Applications of Number Theory CS 202 Epp section

Applications of Number Theory CS 202 Epp section 10. 4 Aaron Bloomfield 1

About this lecture set • I want to introduce RSA – The most commonly used cryptographic algorithm today • Much of the underlying theory we will not be able to get to – It’s beyond the scope of this course • Much of why this all works won’t be taught – It’s just an introduction to how it works 2

Private key cryptography • The function and/or key to encrypt/decrypt is a secret – (Hopefully) only known to the sender and recipient • The same key encrypts and decrypts • How do you get the key to the recipient? 3

Public key cryptography • Everybody has a key that encrypts and a separate key that decrypts – They are not interchangable! • The encryption key is made public • The decryption key is kept private 4

Public key cryptography goals • Key generation should be relatively easy • Encryption should be easy • Decryption should be easy – With the right key! • Cracking should be very hard 5

Is that number prime? • Use the Fermat primality test • Given: – n: the number to test for primality – k: the number of times to test (the certainty) • The algorithm is: repeat k times: pick a randomly in the range [1, n− 1] if an− 1 mod n ≠ 1 then return composite return probably prime 6

Is that number prime? • The algorithm is: repeat k times: pick a randomly in the range [1, n− 1] if an− 1 mod n ≠ 1 then return composite return probably prime • Let n = 105 – Iteration 1: a = 92: 92104 mod 105 = 1 – Iteration 2: a = 84: 84104 mod 105 = 21 – Therefore, 105 is composite 7

Is that number prime? • The algorithm is: repeat k times: pick a randomly in the range [1, n− 1] if an− 1 mod n ≠ 1 then return composite return probably prime • Let n = 101 – – – Iteration 1: a = 55: 55100 mod 101 = 1 Iteration 2: a = 60: 60100 mod 101 = 1 Iteration 3: a = 14: 14100 mod 101 = 1 Iteration 4: a = 73: 73100 mod 101 = 1 At this point, 101 has a (½)4 = 1/16 chance of still 8 being composite

More on the Fermat primality test • Each iteration halves the probability that the number is a composite – Probability = (½)k – If k = 100, probability it’s a composite is (½)100 = 1 in 1. 2 1030 that the number is composite • Greater chance of having a hardware error! – Thus, k = 100 is a good value • However, this is not certain! – There are known numbers that are composite but will always report prime by this test • Source: http: //en. wikipedia. org/wiki/Fermat_primality_test 9

Google’s recruitment campaign 10

RSA • Stands for the inventors: Ron Rivest, Adi Shamir and Len Adleman • Three parts: – Key generation – Encrypting a message – Decrypting a message 11

Key generation steps 1. Choose two random large prime numbers p ≠ q, and n = p*q 2. Choose an integer 1 < e < n which is relatively prime to (p-1)(q-1) 3. Compute d such that d * e ≡ 1 (mod (p-1)(q-1)) – Rephrased: d*e mod (p-1)(q-1) = 1 4. Destroy all records of p and q 12

Key generation, step 1 • Choose two random large prime numbers p ≠ q – In reality, 2048 bit numbers are recommended • That’s 617 digits – From last lecture: chance of a random odd 2048 bit number being prime is about 1/710 • • • We can compute if a number is prime relatively quickly via the Fermat primality test We choose p = 107 and q = 97 Compute n = p*q – n = 10379 13

Key generation, step 1 • Java code to find a big prime number: Big. Integer prime = new Big. Integer (num. Bits, certainty, random); The number of bits of the prime Certainty that the number is a prime The random number generator 14

Key generation, step 1 • Java code to find a big prime number: import java. math. *; import java. util. *; class Big. Prime { static int num. Digits = 617; static int certainty = 100; static final double LOG_2 = Math. log(10)/Math. log(2); static int num. Bits = (int) (num. Digits * LOG_2); public static void main (String args[]) { Random random = new Random(); Big. Integer prime = new Big. Integer (num. Bits, certainty, random); System. out. println (prime); } } 15

