Discrete Mathematics with Applications MATH236 Dr. Hung P. - PowerPoint PPT Presentation

Digital signatures with RSA Digital Signature From Bob’s perspective: 1 Bob receives a message ( M || M ′ ) pub ( Bob ) from someone claiming to be Alice. He begins by using his private key pri ( Bob ) to remove the outer layer of encryption, recovering M || M ′ , which he separates into M and M ′ . 2 Bob now encrypts M ′ with Alice’s public key, pub(Alice), i.e., he finds pub ( Alice ) = ( M ′ ) e mod n and compare it with M , the first half of M ′ the concatenated message he received. There are two possibilities: pub ( Alice ) = M . Then Bob knows that M ′ was encrypted with Alice’s M ′ private key. This proves that the message M is from Alice pub ( Alice ) � = M . Therefore, either M ′ was not encrypted with Alice’s M ′ private key, Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 22

Digital signatures with RSA Digital Signature From Bob’s perspective: 1 Bob receives a message ( M || M ′ ) pub ( Bob ) from someone claiming to be Alice. He begins by using his private key pri ( Bob ) to remove the outer layer of encryption, recovering M || M ′ , which he separates into M and M ′ . 2 Bob now encrypts M ′ with Alice’s public key, pub(Alice), i.e., he finds pub ( Alice ) = ( M ′ ) e mod n and compare it with M , the first half of M ′ the concatenated message he received. There are two possibilities: pub ( Alice ) = M . Then Bob knows that M ′ was encrypted with Alice’s M ′ private key. This proves that the message M is from Alice pub ( Alice ) � = M . Therefore, either M ′ was not encrypted with Alice’s M ′ private key, or some malicious third party altered the text M after Alice added her signature; in either case, Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 22

Digital signatures with RSA Digital Signature From Bob’s perspective: 1 Bob receives a message ( M || M ′ ) pub ( Bob ) from someone claiming to be Alice. He begins by using his private key pri ( Bob ) to remove the outer layer of encryption, recovering M || M ′ , which he separates into M and M ′ . 2 Bob now encrypts M ′ with Alice’s public key, pub(Alice), i.e., he finds pub ( Alice ) = ( M ′ ) e mod n and compare it with M , the first half of M ′ the concatenated message he received. There are two possibilities: pub ( Alice ) = M . Then Bob knows that M ′ was encrypted with Alice’s M ′ private key. This proves that the message M is from Alice pub ( Alice ) � = M . Therefore, either M ′ was not encrypted with Alice’s M ′ private key, or some malicious third party altered the text M after Alice added her signature; in either case, Bob knows that the message was not authorised by Alice. Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 22

