Skip Navigation Links.

Simple RSA Example

With RSA, initially the person picks two prime numbers. For example:

p=11 and q=3

In the following you can either manually add your own values, or generate random ones by pressing the button:

 

p

q

Next, the n value is calculated. Thus:

n = p x q = 11 x 3 = 33

Next PHI is calculated by:

PHI = (p-1)(q-1) = 20

The factors of PHI are 1, 2, 4, 5, 10 and 20. Next the public exponent e is generated so that the greatest common divisor of e and PHI is 1 (e is relatively prime with PHI). Thus, the smallest value for e is:

e = 3

Bob

n

PHI

e



The factors of e are 1 and 3, thus 1 is the highest common factor of them. Thus n (33) and the e (3) values are the public keys. The private key (d) is the inverse of e modulo PHI.

d=e^(-1) mod [(p-1)x(q-1)]

This can be calculated by using extended Euclidian algorithm, to give the =7.

Bob

d

The encryption and decryption keys are then:

 

Public Key n,e)

Private Key (n,d)

 

As a test you can manually put in p=11 and q=3, and get the keys of (n,e)=(33,3) and (n,d)=(33,7).

The PARTY2 can be given the public keys of e and n, so that PARTY2 can encrypt the message with them. PARTY1, using d and n can then decrypt the encrypted message.

For example, if the message value to decrypt is 4, then:

c = m^e mod n = 43 mod 33 = 31

Therefore, the encrypted message (c) is 31.

The encrypted message (c) is then decrypted by PARTY1 with:

m = c^d mod n = 317 mod 33 = 4

which is equal to the message value.

Encryption/Decryption:

Message: 123

Encrypt:

Decrypt: