3.12.17

Proving The Existence Of A Second Private Key That Decrypts a Message Encrypted With The RSA Encryption Algorithm

While taking a course in cryptography 6 years ago I stumbled upon a problem that kept me awake at night for many nights.

In school we learn that the RSA encryption algorithm consists of one public key and one private key. I've asked several professors if there could be a second private key that decrypts all messages in the same way as the intended private key and I've always received a negative answer.

And yet I stumbled upon two private keys in one particular example. I delved into the Mathematics behind the RSA algorithm and I found out that in fact there are at least two keys in every single example. I not only managed to prove the existence of a second private key but I was also able to find two different formulas for obtaining the second key.

One of the formulas requires knowledge of phi(n)/2 where phi(n) is Euler's Totient Function.

If d1 is the intended private key and (n, e) is the public key then the second key d2 can be found using the following formula:

d2*e = phi(n)/2

Here's the full paper: https://marinaibrishimova.net/docs/otherkeys.pdf

Currently, I'm looking into ways of obtaining phi(n)/2 without having to factor n.

4.9.17

Products of Primes Without Selfie Powers

In my previous post I suggested that there are products of primes p and q that are easier (relatively) to factor because there is an integer k such that k = (p*q - 1)/2 and k^k = (phi(p*q))/2 and I called k a selfie power.



Before I talk about such products, I'll talk about products of primes, which have no selfie power and no semi selfie powers* because proper categorization is important.



Theorem: Given a product of primes p and q, if p or q divides phi(p*q) then p*q has no selfie power (ie an integer k such that k = (p*q - 1)/2 and k^k = (phi(p*q))/2 )

Proof: If p or q divides phi(p*q) then GCD(phi(p*q), k) = 1 since by definition of a selfie power k = (p*q - 1)/2 but also by definition of a selfie power k^k = (phi(p*q))/2 so 2*(k^k) = phi(p*q), which is a contradiction.



Example: Let p = 5 and q = 11, then p*q = 55 and phi(p*q) = (p-1)*(q-1) = 4*10 = 40 and clearly p divides phi(p*q). Furthermore, (p*q-1)/2 = 27 and 27^27 (mod 55) = 42 so 27 is not a selfie square as expected.


*to be defined in future blog posts


30.8.17

Selfie Powers

According to some people in Israel, the number 36 is special. According to these people, at any point in time there are 36 people who hold the world together and if one of them dies, then the world will fall apart. I don’t share such esoteric views but I do find the number 35, which is one less than 36, to be quite extraordinary from a mathematical point of view.



The exponentiation table of Z sub 35



It represents a special class of products of two distinct primes with peculiar properties. Such products of primes are really easy to factor. 

Before I talk more about other products of primes like this, first I have to introduce a new definition, which is not new to me but it is new to this blog.

Definition: Selfie powers are integers k mod n such that 2*k = n-1 and k^k = phi(n)/2 where phi(n) is Euler's Totient Function

Example:

Given n = 35 then (n-1)/2 = 17 and 17^17 = 12, which is phi(35)/2

25.7.17

A Few More Conjectures

The subset of safe primes, which I lovingly call "steady" primes are not so steady after all if my latest
conjectures turn out to be true.

As I wrote in a previous post, the set of safe primes is composed of two subsets: the set of steady primes and the set of tough primes.


The first few safe primes where primes in olive green are steady and the rest are tough


Conjecture 1: Given p a steady prime and a < p such that it is not a power of 2 mod p then
a^((p-1)/2)= p-1 else if a is a power of 2 mod p then a^((p-1)/2)= 1 

Conjecture 2: Given p a steady prime then (p-1)/2 is not a power of 2 mod p and
((p-1)/2)^((p-1)/2) = p-1

2.6.17

The Graphing Calculator

When I first moved to Canada I was simultaneously shocked and pleasantly relieved to find out that calculators are mandatory in high schools. 


26.4.17

Subsets of Safe Primes

The set of safe primes is composed of 2 mutually disjoint sets: the set of tough primes and the set of steady primes.


a list of the first few safe primes with tough primes in light blue background; the rest are steady primes


The main difference between these two sets lies in the structure of the exponentiation tables of their elements. For a tough prime p, I conjecture that:

Conjecture: If p is a prime of the form 8n+7, then the order of even powers of 2 mod p is p-1 and the order of odd powers of 2 mod p is (p-1)/2

In contrast, for a steady prime q I conjecture that:

Conjecture: If q is of the form 8n-1 such that 4n-1 and 8n-1 are also primes, then the order of all powers of 2 mod q is equal to (p-1)/2

Example: Let p = 11 and let q  = 7. Below are the corresponding exponentiation tables of Z11 and Z7.
The order of 2 mod 11 = 10, which is equal to the order of 8 mod 11, but the order of 4 mod 11 is half of that. Whereas in  Z7 the order of all powers of 2 is equal to 3.