Digital signatures with RSA Digital Signature From Bob’s perspective: 1 Bob receives a message ( M || M ′ ) pub ( Bob ) from someone claiming to be Alice. He begins by using his private key pri ( Bob ) to remove the outer layer of encryption, recovering M || M ′ , which he separates into M and M ′ . 2 Bob now encrypts M ′ with Alice’s public key, pub(Alice), i.e., he finds pub ( Alice ) = ( M ′ ) e mod n and compare it with M , the first half of M ′ the concatenated message he received. There are two possibilities: pub ( Alice ) = M . Then Bob knows that M ′ was encrypted with Alice’s M ′ private key. This proves that the message M is from Alice pub ( Alice ) � = M . Therefore, either M ′ was not encrypted with Alice’s M ′ private key, or some malicious third party altered the text M after Alice added her signature; in either case, Bob knows that the message was not authorised by Alice. Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and pub(Bob)=(3233,19) and pri(Bob)=2299 Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and pub(Bob)=(3233,19) and pri(Bob)=2299 Alice encodes the message as M = 0715 Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and pub(Bob)=(3233,19) and pri(Bob)=2299 Alice encodes the message as M = 0715 She encrypts M by the rule M ′ = M 157 mod 2773 to produce the message signature M pri ( Alice ) = 2192 Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and pub(Bob)=(3233,19) and pri(Bob)=2299 Alice encodes the message as M = 0715 She encrypts M by the rule M ′ = M 157 mod 2773 to produce the message signature M pri ( Alice ) = 2192 So M || M ′ = 0715 2192 Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and pub(Bob)=(3233,19) and pri(Bob)=2299 Alice encodes the message as M = 0715 She encrypts M by the rule M ′ = M 157 mod 2773 to produce the message signature M pri ( Alice ) = 2192 So M || M ′ = 0715 2192 Alice now encrypts each block B with Bob’s public key, using the rule C = B 19 mod 3233 to obtain Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and pub(Bob)=(3233,19) and pri(Bob)=2299 Alice encodes the message as M = 0715 She encrypts M by the rule M ′ = M 157 mod 2773 to produce the message signature M pri ( Alice ) = 2192 So M || M ′ = 0715 2192 Alice now encrypts each block B with Bob’s public key, using the rule C = B 19 mod 3233 to obtain ( M || M pri ( Alice ) ) pub ( Bob ) = 1718 2330 and she sends this to Bob Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Suppose that Alice wishes to send the signed and encrypted message ‘go’ to Bob and that pub(Alice)=(2773,17) and pri(Alice)=157 and pub(Bob)=(3233,19) and pri(Bob)=2299 Alice encodes the message as M = 0715 She encrypts M by the rule M ′ = M 157 mod 2773 to produce the message signature M pri ( Alice ) = 2192 So M || M ′ = 0715 2192 Alice now encrypts each block B with Bob’s public key, using the rule C = B 19 mod 3233 to obtain ( M || M pri ( Alice ) ) pub ( Bob ) = 1718 2330 and she sends this to Bob Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 22

Digital signatures with RSA Examples Bob receives the ciphertext Y = 1718 2330 First, he uses his private key d = 2299 with the rule C = B 2299 to discover the underline message is X = 0715 2192 Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 22

Digital signatures with RSA Examples Bob receives the ciphertext Y = 1718 2330 First, he uses his private key d = 2299 with the rule C = B 2299 to discover the underline message is X = 0715 2192 He decodes the message to go || 2192 Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 22

Digital signatures with RSA Examples Bob receives the ciphertext Y = 1718 2330 First, he uses his private key d = 2299 with the rule C = B 2299 to discover the underline message is X = 0715 2192 He decodes the message to go || 2192 Bob now verifies the signature, he decrypts the second half of the message with Alice’s public key pub(Alice)=(2773,17) and the rule C = B 17 mod 2773 Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 22

Digital signatures with RSA Examples Bob receives the ciphertext Y = 1718 2330 First, he uses his private key d = 2299 with the rule C = B 2299 to discover the underline message is X = 0715 2192 He decodes the message to go || 2192 Bob now verifies the signature, he decrypts the second half of the message with Alice’s public key pub(Alice)=(2773,17) and the rule C = B 17 mod 2773 The signature becomes 0715 which is translated to ‘go’ Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 22

Digital signatures with RSA Examples Bob receives the ciphertext Y = 1718 2330 First, he uses his private key d = 2299 with the rule C = B 2299 to discover the underline message is X = 0715 2192 He decodes the message to go || 2192 Bob now verifies the signature, he decrypts the second half of the message with Alice’s public key pub(Alice)=(2773,17) and the rule C = B 17 mod 2773 The signature becomes 0715 which is translated to ‘go’ So Bob knows that the message was authorised by Alice Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 22

Digital signatures with RSA Examples Bob receives the ciphertext Y = 1718 2330 First, he uses his private key d = 2299 with the rule C = B 2299 to discover the underline message is X = 0715 2192 He decodes the message to go || 2192 Bob now verifies the signature, he decrypts the second half of the message with Alice’s public key pub(Alice)=(2773,17) and the rule C = B 17 mod 2773 The signature becomes 0715 which is translated to ‘go’ So Bob knows that the message was authorised by Alice Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 22

The mathematics of RSA The mathematics of RSA Lemma Let p and q be distinct primes and let a and b be integers. If a ≡ b (mod p) and a ≡ b (mod q), then a ≡ b (mod pq). Proof. By the hypothesis, we obtain that p | a − b and q | a − b We have that a − b = qm for some integer m Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 22

