Overview Needed Theorems RSA Encryption Caesarian Cipher 1. In Caesar’s times most people, including Romans, were illiterate. 2. But that did not make written communications safe. There were literate barbarians. 3. Caesarian cipher: Scramble the letters of the alphabet. For example, “hello” becomes “ygaap”. 4. To send the message you need to know how to encode the message: h → y , e → g , l → a , o → p . 5. To read the message you need to know how to decode it: y → h , g → e , a → l , p → o . 6. But for this one, as soon as you can encode, you can decode, too. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Caesarian Cipher 1. In Caesar’s times most people, including Romans, were illiterate. 2. But that did not make written communications safe. There were literate barbarians. 3. Caesarian cipher: Scramble the letters of the alphabet. For example, “hello” becomes “ygaap”. 4. To send the message you need to know how to encode the message: h → y , e → g , l → a , o → p . 5. To read the message you need to know how to decode it: y → h , g → e , a → l , p → o . 6. But for this one, as soon as you can encode, you can decode, too. 7. So sender and recipient must keep the code private, which is why this is called “private key encryption”. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption 1. If one of the owners of the key reveals the key, all communications are compromised. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption 1. If one of the owners of the key reveals the key, all communications are compromised. 2. One captured centurion jeopardizes legions. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption 1. If one of the owners of the key reveals the key, all communications are compromised. 2. One captured centurion jeopardizes legions. (“Wind talkers”.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption 1. If one of the owners of the key reveals the key, all communications are compromised. 2. One captured centurion jeopardizes legions. (“Wind talkers”.) 3. One captured operative jeopardizes a spy network. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption 1. If one of the owners of the key reveals the key, all communications are compromised. 2. One captured centurion jeopardizes legions. (“Wind talkers”.) 3. One captured operative jeopardizes a spy network. 4. It does not matter how sophisticated the private key code is. From the encoding process, you can find the decoding process. (Enigma.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption 1. If one of the owners of the key reveals the key, all communications are compromised. 2. One captured centurion jeopardizes legions. (“Wind talkers”.) 3. One captured operative jeopardizes a spy network. 4. It does not matter how sophisticated the private key code is. From the encoding process, you can find the decoding process. (Enigma.) 5. But somehow, even though the encoding mechanism for internet transactions is public, internet transactions are considered safe logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Problems With Private Key Encryption 1. If one of the owners of the key reveals the key, all communications are compromised. 2. One captured centurion jeopardizes legions. (“Wind talkers”.) 3. One captured operative jeopardizes a spy network. 4. It does not matter how sophisticated the private key code is. From the encoding process, you can find the decoding process. (Enigma.) 5. But somehow, even though the encoding mechanism for internet transactions is public, internet transactions are considered safe??? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption 1. The problems with private key encryption would be resolved if the decoding mechanism could not be (easily) obtained from the encoding mechanism. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption 1. The problems with private key encryption would be resolved if the decoding mechanism could not be (easily) obtained from the encoding mechanism. 1.1 One captured centurion’s code would not reveal what the others are sending (did not happen). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption 1. The problems with private key encryption would be resolved if the decoding mechanism could not be (easily) obtained from the encoding mechanism. 1.1 One captured centurion’s code would not reveal what the others are sending (did not happen). 1.2 One captured operative would not be a problem (did not happen until late 1970s). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption 1. The problems with private key encryption would be resolved if the decoding mechanism could not be (easily) obtained from the encoding mechanism. 1.1 One captured centurion’s code would not reveal what the others are sending (did not happen). 1.2 One captured operative would not be a problem (did not happen until late 1970s). 1.3 Internet transmissions could be considered safe. