Chapter 12: Cryptography MAT 320 Spring 2008

Cryptography: Basic Ideas We want to encode information so that no one other than the intended recipient can decode it. Essentially we have two functions: an encoding function E, and a decoding function, D.

Two Functions If x is a message (and it’s not hard to express messages as numbers), then E(x) should be the encoded message Once the message is received, D(E(x)) is the original message x So E and D are inverse functions

Options One option is to keep both functions secret The advantage of this method is that if unintended recipients do not know your functions, they should be unable to decode your message The disadvantage of this method is that the more people who know your functions, the less able you will be to keep them secret

Personal Keys In a more advanced system, each person has their own functions. Anne has her functions E Anne and D Anne Bob has his functions E Bob and D Bob The problem with this method is that so far, Anne and Bob can only send messages to themselves, not to each other

Public Keys To fix this problem, we make everyone’s E function public. So anyone can encode a message using anyone else’s key. However, we need to keep the D functions private, or else our information could be stolen.

How It Works Anne wants to send a message to Bob. Anne knows E Bob, so she sends E Bob (message) to Bob Only Bob knows D Bob, so only Bob can compute D Bob (E Bob (message)) = message

Signatures Another advantage of a public-key is the ability to “sign” messages. Suppose that the bank receives an encoded message claiming to be from Anne. Anyone can send E Bank (message) to the Bank. But only Anne can send E Bank (D Anne (message)) The bank knows D Bank and E Anne, so they can decode the message by applying these functions

Troubles The main problem that comes up with public-key cryptography is that we need to make sure that it’s very difficult to figure out how the D function works from knowing how the E function works. One method that accomplishes this is RSA cryptography.

Intermission Lemma 12.1 (Limited Cancelling) Lemma 12.2 (Fermat’s Little Theorem)

How RSA Works Let p and q be distinct primes. Let n = pq. In practice, we let p and q be quite large, with hundreds of digits. It is difficult to factor large numbers, even by computer, and if someone were able to factor n, they would be able to break our code. Go ahead and choose primes now. For purposes that will become clear soon, make sure that n is at least 270,000. Again, in practice, n is much, much larger.

How RSA Works, continued Let k = (p – 1)(q – 1), and choose d so that (d, k) = 1. Using Bezout’s Theorem, find e so that de  1 (mod k) The numbers e and n are made public, and the number d is kept private. The encoding function is E(x) = x e mod n The decoding function is D(x) = x d mod n

Let’s Try It Following the example on your handout, enter your numbers p, q, n, and k into Mathematica. Choose a number d so that (d, k) = 1. You may have to try a few times to get a d that works. Once you find a value of d, use the ExtendedGCD command to find e so that ed  1 (mod k). If Mathematica gives you a negative value of e, add k to it (since we’re working mod k, this will be congruent)

Sending Messages First we need to convert our message to numbers. Converting the entire message to a single number would make our calculations difficult, so instead we break it up into blocks. Using A = 01, B = 02, …, Z = 26, break your message into 3-letter blocks and convert them to numbers. Now you see why we needed to have n be at least 270,000. Add extra zeros to the end of your message if it doesn’t break up evenly into three-letter blocks.

Encoding Now we are ready to plug these numbers into our encoding function. When x and d are large, computing x d, dividing it by n, and computing the remainder is very time consuming. However, there are many computational shortcuts Mathematica can use, including the PowerMod command. PowerMod[x,d,n] computes x d mod n

One More Proof In order to convince ourselves that RSA always works, we need to prove this theorem: Theorem 12.3 (RSA Works!) Let p and q be distinct primes, and let n = pq and k = (p – 1)(q – 1). If d and e are chosen so that (d, k) = 1 and ed  1 (mod k), then for all integers x, x ed  x (mod n).