The mathematics of RSA The mathematics of RSA Lemma Let p and q be distinct primes and let a and b be integers. If a ≡ b (mod p) and a ≡ b (mod q), then a ≡ b (mod pq). Proof. By the hypothesis, we obtain that p | a − b and q | a − b We have that a − b = qm for some integer m It follows that p | a − b = qm Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 22

The mathematics of RSA The mathematics of RSA Lemma Let p and q be distinct primes and let a and b be integers. If a ≡ b (mod p) and a ≡ b (mod q), then a ≡ b (mod pq). Proof. By the hypothesis, we obtain that p | a − b and q | a − b We have that a − b = qm for some integer m It follows that p | a − b = qm Since p � = q are primes, we have gcd( p , q ) = 1 so that p | m Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 22

The mathematics of RSA The mathematics of RSA Lemma Let p and q be distinct primes and let a and b be integers. If a ≡ b (mod p) and a ≡ b (mod q), then a ≡ b (mod pq). Proof. By the hypothesis, we obtain that p | a − b and q | a − b We have that a − b = qm for some integer m It follows that p | a − b = qm Since p � = q are primes, we have gcd( p , q ) = 1 so that p | m Hence, pq | qm = a − b or equivalently a ≡ b (mod pq ) as wanted. Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 22

The mathematics of RSA The mathematics of RSA Lemma Let p and q be distinct primes and let a and b be integers. If a ≡ b (mod p) and a ≡ b (mod q), then a ≡ b (mod pq). Proof. By the hypothesis, we obtain that p | a − b and q | a − b We have that a − b = qm for some integer m It follows that p | a − b = qm Since p � = q are primes, we have gcd( p , q ) = 1 so that p | m Hence, pq | qm = a − b or equivalently a ≡ b (mod pq ) as wanted. Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 22

The mathematics of RSA The RSA Theorem Theorem Let ( n , e ) be a public key for the RSA cryptosystem and ( n , d ) the corresponding private key, and let E ( M ) = M e mod n and D ( C ) = C d mod n be the encryption and decryption rules, respectively. Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 22

The mathematics of RSA The RSA Theorem Theorem Let ( n , e ) be a public key for the RSA cryptosystem and ( n , d ) the corresponding private key, and let E ( M ) = M e mod n and D ( C ) = C d mod n be the encryption and decryption rules, respectively. Then D ( E ( M )) = M (mod n) . Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 22

The mathematics of RSA The RSA Theorem Theorem Let ( n , e ) be a public key for the RSA cryptosystem and ( n , d ) the corresponding private key, and let E ( M ) = M e mod n and D ( C ) = C d mod n be the encryption and decryption rules, respectively. Then D ( E ( M )) = M (mod n) . Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Since ed ≡ 1 (mod φ ( n )), there exists some integer k such that ed = 1 + k φ ( n ) Hence ( M e ) d (mod n ) D ( E ( M )) ≡ M ed (mod n ) ≡ M k φ ( n )+1 (mod n ) ≡ Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Since ed ≡ 1 (mod φ ( n )), there exists some integer k such that ed = 1 + k φ ( n ) Hence ( M e ) d (mod n ) D ( E ( M )) ≡ M ed (mod n ) ≡ M k φ ( n )+1 (mod n ) ≡ Let p and q be primes such that n = pq Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Since ed ≡ 1 (mod φ ( n )), there exists some integer k such that ed = 1 + k φ ( n ) Hence ( M e ) d (mod n ) D ( E ( M )) ≡ M ed (mod n ) ≡ M k φ ( n )+1 (mod n ) ≡ Let p and q be primes such that n = pq If p does not divide M , then by Fermat’s Little Theorem, we have M p − 1 ≡ 1 (mod p ) Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Since ed ≡ 1 (mod φ ( n )), there exists some integer k such that ed = 1 + k φ ( n ) Hence ( M e ) d (mod n ) D ( E ( M )) ≡ M ed (mod n ) ≡ M k φ ( n )+1 (mod n ) ≡ Let p and q be primes such that n = pq If p does not divide M , then by Fermat’s Little Theorem, we have M p − 1 ≡ 1 (mod p ) Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Raising both sides to the power k ( q − 1) , we have M k ( p − 1)( q − 1) ≡ 1 (mod p) So M k φ ( n ) · M ≡ 1 · M (mod p ) Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Raising both sides to the power k ( q − 1) , we have M k ( p − 1)( q − 1) ≡ 1 (mod p) So M k φ ( n ) · M ≡ 1 · M (mod p ) Finally, we obtain M k φ ( n )+1 ≡ M (mod p ) Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Raising both sides to the power k ( q − 1) , we have M k ( p − 1)( q − 1) ≡ 1 (mod p) So M k φ ( n ) · M ≡ 1 · M (mod p ) Finally, we obtain M k φ ( n )+1 ≡ M (mod p ) The previous equation is trivially true if p | M . So it is true for all M . Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Raising both sides to the power k ( q − 1) , we have M k ( p − 1)( q − 1) ≡ 1 (mod p) So M k φ ( n ) · M ≡ 1 · M (mod p ) Finally, we obtain M k φ ( n )+1 ≡ M (mod p ) The previous equation is trivially true if p | M . So it is true for all M . Similarly, we obtain that M k φ ( n )+1 ≡ 1 (mod q ) Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Raising both sides to the power k ( q − 1) , we have M k ( p − 1)( q − 1) ≡ 1 (mod p) So M k φ ( n ) · M ≡ 1 · M (mod p ) Finally, we obtain M k φ ( n )+1 ≡ M (mod p ) The previous equation is trivially true if p | M . So it is true for all M . Similarly, we obtain that M k φ ( n )+1 ≡ 1 (mod q ) By the previous lemma, we deduce that D ( E ( M )) ≡ M k φ ( n )+1 ≡ M (mod n ) as wanted. Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 22

