Public Key Cryptography 2 RSA

Lemma 1 Let s and t be relatively prime. Then Proof: Let be given by First we show that  actually maps Then we show  is an isomorphism.

Example Let s = 8, t = 15, so that st = 120.  (83) = (83 mod 8, 83 mod 15) = (3,8)  (29) = (29 mod 8, 29 mod 15) = (5,14)  (8329) =  (7) = (7,7) = (35 mod 8, 814 mod 15) =  (83)  (29)

Choose any x in U(st). Then gcd(x,st) = 1. There exist integers a, b with ax + bst = 1. Then 1 is a linear combination of x and s, so gcd(x,s) =1. Hence x mod s is in U(s). Similarly x mod t is in U(t).

 is one-to-one Suppose  (x) =  (y) where 0 ≤ x ≤ y < st. Then (x mod s,x mod t) = (y mod s,y mod t) So x mod s = y mod s and x mod t = y mod t Hence s and t both divide y–x. But s, t are relatively prime, so st divides y–x as well. Also 0 ≤ y–x < st, so y–x = 0. It follows that  is one-to-one.

 is onto Choose any (x s,x t ) in There exist integers a, b with as + bt = 1. Let x = (btx s + asx t ) mod st. In moment, we will show that x is in U(st). Then x = btx s + asx t + stn for some n. So x mod s = (1x s + 0x t + 0n) mod s = x s x mod t = (0x s + 1x t + 0n) mod t = x t So  (x) = (x s, x t ), and  is onto. gcd(x,st) =1

Example: The inverse of   (x) = (x mod 8, x mod 15) Suppose  (x) = (3,8). Find x. First write 28+(-1)15 = 1 Then x = (-115)(3) + (28)(8) = = 83

To show gcd(x,st) = 1: Given x s in U(s), x t in U(t), x = (btx s + asx t ) where as+bt = 1. Set y = (btx s -1 + asx t -1 ). Now xy = (btx s + asx t )(btx s -1 + asx t -1 ), so xy mod s = (1x s + 0)(1 x s ) mod s = 1. xy mod t = (0 + 1x t )(0 + 1x t -1 ) mod t = 1. Now s | xy–1, t |xy–1, and gcd(s,t)=1 implies st | xy–1, so xy mod st = 1. Hence x and st are relatively prime.

 is Operation Preserving  (x)  (y) = (x mod s,x mod t)(y mod s,y mod t) = (xy mod s,xy mod t) =  (xy) Since  is one-to-one, onto, and operation preserving,  is an isomorphism. Therefore,

Theorem: (Gauss) Let p be an odd prime, n > 0. Corollary 1. For odd prime p, Corollary 2. Let p and q be odd primes. Proof:

RSA Recipe Choose (large) odd primes p,q Let N = pq, m = lcm(p-1,q-1) Choose E relatively prime to m Let D = E -1 in U(m) To encode message M: C = M E mod N To decode message C: M = C D mod N Public Key is E, N Private Key is D, N

Will RSA work? M = lcm(p-1,q-1) = h(p-1) = k(q-1) for some integers h, k. ED + sM = 1 for some integer s. So, ED mod (p-1) = ED mod (q-1) = 1 Also, isomorphism Let. Then.

Will RSA work? M = lcm(p-1,q-1) = h(p-1) = k(q-1) for some integers h, k. We claim Let be an isomorphism. Say. Then So as required. Operation Preserving One-to-One

Encoding, Decoding are inverses Recall that E and D are inverses mod M. So ED = 1+sM for some integer s. Let x in U(N) be a message. In U(N), y = x E is the encrypted message. The decrypted message is z = y D = x ED = x 1+sM = x(x M ) s = x RSA works!

How to break RSA Everyone is given E, N. Factor N into pq  Note p and q are large. Let M = lcm(p-1,q-1) = (p-1)(q-1) gcd(p-1,q-1) Let D = E -1 mod M Euclidean Algorithm