Question

Q5. (30 points) The plaintext message can be quickly recovered from a ciphertext message when the decryp- tion key d. an inverse of e modulo (p-1 - 15. is known. (Such an inverse exists because gedie. (p - 16 - 1) = 1.] To see this, note that if de = 1 mod ip - Diq-1))there is an integer k such thal de = 1 kp 11). It follows that C!-(MX- Mile - Ml+htp- 11 mod). By Fermat's little theorem assuming that ged(M.pl - god M.q) = 1. which holds except in rare cases, which we cover in Exercise 28. it follows that M-1 mod p) and M! 1 (mod q). Consequently, el = M (MP 11 = M1 = M (mod pl and c? = M (M4-1}p-1M.1= M (mod). Because gedip. q) = 1. it follows by the Chinese remainder theorem that CM (mod pg). (a) Ife=3, p = 3.9-5. find a decryption key d. explain your answer. (b) If a secret message is 16, use your answer from (a) to decipher the message.

show all the steps in your answer in neat English for a thumbs up

Answer 1

Solution:

(5)

(a)

Given,

=> e = 3, p= 3, q = 5

Explanation:

=>The given algorithm represents RSA.

Calculating value of n:

=>n = p*q

=>n = 3*5

=>n = 15

Calculating value of $\phi$ (n)(Euler totient of n):

=> $\phi$ (n) = $\phi$ (p)*$\phi$(q)

=> $\phi$ (n) = (p-1)*(q-1)

=> $\phi$ (n) = (3-1)*(5-1)

=> $\phi$ (n) = 2*4

=> $\phi$ (n) = 8

Calculating value of d(decryption key):

=>We need to choose d such that e*d mod $\phi$ (n) = 1

=>3*d mod 8 = 1

=>Hence the value of d = 3 because 3*3 mod 8 = 1

=>So the value of d(decryption key) = 3

(b)

Given,

=>Secret message(C) = 16

Explanation:

Finding the value of message(M) or plain text:

=>M = C^d mod n

=>M = 16^3 mod 15

=>M = 4096 mod 15

=>M = 1

=>Hence the value of message or plain text = 1

I have explained each and every part with the help of statements attached to it.