Write a program to encrypt and decrypt your full name using RSA encryption. Use the parameters provided in the example given in the course notes.

RSA.pdf( COurse Notes)

RSA Encryption Algorithm RSA algorithm is named after its inventors (Rivest, Shamir and Adleman). It is a pubic key cryptographic scheme. The public and private keys in this scheme are generated as follows Stepl: Choose two large prime numbers "p" and "q" so that the product is equal to the integer "n", i.e., n-pq. The plaintext to be encrypted, say, "P" is represented as an integer less than "n." (This means "n" must be a large number) Step2: Find a number "e" that is relatively prime to the produce (p-)(q-l). Note that two numbers are said to be relatively prime if they have no common factors except 1. The public key is then given by e,n Step3: Find a number "d" such that the product de = 1 mod ((p-1 )(q-l )). That is, "d" and "e" are multiplicative inverses of each other modulo (p-l)(q-1). The private key is then \d.n). Why does RSA work? From the above steps we see that for any integer Pn, de (mod n P (mod n) RSA uses large binary keys, typically 512 bits long. It takes binary blocks of plaintext of length smaller than the key length and produces a ciphertext that is the same length of the key. If the integer "P" represents a block of plaintext then RSA encrypts "P" as follows: Encryption: Ciphertext, C P(mod n) Note that the ciphertext, C is an integer between 0 and "n." To decrypt, the following procedure is used Decryption: C(mod n)- (Pe) (mod n)-ed (mod n) P (mod n) - P. Example: Suppose we want to encrypt the plaintext "RSA" using RSA encryption. Convert this to integers, "R"-18 (its position in the English alphabet), "S" 19, and "A"-1. Let us choose p-5 and q 11. Then n 55 and (p-l)(q-1) 40. Let e 7 (it is relatively prime to 40). Then, d 23. Therefore, public key is 7,55 and private key is 23,55} Plaintexts, P1 18, P2- 19, P31 ("RSA"). Then, C 18 mod 55 17; C2-19mod 55 24; C 17mod 55 Therefore, the ciphertext is 17,24,1; Decryption: 175 mod 55 18; 2425 mod 55 19; 125 mod 551 Note: To compute mod for large numbers, use this result: (ab) mod n-((a mod n)(b mod n)) mod n Example: 1723 mod 55- 1716+4+2-1 mod 55 ((1716 mod 55 17mod 55X 172 mod 55 17mod 55)) mod 55 18