Steady primes are the only primes where the order of all powers of 2 mod p is the same.  For all other primes the order of different powers of 2 is different.

On a slightly different note, as bad as safe primes may sound for cryptography, they are still not as bad as strong primes.

20.4.17

Why Steady Primes?

In my previous entry I discussed a subset of safe primes with an interesting property.

It appears that when a prime p is of the form 8k - 1 where 4k - 1 and 8k - 1 are both primes, then the order of each power of 2 mod p is phi(p)/2 where phi() is the Euler's Totient Function.

This is only one part of my conjecture. Here's another part of it:

Conjecture: If p is a prime of the form 8k - 1 where 4k - 1 and 8k - 1 are both primes and n is an integer smaller than q such that it is not a power of 2 mod p, then the order of n mod p is phi(p)= p-1

Example: Below is an image of the exponentiation tables of Z23 and Z7.



In my previous entry I called such primes "steady primes". The reason why I chose the name "steady" is because of the "steady", uniform structure of the exponentiation table of Zp when p is a steady prime.

All elements of Zp when p is a steady prime have order that is equal to or greater than phi(p)/2. This is also true for the rest of the safe primes, which are the tough primes, but the structure of the exponentiation table of Zp when p is a tough prime is different.  With tough primes every odd power of 2 mod p has order p-1 and every even power of 2 mod p has order phi(p)/2.

Conjecture: If p is a steady prime, then phi(p)/2 is also a prime number.

Example: The first few steady primes are: 7,23,47,167,263,359,383,479,503

phi(p)/2 when p is the first few steady primes is equal to: 3,11, 23, 83, 131, 179, 191, 239, 261

Unfortunately, the curious properties of steady primes are not applicable to all products of steady primes. There are some products of steady primes pq with such uniform structure of the exponentiation table of Zpq but they are kind of rare.

Below is a calculator that can be used to generate powers of powers of integers modulo other integers, the algorithm for which I described here.

Steady Primes

Another interesting subset of the set of safe primes is the set of steady primes as defined below:

Definition: A steady prime is a safe prime p such that all powers of 2 mod p have order phi(p)/2 where phi() is Euler's Totient Function.

Example:  Let p = 23.

Using a calculator, it is easy to see that all unique powers of 2 mod 23 are 2,4,8,16,9,18,13,3,6,12,1 

Using another calculator, it is also easy to verify that the order of 2 mod 23, or the smallest integer k for which 2^k = 1 mod 23 is 11, the order of 4 mod 23 is also 11,  and so is the order of 8 mod 23, and this is the case for every other power of 2 mod 23 in the list above.

The first steady primes are 7,23,47,167,263,359,383,479,503,...

Conjecture: Steady primes are primes of the form 8p-1 such that 4p-1 and 8p-1 are also primes.

Claim: Given two steady primes p and q, the order of every power of 2 mod p*q is phi(p*q)/4

Claim: For all other safe primes q that are not steady primes, the order of at least one power of 2 mod q is at least 2 times less than the order of 2 mod q itself.

Below is a calculator for finding powers of any integer a mod n such that a < n.

26.3.17

Other Collatz-like Algorithms

I wrote about the Collatz conjecture in relation to finite fields and I came across a few interesting sequences but I didn't discuss the possibility of other Collatz-like algorithms that always reach 1.

Once again, the Collatz problem is defined as follows:

Collatz Algorithm: Given any positive integer n, let a0 = n. For i > 0, let:

ai = ai-1/2 if ai-1 is even
ai = 3*ai-1 + 1 if ai-1 is odd

Conjecture: Given any positive integer, the above recurrence relation returns 1.

This conjecture seems to hold although it hasn't been proven yet.

While looking at a different problem related to generating powers of 2 backwards, I stumbled upon a Collatz-like algorithm, which I generalize below for all positive integers.

Collatz-like Algorithm #1: Given any positive integer n, let a0 = (n+1)/2 if n is odd or a0 = n/2 if n is even. For i > 0, let:

ai = ai-1/2 if ai-1 is even
ai = (ai-1 - 1)/2 +  a0 if ai-1 is odd

Conjecture: Given any positive integer n, the above recurrence relation returns 1, and all integers it reaches before it reaches 1 are strictly less than n

In fact, I claim that:

Claim: If n is even then the Collatz-like Algorithm #1 generates all powers of 2 mod (n-1) in reverse order and if n is odd then it generates all powers of 2 mod n in reverse order.

Example: let n = 36 so a0 = 18
Then:
a1 = 18/2 = 9
a2 = ((9-1)/2) + 18 = 22
a3= 22/2 = 11
a4 = ((11-1)/2) + 18 = 23
a5 = ((23-1)/2) + 18 = 29
a6 = ((29-1)/2) + 18 = 32
a7 = 32/2 = 16
a8 = 16//2 = 8
a9 = 8/2 = 4
a10 = 4/2 = 2
a11 = 2/2 = 1