Key generation, step 1 • How long does this take? – Keep in mind this is Java! – These tests done on a 850 Mhz Pentium machine – Average of 100 trials (certainty = 100) – 200 digits (664 bits): about 1. 5 seconds – 617 digits (2048 bits): about 75 seconds 16

Key generation, step 1 • Practical considerations – p and q should not be too close together – (p-1) and (q-1) should not have small prime factors – Use a good random number generator 18

Key generation, step 2 • Choose an integer 1 < e < n which is relatively prime to (p-1)(q-1) • There algorithms to do this efficiently – We aren’t going over them in this course • Easy way to do this: make e be a prime number – It only has to be relatively prime to (p-1)(q-1), but can be fully prime 19

Key generation, step 2 • Recall that p = 107 and q = 97 – (p-1)(q-1) = 106*96 = 10176 = 26*3*53 • We choose e = 85 – 85 = 5*17 – gcd (85, 10176) = 1 – Thus, 85 and 10176 are relatively prime 20

Key generation, step 3 • Compute d such that: d * e ≡ 1 (mod (p-1)(q-1)) – Rephrased: d*e mod (p-1)(q-1) = 1 • There algorithms to do this efficiently – We aren’t going over them in this course • We choose d = 4669 – 4669*85 mod 10176 = 1 • Use the script at http: //www. cs. virginia. edu/cgi-bin/cgiwrap/asb/modpow 21

Key generation, step 3 • Java code to find d: import java. math. *; class Find. D { public static void main (String args[]) { Big. Integer pq = new Big. Integer("10176"); Big. Integer e = new Big. Integer ("85"); System. out. println (e. mod. Inverse(pq)); } } • Result: 4669 22

Key generation, step 4 • Destroy all records of p and q • If we know p and q, then we can compute the private encryption key from the public decryption key d * e ≡ 1 (mod (p-1)(q-1)) 23

The keys • We have n = p*q = 10379, e = 85, and d = 4669 • The public key is (n, e) = (10379, 85) • The private key is (n, d) = (10379, 4669) • Thus, n is not private – Only d is private • In reality, d and e are 600 (or so) digit numbers – Thus n is a 1200 (or so) digit number 24

Encrypting messages • To encode a message: 1. Encode the message m into a number 2. Split the number into smaller numbers m < n 3. Use the formula c = me mod n • • c is the ciphertext, and m is the message Java code to do the last step: – – m. mod. Pow (e, n) Where the object m is the Big. Integer to encrypt 25

Encrypting messages example 1. Encode the message into a number – – String is “Go Cavaliers!!” Modified ASCII codes: • 2. 41 81 02 37 67 88 67 78 75 71 84 85 03 03 Split the number into numbers < n – – 3. Recall that n = 10379 4181 0237 6788 6778 7571 8485 0303 Use the formula c = me mod n – – • 418185 mod 10379 = 4501 023785 mod 10379 = 2867 678885 mod 10379 = 4894 Etc… Encrypted message: – 4501 2867 4894 0361 3630 4496 6720 26

Encrypting RSA messages n Formula is c = me mod n 27

Decrypting messages 1. Use the formula m = cd mod n on each number 2. Split the number character numbers into individual ASCII 3. Decode the message into a string 28

Decrypting messages example • Encrypted message: – 1. 4501 2867 4894 0361 3630 4496 6720 Use the formula m = cd mod n on each number – – 2. 45014669 mod 10379 = 4181 28674669 mod 10379 = 0237 48944669 mod 10379 = 6788 Etc… Split the numbers into individual characters – 3. 41 81 02 37 67 88 67 78 75 71 84 85 03 03 Decode the message into a string – Modified ASCII codes: • – 41 81 02 37 67 88 67 78 75 71 84 85 03 03 Retrieved String is “Go Cavaliers!!” 29

mod. Pow computation 1. How to compute c = me mod n or m = cd mod n? – • Example: 45014669 mod 10379 = 4181 Use the script at http: //www. cs. virginia. edu/cgi-bin/cgiwrap/asb/modpow • Other means: – – – Java: use the Big. Integer. mod. Pow() method Perl: use the bmodpow function in the Big. Int library Etc… 30

