Applied Symbolic Computation CS 300 Modular Arithmetic and

Applied Symbolic Computation (CS 300) Modular Arithmetic and the RSA Public Key Cryptosystem Jeremy R. Johnson 1

Introduction • Objective: To understand what a public key cryptosystem is and how the RSA algorithm works. To review the number theory behind the RSA algorithm. – – – Public Key Cryptosystems RSA Algorithm Modular Arithmetic Euler’s Identity Chinese Remainder Theorem References: Rivest, Shamir, Adelman. 2

Modular Arithmetic (Zn) Definition: a b (mod n) n | (b - a) Alternatively, a = qn + b Properties (equivalence relation) – a a (mod n) [Reflexive] – a b (mod n) b a (mod n) [Symmetric] – a b (mod n) and b c (mod n) a c (mod n) [Transitive] Definition: An equivalence class mod n [a] = { x: x a (mod n)} = { a + qn | q Z} 3

Modular Arithmetic (Zn) It is possible to perform arithmetic with equivalence classes mod n. – [a] + [b] = [a+b] – [a] * [b] = [a*b] In order for this to make sense, you must get the same answer (equivalence) class independent of the choice of a and b. In other words, if you replace a and b by numbers equivalent to a or b mod n you end of with the sum/product being in the same equivalence class. a 1 a 2 (mod n) and b 1 b 2 (mod n) a 1+ b 1 a 2 + b 2 (mod n) a 1* b 1 a 2 * b 2 (mod n) (a + q 1 n) + (b + q 2 n) = a + b + (q 1 + q 2)n (a + q 1 n) * (b + q 2 n) = a * b + (b*q 1 + a*q 2 + q 1* q 2)n 4

Representation of Zn The equivalence classes [a] mod n, are typically represented by the representatives a. • Positive Representation: Choose the smallest positive integer in the class [a] then the representation is {0, 1, …, n-1}. • Symmetric Representation: Choose the integer with the smallest absolute value in the class [a]. The representation is {- (n-1)/2 , …, n/2 }. When n is even, choose the positive representative with absolute value n/2. • E. G. Z 6 = {-2, -1, 0, 1, 2, 3}, Z 5 = {-2, -1, 0, 1, 2} 5

Modular Inverses Definition: x is the inverse of a mod n, if ax 1 (mod n) The equation ax 1 (mod n) has a solutioniff gcd(a, n) = 1. By the Extended Euclidean Algorithm, there exist x and y such that ax + ny = gcd(a, n). When gcd(a, n) = 1, we get ax + ny = 1. Taking this equation mod n, we see thatax 1 (mod n) By taking the equation mod n, we mean applying the mod n homomorphism: m Z Zm, which maps the integer a to the equivalence class [a]. This mapping preserves sums and products. I. E. m(a+b) = m(a) + m(b), m(a*b) = m(a) * m(b) 6

Fermat’s Theorem: If a 0 Zp, then ap-1 1 (mod p). More generally, if a Zp, then ap a (mod p). Proof: Assume thata 0 Zp. Then a * 2 a * … (p-1)a = (p-1)! * ap-1 Also, since a*i a*j (mod p) i j (mod p), the numbers a, 2 a, …, (p-1)a are distinct elements of Zp. Therefore they are equal to 1, 2, …, (p-1) and their product is equal to (p-1)! mod p. This implies that (p-1)! * ap-1 (p-1)! (mod p) ap-1 1 (mod p). 7

Euler phi function • Definition: phi(n) = #{a: 0 < a < n andgcd(a, n) = 1} • Properties: – – (p) = p-1, for prime p. (p^e) = (p-1)*p^(e-1) (m*n) = (m)* (n) for gcd(m, n) = 1. (p*q) = (p-1)*(q-1) • Examples: – (15) = (3)* (5) = 2*4 = 8. = #{1, 2, 4, 7, 8, 11, 13, 14} – (9) = (3 -1)*3^(2 -1) = 2*3 = 6 = #{1, 2, 4, 5, 7, 8} 8

