Comments on: RSA Example

By: Thomas King

Thomas King — Thu, 07 Mar 2024 11:51:55 +0000

I have a public key (“E”) that also decrypts my cipher text. The numbers to generate E follow all the rules. This is what they are:

P = 5 (must be prime number)
Q = 7 (must be prime number)
N = 35 = P*Q
T = 24 = (P-1)*(Q-1)
E = 11 (must be: 1. a prime number; 2. less than T; 3. not be a factor of T)

If my original message is 10 (“M”), I encrypt it with the public key as follows:

Cipher text = M^E MOD N
= 10^11 MOD 35
= 5

Here’s the problem: I can decrypt this cipher text (“C”) with the public key as well:

C^E MOD 35
= 5^11 MOD 35
= 10

I have noticed that if I try a different value for the original message, then I cannot decrypt with the public key.

e.g. if M = 59

Cipher text (“C”) = 59^11 MOD 35
= 19

C^E MOD N
= 19^11 MOD 35
= 24 (not the original message, 59)

By: K.lem

K.lem — Fri, 19 Jan 2024 01:52:15 +0000

Shouldn't the formulas for decryption read "Original Message = C^D MOD N" and "Original Message = S^E MOD N" respectively?As they're written in the article, you're decrypting the original message instead of the cypher text.

By: MMG

MMG — Sun, 19 Nov 2023 22:04:33 +0000

In reply to Ed Harmoush. How the encryptor or decryptor knows the value of N: Encryption (using the public key):The value of N is included in the recipient's public key, which is made public. Anyone who wants to send an encrypted message to the recipient can obtain the public key, including N, and use it to perform encryption. Decryption (using the private key):The recipient knows the value of N because it is a part of their private key. The private key is kept secret and is only known to the recipient. When someone sends an encrypted message, the recipient uses their private key, which includes N, to decrypt the message.In summary, the knowledge of N is distributed through the public and private key pairs, with N being included in the public key for encryption and in the private key for decryption.

By: Daniel

Daniel — Mon, 06 Nov 2023 23:08:49 +0000

There are a few values that would work for the Private Key as well. But again, for the sake of our example, we will select 41 So there is more private keys to one public key???

By: mustafa

mustafa — Mon, 17 Apr 2023 22:04:31 +0000

13 = m^23 mod 55
please help me

By: Ed Harmoush

Ed Harmoush — Fri, 14 Oct 2022 23:52:28 +0000

In reply to MikeH.

Glad you enjoyed it. Great question. When doing RSA, the “Public Key” value is actually both the “E” value and the Modulus. More details here: https://www.youtube.com/watch?v=om9-9BCZr5M

By: MikeH

MikeH — Fri, 14 Oct 2022 03:31:09 +0000

I love this. this is PRECISELY what I was looking for. sorry for posting on an old post.

I have one question. In the example, how does the encryptor or decryptor know the value of “N” when they are decrypting or decrypting.

At the time of encrypting I know:
1) the message
2) and the public Key

which I can:

M^E MOD ( unknown N? ) = cipher text?
99^29 MOD? = 92

This is the last piece of the puzzle for me ( in a long journey to understand how this works! )

By: waell kateb

waell kateb — Tue, 29 Mar 2022 14:57:15 +0000

In reply to Indra. you should to chinge it to binery

By: Leroy van Logchem

Leroy van Logchem — Thu, 10 Sep 2020 06:41:26 +0000

Thank you for the clear explanation! Bringing it down to small numbers really helped to follow along.

By: Indra

Indra — Mon, 12 Aug 2019 13:54:54 +0000

Using rsa can’t we encrypt any text messege taking text as plain text. If yes then please help me with an example of encrypting any text(take “practicalnetworking” as example) with rsa. Thanking you