Given will be the p and q which is the prime numbers. lets take:
p=3 and q=11——– prime numbers.
First step is Generation of keys
since RSA uses two keys that is public and private so,
using this we will get the first public key i.e.
N=p*q= 3*11=33 where N is the first public key.
we will also find z by the formula :
choose ‘K’ a prime number such that z is not divided by k.
so, k will be 7
// we choose this by seeing prime number as 1,3,5,7,11,13,17,………
//we don’t choose 1 mostly. 3 is already used. we take 5 but it is divisible by z so, we cannot take 5. That’s why we choose k=7.
Now we will find the Private key. To find the private key we use this function:
k * j= 1 |mod (z)| where k is the public key, j is the private key.
7 * j= 1 mod|20| // this says that whatever the quotient be remainder should be 1.
(7 * j)/20= 1
so, substituting the value of j as 1,2,3 we get
7*1/20=7/20 != reminder 1.
7*2/20=14/20 != reminder 1.
7*3/20 =21/20 = reminder 1.
so, j will be 3 i.e
j=3—– private key.
Second method is Encryption
For encryption the equation will be P^k= e (mod n) where P is the plain text, k is the public key, n is the public key, e is the cipher text.
given plain text is P=14.
substituting the value in the equation we get:
14^7=e (mod 33)
or, 3194348.60=e //it is not the e coz we need to get the remainder of this. so ,we take only the decimal number that is 3194348 and multiply by 33
Now to get e we have to subtract 105413504 with 105413484 and we get e =20 ( 105413504-10541384)
e is the cipher text that is e= 20.