The mathematics of RSA Proof of the RSA Theorem Proof. Raising both sides to the power k ( q − 1) , we have M k ( p − 1)( q − 1) ≡ 1 (mod p) So M k φ ( n ) · M ≡ 1 · M (mod p ) Finally, we obtain M k φ ( n )+1 ≡ M (mod p ) The previous equation is trivially true if p | M . So it is true for all M . Similarly, we obtain that M k φ ( n )+1 ≡ 1 (mod q ) By the previous lemma, we deduce that D ( E ( M )) ≡ M k φ ( n )+1 ≡ M (mod n ) as wanted. Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation This system was first published by Taher El Gamal in 1985 Suppose that Alice wants to use the El Gamal system Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation This system was first published by Taher El Gamal in 1985 Suppose that Alice wants to use the El Gamal system She must first generate a key-pair: pri(Alice) and pub(Alice) Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation This system was first published by Taher El Gamal in 1985 Suppose that Alice wants to use the El Gamal system She must first generate a key-pair: pri(Alice) and pub(Alice) This is accomplished as follows: Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation This system was first published by Taher El Gamal in 1985 Suppose that Alice wants to use the El Gamal system She must first generate a key-pair: pri(Alice) and pub(Alice) This is accomplished as follows: Alice chooses a large random prime p and a generator α of Z ∗ 1 p Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation This system was first published by Taher El Gamal in 1985 Suppose that Alice wants to use the El Gamal system She must first generate a key-pair: pri(Alice) and pub(Alice) This is accomplished as follows: Alice chooses a large random prime p and a generator α of Z ∗ 1 p She next choses a random integer a ∈ { 2 , 3 , · · · , p − 2 } and computes 2 α a (mod p) Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation This system was first published by Taher El Gamal in 1985 Suppose that Alice wants to use the El Gamal system She must first generate a key-pair: pri(Alice) and pub(Alice) This is accomplished as follows: Alice chooses a large random prime p and a generator α of Z ∗ 1 p She next choses a random integer a ∈ { 2 , 3 , · · · , p − 2 } and computes 2 α a (mod p) She sets pub ( Alice ) = ( p , α, α a ) and pri ( Alice ) = a 3 Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation This system was first published by Taher El Gamal in 1985 Suppose that Alice wants to use the El Gamal system She must first generate a key-pair: pri(Alice) and pub(Alice) This is accomplished as follows: Alice chooses a large random prime p and a generator α of Z ∗ 1 p She next choses a random integer a ∈ { 2 , 3 , · · · , p − 2 } and computes 2 α a (mod p) She sets pub ( Alice ) = ( p , α, α a ) and pri ( Alice ) = a 3 Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Suppose that Alice chooses the prime p = 149 She now find a generator α of Z ∗ 149 (Using Theorem 40) Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Suppose that Alice chooses the prime p = 149 She now find a generator α of Z ∗ 149 (Using Theorem 40) She decides to try α = 5 Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Suppose that Alice chooses the prime p = 149 She now find a generator α of Z ∗ 149 (Using Theorem 40) She decides to try α = 5 Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Since 148 = 2 2 · 37 , she must compute 5 148 / 2 = 5 74 mod 149 and 5 148 / 37 = 5 4 mod 149 She finds 5 74 ≡ 1 mod 149 so 5 is not a generator Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Since 148 = 2 2 · 37 , she must compute 5 148 / 2 = 5 74 mod 149 and 5 148 / 37 = 5 4 mod 149 She finds 5 74 ≡ 1 mod 149 so 5 is not a generator She next tries α = 12 and finds that 12 74 ≡ 148 mod 149 and 12 4 ≡ 25 mod 149 so α = 12 is a generator for Z ∗ 149 Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Since 148 = 2 2 · 37 , she must compute 5 148 / 2 = 5 74 mod 149 and 5 148 / 37 = 5 4 mod 149 She finds 5 74 ≡ 1 mod 149 so 5 is not a generator She next tries α = 12 and finds that 12 74 ≡ 148 mod 149 and 12 4 ≡ 25 mod 149 so α = 12 is a generator for Z ∗ 149 She pick a = 37 and calculate α a = 12 37 ≡ 105 mod 149 Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Since 148 = 2 2 · 37 , she must compute 5 148 / 2 = 5 74 mod 149 and 5 148 / 37 = 5 4 mod 149 She finds 5 74 ≡ 1 mod 149 so 5 is not a generator She next tries α = 12 and finds that 12 74 ≡ 148 mod 149 and 12 4 ≡ 25 mod 149 so α = 12 is a generator for Z ∗ 149 She pick a = 37 and calculate α a = 12 37 ≡ 105 mod 149 Thus pub(Alice)=(149 , 12 , 105) and pri(Alice)=37 Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 22

