Elgamal Public Key Encryption CSCI 5857: Encoding and Encryption

Outline Background and uses of Elgamal Primitive roots Public and private key generation in ElGamal ElGamal encryption and decryption Mathematical justification Implementation as a block cipher

Elgamal Public Key Encryption Structure similar to RSA (discrete logs) Component of many secure systems –Digital Signature Standard (DSS) –S/MIME ( )

Components of Elgamal Components: –Large prime q –  = some primitive root of q  is primitive root of q if –For all 0 < p < q there exists some n < q such that  n mod q = p That is, powers of  “generate” all integers mod q –Necessary to make sure encryption has unique inverse, as this insures that (  n mod q) ≠ (  m mod q) for n ≠ m

Primitive Roots Example: q = 19 Only primitive roots:

Public and Private Keys Example: q = 19,  = 10 Generate random integer X A –1 < X A < q-1 X A = 5 Compute Y A =  X A mod q Y A = 3 Public key: {q, , Y A } Private key: {X A } Security similar to RSA –Must be able to solve modular logarithm X A = log α Y A to crack

Encryption Sender creates “one time key” for encryption –Message encrypted using modular exponentiation on one time key and public keys –Key “encrypted” using modular exponentiation and public keys –Recipient “decrypts” one time key using their private key –Recipient then decrypts message using that key 7

Encryption Sender generates random integer k < q Sender computes one-time key K = (Y A ) k mod q Message M encrypted as two integers (C 1, C 2 ) C 1 = α k mod q C 2 = KM mod q Example: Plaintext message M = 17 Choose random k = 6 K = 3 6 mod 19 = 7 C 1 = 10 6 mod 19 = 11 C 2 = 7 x 17 mod 19 = 5 8

Decryption Recipient recovers K = C 1 X A mod q Example: K = 11 5 mod 19 = 7 Recipient then recovers M = (C 2 K -1 ) mod q –K -1 is inverse of K mod q Example: 7 -1 mod 19 = 11 (77 = 4 x ) M = 5 x 11 mod 19 = 17 9

Why Does This Work? K = (Y A ) k mod qDefinition of K = (  X A mod q) k mod qDefinition of Y A = (  X A k ) mod q Rules of modular exponentiation = (  k X A ) mod q = (C 1 X A ) mod qDefinition of C 1 10

Why Does This Work? M = (C 2 K -1 ) mod q = (KM mod q K -1 ) mod q Definition of C 2 = (M K K -1 ) mod q Rules of modular exponentiation = (M) mod q = MIf M < q 11

Implementation Details If M > q must break into smaller blocks M 1, M 2, M 3 … < q Must use different k for each block –Otherwise single known M i allows all blocks to be decrypted by factoring out K C 21 = KM 1 mod q = M 1 mod q = M 1 C 22 KM 2 mod q M 2 mod q M 2 If M 1 known then M 2 = C C 22 M 1 mod q 12