are also only 3.

It seems 5 is the most commonly occurring Collatz exit point for all odd integers below 1000.

What about below 10000?

What about below 100000?

...



12.1.19

Non-Complimentary Primes

On the other side of complimentary primes are the non-complimentary primes, which are primes p for which the order of 2 mod p is odd.

The first few primes are 2,7,23,31,47,71,73,79,89,103,127,151 and more can be generated here using the following Mathematica line
Select[Prime[Range[800]], OddQ[MultiplicativeOrder[2, #/(2^IntegerExponent[ #, 2])]]&]

In OEIS these primes are known as primes p that do not divide 2^x+1 for any x>=1.  What's interesting to me about these primes is that their exponentiation table follows a very different pattern than that of complimentary primes, and it all seems to be dictated by the order of 2 mod p.

5.1.19

Complimentary Primes

Definition: Complimentary primes are prime integers p such that 2k/2 mod p = p-1 where k is the order of 2 mod p.

Example: Let p = 43. It can be verified  that k = 14 since 214 mod 43 = 1 and 14 is the smallest such integer so 14 is the order of 2 mod 43. Since 27 mod 43 = 42 then 43 is a complimentary prime.

Verified result

The first few complimentary primes are 5,11,13,17,19,29,37,41,43,53,59,61,67,83,97,101.

This is a sequence documented in oeis.org (more or less) as the sequence of primes p where the order of 2 mod p is even. What isn't documented there and what I find special about complimentary primes is the following conjecture that I made:

Conjecture: If p is a complimentary prime and k is the order of 2 mod p then
(2i (mod p)) + (2(k/2)+i(mod p)) = p for all i greater or equal to 0

Example: Let p = 43. Since k is the order of 2 and k = 14, then k/2 = 7
20 mod 43 = 1  and  27 mod 43 = 42,
21 mod 43 = 2 and 28 mod 43 = 41,
22 mod 43 = 4 and 29 mod 43 = 39
23 mod 43 = 8 and 210 mod 43 = 35,
24 mod 43 = 16 and 211 mod 43 = 27
25 mod 43 = 32 and 212 mod 43 = 11,
26 mod 43 = 21 and 213 mod 43 = 22
27 mod 43 = 42 and 214 mod 43 = 1

You can use the iFrame below to find powers of 2 modulo an odd integer.