1. Public Key Encryption Engineering & Analysis Operation-Part2
James C. Bradas, Ph.D.
18 June 2009

2. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( n , e )
Public Key
5. Compute d such that
( n , d )
Private Key

3. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( n , e )
Public Key
5. Compute d such that
( n , d )
Private Key
For this example, we’ll use small prime numbers –
the principal is exactly the same….

4. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
let p = 11, q = 3
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( n , e )
Public Key
5. Compute d such that
( n , d )
Private Key

5. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
let p = 11, q = 3
n = 11 x 3 = 33
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( 33 , e )
Public Key
5. Compute d such that
( 33 , d )
Private Key

6. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
let p = 11, q = 3
n = 11 x 3 = 33
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( 33 , e )
Public Key
5. Compute d such that
( 33 , d )
Private Key

7. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
let p = 11, q = 3
n = 11 x 3 = 33
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( 33 , 3 )
Public Key
5. Compute d such that
( 33 , d )
Private Key

8. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
let p = 11, q = 3
n = 11 x 3 = 33
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( 33 , 3 )
Public Key
5. Compute d such that
( 33 , d )
Private Key
Check

9. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
let p = 11, q = 3
n = 11 x 3 = 33
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( 33 , 3 )
Public Key
5. Compute d such that
( 33 , 7 )
Private Key
Check

10. Let’s Try An Example to See How This Works
RSA Public Key Encryption Scheme
Alice
let p = 11, q = 3
n = 11 x 3 = 33
1. Select two large prime numbers, p & q
2. Compute their product – the “modulus” n:
3. Compute the “totient” F
Public Key
4. Choose e, 1 < e < F
such that greatest common divisor (gcd) ( e , F ) = 1
e is the “public key exponent”
( Common choices are e = 3, & )
Private Key
( 33 , 3 )
Public Key
5. Compute d such that
( 33 , 7 )
Private Key
Check

11. Suppose Bob Wants to Send Alice the Letter “Z”
Let’s say that in the agreed-upon reversible padding scheme, “Z” equals the number 7.
Therefore, we want to encrypt m = 7 n e
( 33 , 3 )
Public Key
Recall
is a solution to

12. So….
Check

13. After Computing
Alice
Bob
“Z”
Bob sends c = 13 to Alice

14. Alice Receives “13”
Alice
Bob
( 33 , 7 )
Private Key n d

15. Reversible Padding Scheme
Alice Receives “13”
Alice
Bob
( 33 , 7 )
Private Key n d
Reversible
Padding
Scheme

16. “Z”
Alice
Bob
( 33 , 7 )
Private Key n d
Reversible
Padding
Scheme
“Z”

17. How does this Work? So What’s Going On Here? M m Encrypt Alice Bob
Bob’s Message
Public Key
Private Key
Decrypt
How does this Work?

18. Some more properties we need to knowIf
and
Then
let

19. … and more properties or
Fermat’s Little Theorem: If p is a prime number, then for ANY integer a,
will be evenly divisible by p. or
Euler’s Theorem (An Extension of Fermat’s Little Theorem)
gcd(a,n)=1
Φ(n) = (p-1)(q-1)
is Euler’s “Totient”

20. Here’s What I Want to Prove
Given: 1. 2. 3.
If:
I can recover m via
Then:

21. Here’s The Details
Start With:
Let’s raise c to the d power and use

22. Here’s The Details

23. Here’s The Details

24. Here’s The Details

25. Here’s The Details
Now, recall that
Euler’s Theorem

26. Here’s The Details
(cont’d)

27. Here’s The Details
(cont’d)
Which Can Be Written

28. Here’s The Details
(cont’d)
Which is What I Wanted to Prove

29. Here’s The Details (cont’d)
Which is What I Wanted to Prove
This is because:
Except for the change in sign, the two terms are equivalent.

30. So, Where Does RSA Encryption Stand?
For now, RSA PKE is still secure
In 1991, RSA Laboratories published 54 large semiprimes (numbers
with exactly two prime factors) and issued cash prizes for successful
factorization.
According to Wikepedia, 12 of the 54 listed numbers had been factored
by March 2008
The RSA challenge officially ended in 2007
Fastest Published Integer Factorization Algorithms:
General Number Field Sieve
Quadratic Sieve
Development of a large Q-Bit Quantum Computer MIGHT make RSA
vulnerable, although this is not certain
Fundamental breakthroughs in Number Theory (such as solving the
Riemann Hypothesis) still required before RSA becomes vulnerable

31. "The whole of e-commerce depends on prime numbers
"The whole of e-commerce depends on prime numbers. I have described the primes as atoms: what mathematicians are missing is a kind of mathematical prime spectrometer. Chemists have a machine that, if you give it a molecule, will tell you the atoms that it is built from. Mathematicians haven't invented a mathematical version of this. That is what we are after. If the Riemann hypothesis is true, it won't produce a prime number spectrometer. But the proof should give us more understanding of how the primes work, and therefore the proof might be translated into something that might produce this prime spectrometer. If it does, it will bring the whole of e-commerce to its knees, overnight. So there are very big implications." - Marcus du Sautoy (“The Music of the Primes”)

32. Questions