The El Gamal public-key cryptosystem El Gamal: Key generation-Examples Example Since 148 = 2 2 · 37 , she must compute 5 148 / 2 = 5 74 mod 149 and 5 148 / 37 = 5 4 mod 149 She finds 5 74 ≡ 1 mod 149 so 5 is not a generator She next tries α = 12 and finds that 12 74 ≡ 148 mod 149 and 12 4 ≡ 25 mod 149 so α = 12 is a generator for Z ∗ 149 She pick a = 37 and calculate α a = 12 37 ≡ 105 mod 149 Thus pub(Alice)=(149 , 12 , 105) and pri(Alice)=37 Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Suppose that Bob wants to send Alice a message using the El Gamal cryptosystem Bob first looks up Alice’s public key pub(Alice)= ( p , α, α a ) Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Suppose that Bob wants to send Alice a message using the El Gamal cryptosystem Bob first looks up Alice’s public key pub(Alice)= ( p , α, α a ) He then represents the message as an integer M in the range { 0 , 1 , · · · , p − 1 } Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Suppose that Bob wants to send Alice a message using the El Gamal cryptosystem Bob first looks up Alice’s public key pub(Alice)= ( p , α, α a ) He then represents the message as an integer M in the range { 0 , 1 , · · · , p − 1 } Bob selects a random integer k ∈ { 1 , 2 , · · · , p − 2 } Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Suppose that Bob wants to send Alice a message using the El Gamal cryptosystem Bob first looks up Alice’s public key pub(Alice)= ( p , α, α a ) He then represents the message as an integer M in the range { 0 , 1 , · · · , p − 1 } Bob selects a random integer k ∈ { 1 , 2 , · · · , p − 2 } He then computes γ = α k mod p and δ = M ( α a ) k mod p Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Suppose that Bob wants to send Alice a message using the El Gamal cryptosystem Bob first looks up Alice’s public key pub(Alice)= ( p , α, α a ) He then represents the message as an integer M in the range { 0 , 1 , · · · , p − 1 } Bob selects a random integer k ∈ { 1 , 2 , · · · , p − 2 } He then computes γ = α k mod p and δ = M ( α a ) k mod p Finally, Bob sends ( γ, δ ) to Alice Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Suppose that Bob wants to send Alice a message using the El Gamal cryptosystem Bob first looks up Alice’s public key pub(Alice)= ( p , α, α a ) He then represents the message as an integer M in the range { 0 , 1 , · · · , p − 1 } Bob selects a random integer k ∈ { 1 , 2 , · · · , p − 2 } He then computes γ = α k mod p and δ = M ( α a ) k mod p Finally, Bob sends ( γ, δ ) to Alice Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption To decrypt the message that Bob sends her, Alice follows a two-step procedure: 1 She uses her private key pri(Alice)=a to compute γ p − 1 − a mod p Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption To decrypt the message that Bob sends her, Alice follows a two-step procedure: 1 She uses her private key pri(Alice)=a to compute γ p − 1 − a mod p 2 Now she can recover the message M by finding δγ p − 1 − a = M · α ak α − ak ≡ M mod p . Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption To decrypt the message that Bob sends her, Alice follows a two-step procedure: 1 She uses her private key pri(Alice)=a to compute γ p − 1 − a mod p 2 Now she can recover the message M by finding δγ p − 1 − a = M · α ak α − ak ≡ M mod p . Tong-Viet (UKZN) MATH236 Semester 1, 2013 17 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Key generation) Suppose that Alice chooses p = 2579 She writes 2578 = 2 · 1289 Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Key generation) Suppose that Alice chooses p = 2579 She writes 2578 = 2 · 1289 She tries to find a generator for Z ∗ 2579 Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Key generation) Suppose that Alice chooses p = 2579 She writes 2578 = 2 · 1289 She tries to find a generator for Z ∗ 2579 She tries α = 2; She computes α 2 = 4 mod 2579 and α 1289 = 2578 mod 2579 Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Key generation) Suppose that Alice chooses p = 2579 She writes 2578 = 2 · 1289 She tries to find a generator for Z ∗ 2579 She tries α = 2; She computes α 2 = 4 mod 2579 and α 1289 = 2578 mod 2579 She picks a = 956 and finds 2 956 ≡ 1272 mod 2579 Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Key generation) Suppose that Alice chooses p = 2579 She writes 2578 = 2 · 1289 She tries to find a generator for Z ∗ 2579 She tries α = 2; She computes α 2 = 4 mod 2579 and α 1289 = 2578 mod 2579 She picks a = 956 and finds 2 956 ≡ 1272 mod 2579 She publishes the information pub(Alice)=(2579 , 2 , 1272) Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Key generation) Suppose that Alice chooses p = 2579 She writes 2578 = 2 · 1289 She tries to find a generator for Z ∗ 2579 She tries α = 2; She computes α 2 = 4 mod 2579 and α 1289 = 2578 mod 2579 She picks a = 956 and finds 2 956 ≡ 1272 mod 2579 She publishes the information pub(Alice)=(2579 , 2 , 1272) She keeps pri(Alice)=956 Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Key generation) Suppose that Alice chooses p = 2579 She writes 2578 = 2 · 1289 She tries to find a generator for Z ∗ 2579 She tries α = 2; She computes α 2 = 4 mod 2579 and α 1289 = 2578 mod 2579 She picks a = 956 and finds 2 956 ≡ 1272 mod 2579 She publishes the information pub(Alice)=(2579 , 2 , 1272) She keeps pri(Alice)=956 Tong-Viet (UKZN) MATH236 Semester 1, 2013 18 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) Bob decides to send the message ‘nuts’ to Alice He encodes ‘nuts’ as 14212019 Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) Bob decides to send the message ‘nuts’ to Alice He encodes ‘nuts’ as 14212019 He looks up Alice’s public key and determines that p = 2579 so he decides to split the message up into two blocks, each of length 4 : 1421 2019 Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) Bob decides to send the message ‘nuts’ to Alice He encodes ‘nuts’ as 14212019 He looks up Alice’s public key and determines that p = 2579 so he decides to split the message up into two blocks, each of length 4 : 1421 2019 For additional security, he will seclect a different value of k for each block Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) Bob decides to send the message ‘nuts’ to Alice He encodes ‘nuts’ as 14212019 He looks up Alice’s public key and determines that p = 2579 so he decides to split the message