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption 1. The problems with private key encryption would be resolved if the decoding mechanism could not be (easily) obtained from the encoding mechanism. 1.1 One captured centurion’s code would not reveal what the others are sending (did not happen). 1.2 One captured operative would not be a problem (did not happen until late 1970s). 1.3 Internet transmissions could be considered safe. (We do consider them as safe as can be.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption 1. The problems with private key encryption would be resolved if the decoding mechanism could not be (easily) obtained from the encoding mechanism. 1.1 One captured centurion’s code would not reveal what the others are sending (did not happen). 1.2 One captured operative would not be a problem (did not happen until late 1970s). 1.3 Internet transmissions could be considered safe. (We do consider them as safe as can be.) 2. But how do you get something like that? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Public Key Encryption 1. The problems with private key encryption would be resolved if the decoding mechanism could not be (easily) obtained from the encoding mechanism. 1.1 One captured centurion’s code would not reveal what the others are sending (did not happen). 1.2 One captured operative would not be a problem (did not happen until late 1970s). 1.3 Internet transmissions could be considered safe. (We do consider them as safe as can be.) 2. But how do you get something like that? 3. Make breaking the code depend on being able to solve a hard problem, like the factorization of a large number. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that xy 1 �≡ xy 2 ( mod m ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements and [ 0 ] m is not one of them. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements and [ 0 ] m is not one of them. There are exactly m − 1 equivalence classes modulo m that are not [ 0 ] m . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements and [ 0 ] m is not one of them. There are exactly m − 1 equivalence classes modulo m that are not � � [ 0 ] m . So A = [ z ] m : z = 1 ,..., m − 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements and [ 0 ] m is not one of them. There are exactly m − 1 equivalence classes modulo m that are not � � [ 0 ] m . So A = [ z ] m : z = 1 ,..., m − 1 and [ 1 ] m ∈ A . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements and [ 0 ] m is not one of them. There are exactly m − 1 equivalence classes modulo m that are not � � [ 0 ] m . So A = [ z ] m : z = 1 ,..., m − 1 and [ 1 ] m ∈ A . Hence there is a y ∈ { 1 ,..., m − 1 } so that [ x ] m · [ y ] m = [ 1 ] m logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements and [ 0 ] m is not one of them. There are exactly m − 1 equivalence classes modulo m that are not � � [ 0 ] m . So A = [ z ] m : z = 1 ,..., m − 1 and [ 1 ] m ∈ A . Hence there is a y ∈ { 1 ,..., m − 1 } so that [ x ] m · [ y ] m = [ 1 ] m , that is, so that xy ≡ 1 ( mod m ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Proposition. Let x , m ∈ N have no common factors. Then for all c ∈ N that are not divisible by m, we have that cx �≡ 0 ( mod m ) . Moreover, there is a y ∈ N so that xy ≡ 1 ( mod m ) . Proof. Let b ∈ N be so that bx ≡ 0 ( mod m ) . Then m | bx . Because x and m do not have any common factors, m | b . Hence, if c is not divisible by m , then cx �≡ 0 ( mod m ) . Now consider { xy : y = 1 ,..., m − 1 } . For any two distinct y 1 , y 2 ∈ { 1 ,..., m − 1 } with y 1 < y 2 we have that m ∤ ( y 2 − y 1 ) . Hence, x ( y 2 − y 1 ) �≡ 0 ( mod m ) , which means that � � xy 1 �≡ xy 2 ( mod m ) . But then A : = [ xy ] m : y = 1 ,..., m − 1 has m − 1 distinct elements and [ 0 ] m is not one of them. There are exactly m − 1 equivalence classes modulo m that are not � � [ 0 ] m . So A = [ z ] m : z = 1 ,..., m − 1 and [ 1 ] m ∈ A . Hence there is a y ∈ { 1 ,..., m − 1 } so that [ x ] m · [ y ] m = [ 1 ] m , that is, so that xy ≡ 1 ( mod m ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. ( a + 1 ) p logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p � p � a k 1 p − k ( a + 1 ) p ∑ = k k = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ 1 + a ( mod p ) ≡ logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ 1 + a ( mod p ) ≡ Now let a ∈ N be so that p ∤ a . