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

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, 17 & 65537 ) Private Key ( n , e ) Public Key 5. Compute d such that ( n , d ) Private Key

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, 17 & 65537 ) 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….

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, 17 & 65537 ) Private Key ( n , e ) Public Key 5. Compute d such that ( n , d ) Private Key

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, 17 & 65537 ) Private Key ( 33 , e ) Public Key 5. Compute d such that ( 33 , d ) Private Key

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, 17 & 65537 ) Private Key ( 33 , e ) Public Key 5. Compute d such that ( 33 , d ) Private Key

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, 17 & 65537 ) Private Key ( 33 , 3 ) Public Key 5. Compute d such that ( 33 , d ) Private Key

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, 17 & 65537 ) Private Key ( 33 , 3 ) Public Key 5. Compute d such that ( 33 , d ) Private Key Check

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, 17 & 65537 ) Private Key ( 33 , 3 ) Public Key 5. Compute d such that ( 33 , 7 ) Private Key Check

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, 17 & 65537 ) Private Key ( 33 , 3 ) Public Key 5. Compute d such that ( 33 , 7 ) Private Key Check

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

So…. Check

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

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

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

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

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?

Some more properties we need to know If and Then let

… 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”

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

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

Here’s The Details

Here’s The Details

Here’s The Details

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

Here’s The Details (cont’d)

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

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

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.

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

"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”)

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