Euler’s Identity • The number of elements in Zn that have multiplicative inverses is equal to phi(n). • Theorem: Let (Zn)* be the elements of Zn with inverses (called units). If a (Z n)*, then a (n) 1 (mod n). Proof. The same proof presented for. Fermat’s theorem can be used to prove this theorem. 9

Chinese Remainder Theorem: If gcd(m, n) = 1, then given a and b there exist an integer solution to the system: x a (mod m) and x = b (mod n). Proof: Consider the map x (x mod m, x mod n). This map is a 1 -1 map from. Zmn to Zm Zn, since if x and y map to the same pair, then x y (mod m) and x y (mod n). Since gcd(m, n) = 1, this implies thatx y (mod mn). Since there are mn elements in both. Zmn and Zm Zn, the map is also onto. This means that for every pair (a, b) we can find the desired x. 10

Alternative Interpretation of CRT • Let Zm Zn denote the set of pairs (a, b) where a Zm and b Zn. We can perform arithmetic on. Zm Zn by performing componentwisemodular arithmetic. – (a, b) + (c, d) = (a+b, c+d) – (a, b)*(c, d) = (a*c, b*d) • Theorem: Zmn Zm Zn. I. E. There is a 1 -1 mapping from Zmn onto Zm Zn that preserves arithmetic. – (a*c mod m, b*d mod n) = (a mod m, b mod n)*(c mod m, d mod n) – (a+c mod m, b+d mod n) = (a mod m, b mod n)+(c mod m, d mod n) – The CRT implies that the map is onto. I. E. for every pair (a, b) there is an integer x such that (x mod m, x mod n) = (a, b). 11

Constructive Chinese Remainder Theorem: If gcd(m, n) = 1, then there exist em and en (orthogonal idempotents) – – em en 1 (mod m) 0 (mod n) 0 (mod m) 1 (mod n) It follows that a*em + b* en a (mod m) and b (mod n). Proof. Since gcd(m, n) = 1, by the Extended Euclidean Algorithm, there exist x and y with m*x + n*y = 1. Setem = n*y and en = m*x 12

Public Key Cryptosystem Let M be a message and let C be the encrypted message (ciphertext). A public key cryptosystem has a separate method E() for encrypting and D() decrypting. • D(E(M)) = M • Both E() and D() are easy to compute • Publicly revealing E() does not make it easy to determine D() • E(D(M)) = M - needed for signatures The collection of E()’s are made publicly available but the D()’s remain secret. Called a one-way trap-door function(hard to invert, but easy if you have the secret information) 13

RSA Public Key Cryptosystem Based on the idea that it is hard to factor large numbers. First encode M as an integer (e. g. use ASCII). Large messages will need to be blocked. • Choose n = p*q, the product of two large prime numbers. • Choose e such thatgcd(e, phi(n)) = 1. • Choose d such that de 1 (mod (n)) • E = (e, n) and E(M) = M e mod n • D = (d, n) and D(M) = Md mod n 14

Correctness of the RSA Algorithm Theorem: D(E(M)) = E(D(M)) = M. Proof. D(E(M)) = (Me)d (mod n) = Med (mod n). Since ed 1 (mod (n)), ed = k* (n) + 1, for some integer k. Mk* (n)+1 (M k* (n)+1 mod p, Mk* (n)+1 mod q) = (M k* (n) * M mod p, Mk* (n) * M mod q) = (M (p-1)*(q-1)*k * M mod p, M(q-1)*(p-1)*k * M mod q) [since n = pq] = ((M (p-1))(q-1)*k * M mod p, (M(q-1))(p-1)*k * M mod q) = (M mod p, M mod q) [By. Fermat’s theorem] Therefore, by the CRT, Mk* (n)+1 M (mod n). 15