up into two blocks, each of length 4 : 1421 2019 For additional security, he will seclect a different value of k for each block For the first block, M 1 = 1421 , he picks k 1 = 318 , while for the second block, M 2 = 2019 , he will use k 2 = 1905 Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) Bob decides to send the message ‘nuts’ to Alice He encodes ‘nuts’ as 14212019 He looks up Alice’s public key and determines that p = 2579 so he decides to split the message up into two blocks, each of length 4 : 1421 2019 For additional security, he will seclect a different value of k for each block For the first block, M 1 = 1421 , he picks k 1 = 318 , while for the second block, M 2 = 2019 , he will use k 2 = 1905 Tong-Viet (UKZN) MATH236 Semester 1, 2013 19 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) For the first block: γ 1 = α k 1 mod p = 2 318 δ 1 = M 1 ( α a ) k 1 = 1421 · 1272 318 mod 2579 = 590 Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) For the first block: γ 1 = α k 1 mod p = 2 318 δ 1 = M 1 ( α a ) k 1 = 1421 · 1272 318 mod 2579 = 590 For the second block: γ 2 = α k 2 mod p = 2 1905 mod 2579 = 1035 Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) For the first block: γ 1 = α k 1 mod p = 2 318 δ 1 = M 1 ( α a ) k 1 = 1421 · 1272 318 mod 2579 = 590 For the second block: γ 2 = α k 2 mod p = 2 1905 mod 2579 = 1035 δ 2 = M 2 ( α a ) k 2 = 2019 · 1272 1905 mod 2579 = 1684 Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) For the first block: γ 1 = α k 1 mod p = 2 318 δ 1 = M 1 ( α a ) k 1 = 1421 · 1272 318 mod 2579 = 590 For the second block: γ 2 = α k 2 mod p = 2 1905 mod 2579 = 1035 δ 2 = M 2 ( α a ) k 2 = 2019 · 1272 1905 mod 2579 = 1684 Bob sends the message (792 , 590) , (1035 , 1684) to Alice Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) For the first block: γ 1 = α k 1 mod p = 2 318 δ 1 = M 1 ( α a ) k 1 = 1421 · 1272 318 mod 2579 = 590 For the second block: γ 2 = α k 2 mod p = 2 1905 mod 2579 = 1035 δ 2 = M 2 ( α a ) k 2 = 2019 · 1272 1905 mod 2579 = 1684 Bob sends the message (792 , 590) , (1035 , 1684) to Alice He could concatenate everything and sends 0792059010351684 to Alice Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Encryption) For the first block: γ 1 = α k 1 mod p = 2 318 δ 1 = M 1 ( α a ) k 1 = 1421 · 1272 318 mod 2579 = 590 For the second block: γ 2 = α k 2 mod p = 2 1905 mod 2579 = 1035 δ 2 = M 2 ( α a ) k 2 = 2019 · 1272 1905 mod 2579 = 1684 Bob sends the message (792 , 590) , (1035 , 1684) to Alice He could concatenate everything and sends 0792059010351684 to Alice Tong-Viet (UKZN) MATH236 Semester 1, 2013 20 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Decryption) Alice receives the message 0792059010351684 from Bob, and recovers γ 1 = 792 , δ 1 = 590 and γ 2 = 1035 , δ 2 = 1684 Tong-Viet (UKZN) MATH236 Semester 1, 2013 21 / 22

The El Gamal public-key cryptosystem El Gamal: Encryption and Decryption Example (Decryption) Alice receives the message 0792059010351684 from Bob, and recovers γ 1 = 792 , δ 1 = 590 and γ 2 = 1035 , δ 2 = 1684 Using her private key pri(Alice)= a = 956 , she decrypts the first block Tong-Viet (UKZN) MATH236 Semester 1, 2013 21 / 22