Why this works • m = cd mod n • c = me mod n • cd ≡ (me)d ≡ med (mod n) • Recall that: – ed ≡ 1 (mod p-1) – ed ≡ 1 (mod q-1) • Thus, – med ≡ m (mod p) – med ≡ m (mod q) • med ≡ m (mod pq) • med ≡ m (mod n) 31

Cracking a message • In order to decrypt a message, we must compute m = cd mod n – n is known (part of the public key) – c is known (the ciphertext) – e is known (the encryption key) • Thus, we must compute d with no other information – Recall: choose an integer 1 < e < n which is relatively prime to (p-1)(q-1) – Recall: Compute d such that: d*e mod (p-1)(q-1) = 1 • Thus, we must factor the composite n into it’s component primes – There is no efficient way to do this! – We can, very easily, tell that n is composite, but we can’t tell what its factors are • Once n is factored into p and q, we compute d as above – Then we can decrypt c to obtain m 32

Cracking a message example • In order to decrypt a message, we must compute m = cd mod n – n = 10379 – c is the ciphertext being cracked – e = 85 • In order to determine d, we need to factor n – – • d*e mod (p-1)(q-1) = 1 We factor n into p and q: 97 and 107 This would not have been feasible with two large prime factors!!! d * 85 (mod (96)(106)) = 1 We then compute d as above, and crack the message 33

Signing a message • Recall that we computed: d*e mod (p-1)(q-1) = 1 • Note that d and e are interchangable! – You can use either for the encryption key • You can encrypt with either key! – Thus, you must use the other key to decrypt 34

Signing a message • To “sign” a message: 1. Write a message, and determine the MD 5 hash 2. Encrypt the hash with your private (encryption) key 3. Anybody can verify that you created the message because ONLY the public (encryption) key can decrypt the hash 4. The hash is then verified against the message 35

PGP and Gnu. PG • Two applications which implement the RSA algorithm – Gnu. PG Is open-source (thus it’s free) – PGP was first, and written by Phil Zimmerman • The US gov’t didn’t like PGP… 36

The US gov’t and war munitions 37

How to “crack” PGP • Factoring n is not feasible • Thus, “cracking” PGP is done by other means – Intercepting the private key • “Hacking” into the computer, stealing the computer, etc. – Man-in-the-middle attack (next 2 slides) – Etc. 38

Man-in-the-middle attack: “Normal” RSA communication What is your public key? My public key is 12345… What is your public key? My public key is 67890… Here’s message encrypted with 12345… Here’s a response encrypted with 67890… 39

What is your public key? My public key is abcde… What Blackishas yourthe public key? private decryption My key is 67890… keypublic for abcde… What is your public key? My public key is 12345… What is your public key? My public key is vwxyz… Here’s message encrypted w/ abcde… Black has the Decrypts message with correspondingprivatedecryption key to abcde…; key for vwxyz… re-encrypts message with blue’s public key (12345…) Here’s message encrypted w/ 12345… Here’s response encrypted w /vwxyz… Decrypts message with corresponding private key to vwxyz…; re-encrypts message with yellow’s public key (67890…) Here’s response encrypted w/ 67890… 40

Other public key encryption methods • Modular logarithms – Developed by the US government, therefore not widely trusted • Elliptic curves 41

Quantum computers • A quantum computer could (in principle) factor n in reasonable time – This would make RSA obsolete! – Shown (in principle) by Peter Shor in 1993 – You would need a new (quantum) encryption algorithm to encrypt your messages • This is like saying, “in principle, you could program a computer to correctly predict the weather” • A few years ago, IBM created a quantum computer that successfully factored 15 into 3 and 5 • I bet the NSA is working on such a computer, also 42

Sources • Wikipedia article has a lot of info on RSA and the related algorithms – Those articles use different variable names – Link at http: //en. wikipedia. org/wiki/RSA 43