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ 1 + a ( mod p ) ≡ Now let a ∈ N be so that p ∤ a . There is a b ∈ N with ab ≡ 1 ( mod p ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ 1 + a ( mod p ) ≡ Now let a ∈ N be so that p ∤ a . There is a b ∈ N with ab ≡ 1 ( mod p ) . Hence a p ≡ a ( mod p ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ 1 + a ( mod p ) ≡ Now let a ∈ N be so that p ∤ a . There is a b ∈ N with ab ≡ 1 ( mod p ) . Hence a p ≡ a ( mod p ) implies a p b ≡ ab ( mod p ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ 1 + a ( mod p ) ≡ Now let a ∈ N be so that p ∤ a . There is a b ∈ N with ab ≡ 1 ( mod p ) . Hence a p ≡ a ( mod p ) implies a p b ≡ ab ( mod p ) , which implies a p − 1 ≡ 1 ( mod p ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Theorem. Fermat’s Little Theorem . Let p be a prime number. Then for every a ∈ N we have that a p ≡ a ( mod p ) . Moreover, for every a ∈ N that is not divisible by p we have that a p − 1 ≡ 1 ( mod p ) . Proof. Let a ∈ N . We prove a p ≡ a ( mod p ) by induction on a . Base step a = 1: obvious. Induction step. p − 1 p � p � � p � a k 1 p − k = 1 + a p + ( a + 1 ) p ∑ ∑ a k = k k k = 0 k = 1 1 + a p ( mod p ) ≡ 1 + a ( mod p ) ≡ Now let a ∈ N be so that p ∤ a . There is a b ∈ N with ab ≡ 1 ( mod p ) . Hence a p ≡ a ( mod p ) implies a p b ≡ ab ( mod p ) , which implies a p − 1 ≡ 1 ( mod p ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) 4. ϕ ( n ) : = ( p − 1 )( q − 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) 4. ϕ ( n ) : = ( p − 1 )( q − 1 ) � � � � 5. e ∈ 2 ,..., ϕ ( n ) − 1 must be so that e , ϕ ( n ) = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) 4. ϕ ( n ) : = ( p − 1 )( q − 1 ) � � � � 5. e ∈ 2 ,..., ϕ ( n ) − 1 must be so that e , ϕ ( n ) = 1 (there is an efficient algorithm to check e ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) 4. ϕ ( n ) : = ( p − 1 )( q − 1 ) � � � � 5. e ∈ 2 ,..., ϕ ( n ) − 1 must be so that e , ϕ ( n ) = 1 (there is an efficient algorithm to check e ) � � 6. d is so that de ≡ 1 mod ϕ ( n ) , logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) 4. ϕ ( n ) : = ( p − 1 )( q − 1 ) � � � � 5. e ∈ 2 ,..., ϕ ( n ) − 1 must be so that e , ϕ ( n ) = 1 (there is an efficient algorithm to check e ) � � 6. d is so that de ≡ 1 mod ϕ ( n ) , (there is an efficient algorithm to find d ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) 4. ϕ ( n ) : = ( p − 1 )( q − 1 ) � � � � 5. e ∈ 2 ,..., ϕ ( n ) − 1 must be so that e , ϕ ( n ) = 1 (there is an efficient algorithm to check e ) � � 6. d is so that de ≡ 1 mod ϕ ( n ) , (there is an efficient algorithm to find d ) 7. ( n , e ) is the public key (disseminated) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption RSA Encryption 1. R. Rivest, A. Shamir, L. Adleman, (1978) A Method for Obtaining Digital Signatures and Public-Key Cryptosystems, Communications of the ACM 21, 120-126 2. p , q : fixed, distinct prime numbers 3. n : = pq (must be hard to factor, so large, proprietary prime numbers are used) 4. ϕ ( n ) : = ( p − 1 )( q − 1 ) � � � � 5. e ∈ 2 ,..., ϕ ( n ) − 1 must be so that e , ϕ ( n ) = 1 (there is an efficient algorithm to check e ) � � 6. d is so that de ≡ 1 mod ϕ ( n ) , (there is an efficient algorithm to find d ) 7. ( n , e ) is the public key (disseminated) 8. d is the private key (kept secret) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Sending Messages logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Sending Messages 1. The message is a large number m smaller than n (or a string of numbers). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Sending Messages 1. The message is a large number m smaller than n (or a string of numbers). Group letters in blocks and encode them with numbers. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Sending Messages 1. The message is a large number m smaller than n (or a string of numbers). Group letters in blocks and encode them with numbers. Make sure a cryptoquote style approach is unlikely to break the code. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Sending Messages 1. The message is a large number m smaller than n (or a string of numbers). Group letters in blocks and encode them with numbers. Make sure a cryptoquote style approach is unlikely to break the code. 2. Encrypted message: c : ≡ m e ( mod n ) (use the positive representative smaller than n for convenience). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Sending Messages 1. The message is a large number m smaller than n (or a string of numbers). Group letters in blocks and encode them with numbers. Make sure a cryptoquote style approach is unlikely to break the code. 2. Encrypted message: c : ≡ m e ( mod n ) (use the positive representative smaller than n for convenience). � c d � 3. Decrypted message: Representative of n that is in { 0 ,..., n − 1 } . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
Overview Needed Theorems RSA Encryption Sending Messages 1. The message is a large number m smaller than n (or a string of numbers). Group letters in blocks and encode them with numbers. Make sure a cryptoquote style approach is unlikely to break the code. 2. Encrypted message: c : ≡ m e ( mod n ) (use the positive representative smaller than n for convenience). � c d � 3. Decrypted message: Representative of n that is in { 0 ,..., n − 1 } . Why does this work? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Public Key (RSA) Encryption
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