Question

(c) In RSA, Given n = 143, compute the public and private keys. Do you observe anything special from the result? If plaintext m=2, what would be the cipher? [6 marks]

Answer 1

Solution:

Given,

=>n = 143

=>Plain text(m) = 2

Explanation:

=>n = p*q where p and q are prime number

=>p*q = 143

=>11*13 = 143

=>p = 11 and q = 13

=>$\phi$(n) = $\phi$ (p)*$\phi$(q) where $\phi$ (n) is euler totient function of n , $\phi$ (p) is euler totient function of p and $\phi$ (q) is euler totient function of q

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

=>$\phi$(n) = (11-1)*(13-1)

=>$\phi$(n) = 10*12

=>$\phi$(n) = 120

Finding value of e:

=>Choosing e such that 1<e<$\phi$(n) and gcd(e,$\phi$(n)) = 1 where gcd means highest common factor

=>let say e = 11 as it satisfies both the given condition

Finding value of d:

=>Choosing d such that e*dmod$\phi$(n) = 1

=>11*dmod120 = 1

=>Let say d = 11 as it satifies the above condition

Something special:

=>We have observed that e = d = 11

Finding public key and private key:

=>Pair (e,n) represents the public key

=>Public key = (11,143)

=>Pair (d,n) represents the private key

=>Private key = (11,143)

Finding ciphertext(C):

=>C = m^e mod n

=>C = 2^11 mod 143

=>C = 46

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