2.  Introduction  Requirements for RSA  Ingredients for RSA  RSA Algorithm  RSA Example  Problems on RSA

3.  Developed by Rivest, Shamir, and Adleman in 1977.  Public-key encryption technique : - Public-key/two-key/asymmetric cryptography involves the use of two keys: - a public key, which may be known by anybody, and can be used to encrypt messages, and verify signatures. - a private key, known only to the recipient, used to decrypt messages, and sign (create) signatures. - is asymmetric because those who encrypt messages or verify signatures cannot decrypt messages or create signatures.

4.  Block cipher scheme - A block cipher is one in which a block of plaintext is treated as a whole and used to produce a ciphertext block of equal length. - Each block having a binary value less than some number n (0 to n-1). - Typical size of block is 1024 bits or 309 decimal digits. - The block size is i bits, where 2^i < n <= 2^(i+1)

5.  Encryption and Decryption are of the following form: Plaintext M Ciphertext C C=M e mod n M=C d mod n = (M e ) d mod n= M ed mod n - Both sender and receiver must know the value of n. - The sender knows the value of e, and only the receiver knows the value of d. - Public key = {e, n} - Private key = {d, n}

6.  We need to find a relationship of the form M ed mod n - This relationship holds if e and d are multiplicative inverse modulo ø(n), where ø(n) is the Euler totient function. - Euler totient function ø(n), defined as number of the integers less than n and relatively prime to n. - The relationship between e and d can be expressed as e.d mod ø(n) =1 - d(and therefore e) is relatively prime to ø(n).{by the rules of moduler arithmetic} gcd(ø(n),d)=1

7.  For this algorithm to be satisfactory for public key encryption, the following requirements must be met: - It is possible to find values of e, d, n such that M ed mod n = M for all M<n. - It is relatively easy to calculate M e mod n and C d mod n for all values of M<n. - It is infeasible to determine d given e and n.

8.  p, q two prime numbers (private, chosen)  n=pq (public, calculated)  e, with gcd(ø(n),e)=1; 1<e<ø(n) (public, chosen)  e.d=1 mod ø(n) (private, calculated)

9. Key Generation  Select two large primes at random - p, q  Calculate n=p.q  Calculate ø(n)=(p-1)(q-1)  Selecting at random the encryption key e  where 1<e<ø(n), gcd(e,ø(n))=1  Solve following equation to find decryption key d e.d=1 mod ø(n) and 0≤d≤n  Publish their Public key: PU={e,n}  keep secret Private key: PR={d,n}

10. Encryption  Plaintext M, 0≤M<n  Ciphertext C=M e mod n Decryption  Ciphertext C  Plaintext M=C d mod n

11. 1. Select primes: p=17 & q=11 2. Compute n = pq =17×11=187 3. Compute ø(n)=(p–1)(q-1)=16×10=160 4. Select e : gcd(e,160)=1; choose e=7 5. Determine d: de=1 mod 160 and d < 160 Value is d=23 since 23×7=161= 1×160+1 6. Publish public key PU={7,187} 7. Keep secret private key PR={23,187}

12. Given message M = 88  Encryption: C = 88 7 mod 187 = 11  Decryption: M = 11 23 mod 187 = 88

13. 1. Perform encryption and decryption using RSA algorithm, for the following: ① p = 3; q = 11, e = 7; M = 5(Ans. d=3, C=14) ② p = 5; q = 11, e = 3; M = 9(Ans. d=27, C=14) 2. In a public-key system using RSA, you intercept the ciphertext C = 10 sent to a user whose public key is e = 5, n = 35. What is the plaintext M?(Ans. M=5)