Simplest explanation of the math behind Public Key Cryptography (2013)

I was trying to explain public key cryptography today and totally failed at it. Here are notes to myself based on various Wikipedia pages. There's a lot more to it than this (like padding ) but this is the gist of it.

1. Pick two prime numbers:

p = 7
q = 13

2. Multiply them together:

n = p * q
n = 7 * 13
n = 91

3. Compute Euler's " totient " function of n :

φ(n) = (p - 1) * (q - 1)
φ(91) = (7 - 1) * (13 - 1)
φ(91) = 6 * 12
φ(91) = 72

4. Pick a random number e such that:

1 < e < φ(n)
1 < e < φ(91)
1 < e < 72

and e is " coprime " with φ(n), meaning it has no common factors.

e = 23  (because I said so)

5. Compute d , the modular multiplicative inverse of e (mod φ(n)):

e^-1 = d (mod φ(n))
23^-1 = d (mod φ(91))
23^-1 = d (mod 72)
23 * d = 1 (mod 72)
23 * 47 = 1 (mod 72)
d = 47

Now you have all the magic numbers you need:

public key  = (n = 91, e = 23)
private key = (n = 91, d = 47)

Secret messages to you

1. Have someone else encrypt a message m using your public key (n, e):

m = 60
c(m) = m^e mod n
c(60) = 60^23 mod 91
c(60) = 44
c = 44  (they send this to you)

2. Decrypt the message c using your private key (n, d):

c = 44
m(c) = c^d mod n
m(44) = 44^47 mod 91
m(44) = 60
m = 60  (now you have the secret)

Prove you wrote something

1. Hash the message to broadcast (this is Knuth's insecure hash function):

broadcast = 102
hash(broadcast) = broadcast * K mod 2^32
hash(102) = 102 * 2654435761 mod 2^32
hash(102) = 169507974

2. Compute the signature with your private key (n, d):

signature = hash(broadcast) ^ d mod n
signature = hash(102) ^ 47 mod 91
signature = 169507974 ^ 47 mod 91
signature = 90

3. People who receive the broadcast message (102) and the signature (90) can confirm with your public key (n, e):

broadcast = 102
hash(broadcast) = broadcast * K mod 2^32
hash(102) = 102 * 2654435761 mod 2^32
hash(102) = 169507974
confirmation = hash(broadcast)^e mod n
confirmation = 169507974^23 mod 91
confirmation = 90  (matches your signature)

Why is this safe?

In this simple example it's totally not. You can trivially take n = 91 and guess p and q with simple factorization :

>>> set((91 / i) for i in xrange(1, 91) if 91 % i == 0)
set([91, 13, 7])

Redo all the math above and you have the private key. Doh~

Now imagine if n was such a large number that it took a very long time to guess p and q :

>>> set((12031294782491844L / i) for i in xrange(1, 12031294782491844L) if 12031294782491844L % i == 0)

This takes a really long time. Even fancy solutions on the fastest computer on Earth would take until the end of the universe. That's why transmitting secrets with public key cryptography is safe. That's also why great leaps in prime number theory are frightening / exciting.