Question

Networking class question...

What is the relationship between the private key and public key? Who uses the public key and who uses the private key?

Answer 1

`Hey,

Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

In Private key, the same key (secret key) is used for encryption and decryption. In this key is symmetric because the only key is copy or share by another party to decrypt the cipher text. It is faster than the public key cryptography.

In Public key, two keys are used one key is used for encryption and another key is used for decryption. One key (public key) is used for encrypt the plain text to convert it into cipher text and another key (private key) is used bu receiver to decrypt the cipher text to read the message.

In RSA, you have a public key, consisting of exponent "e" and modulus "m". If you want to encrypt a message (remember that message is always a number), you raise it to the power of e, divide the result by modulus m, and the remainder after division is your encrypted message.

For example, your public key can be e=5, m=299. If you want to encrypt the message "75", you raise it to the power of 5 and give remainder after division by 299, i. e. 75^5 mod 299 = 186. So your message after encryption is 186.

How does the recipient decrypt it? He has his private key, his secret (or decryption) exponent "d", and the same modulus "m". "d" in this case is 53. So he takes the encrypted message 186, raise it to power of 53, and modulates by 299, i. e. 186^53 mod 299 = 75... thus he recovers the original message.

But how does he know what his private exponent "d" is?

You might have noted that the modulus "m" is made by multiplying two prime numbers - in this case 13 * 23 = 299. These two prime numbers are customarily called "p" and "q".

The private and public exponents e and d must satisfy following rule:

e * d = (p-1) * (q-1) * k +1

The coefficient "k" must be a whole number, which means that e multiplied by d gives a remainder of 1 when divided by (p-1) * (q-1).

Lets see if this holds in our example:

5 * 53 = 265

(13-1) * (23-1) = 12 * 22 = 264

So yes, 265 = 264 * 1 + 1.

For small numbers like 299 it is easy to recover the p and q that were multiplied to obtain the modulus m. But in real life applications, the p and q numbers are each about 300 digits long, and the resulting modulus is more than 600 digits long. Multiplying large p and q is easy for computers, but factoring m it into its prime factors in computationaly infeasible.

We can derive the private key from the public key, but in order to do that, we would need to factor the modulus (which is publicly available) into its prime factors p and q, which were used to generate it. This is at present time computationally infeasible for large m (600 decimal digits or more).

Kindly revert for any queries

Thanks.