Score:-1

Crypto

Am I Doing RSA Correctly?

Jame

10/16/22, 2:52 PM

I am trying to figure out how to complete RSA manually. I am trying to encode a simple block message (Mi). I used CrypTool to determine the encryption. When I "manually" computed the plaintext, I obtained a different number than what CrypTool provided. Can someone guide me? Am I doing the manual encryption for RSA correct?

188

0 + 6

rsa

encryption

public-key

keys

decryption

Ievgeni

10/16/22, 3:03 PM

What is the integer $A$?

Ievgeni

10/16/22, 3:09 PM

It could be also $Z$...

SSA

10/16/22, 3:24 PM

there are many textbook RSA example on internet, please refer to them. your Z is called ${\phi(n)}$ . and public exponent e is chosen as ${1<e<\phi(n)}$, and private exponent d is selected such that ${ed \equiv 1(mod(\phi(n))}$

kelalaka

10/16/22, 6:20 PM

Just used the [Cryptool Online RSA](https://www.cryptool.org/en/cto/rsa-step-by-step) with your setup and got the $\color{red}{Red}$ numbers. What is your problem, then?

Jame

10/16/22, 11:50 PM

@kelalaka Thank you for the online resource! When I used CrypTool RSA Encryption and input the text MI, I receive the following: Numbers input in base 10 format: 360 / Encryption into ciphertext: 11,807. Why is there only one number?

kelalaka

10/17/22, 5:54 AM

That is input handling. The online tool uses to encrypt character by character. The other combines them into a large integer. Note that there is a limit on this due to the modulus and remember that RSA is not preferred for encryption.

Yves

5/20/23, 2:19 PM

I checked it with version 1.4.42: When using the default settings of CrypTool 1 it works well and shows the results you expected. This fits with the answer of Peregrinus.

fgrieu

6/22/23, 5:37 AM

This is an exercise done correctly, or not, depending on how it was asked to encode the plaintext. But this is not RSA done correctly. For one, the factors of $n$ are much too small, making it possible to factor $n$. For another, the encoding of message `MI` to representative integer(s) $m$ does not map to an essentially random and unpredictable element of $\mathbb Z_n$, as it should.

Score:1

Crypto

meshcollider

5/22/23, 11:33 PM

As kelalaka mentioned in the comments, the reason for the difference is that the tool turns the message "MI" into a single number 360, while you have encrypted each character "M", "I" individually in your question.

The way that the tool turns the message "MI" into the number 360 is because it uses the alphabet of 26 capital letters plus the space character (number 0), giving 27 in total. So it turns each character into a digit between 0 and 26, and then convert from base 27 to base 10. In your case this gives: $$ \text{"M"} = 13 \\ \text{"I"} = 9\\ 13*27 + 9 = 360$$

It then encrypts this number 360 in the same way, using $$360^{11} \pmod{40741} = 11807.$$ You can decode the base 10 number (360) back into a message by turning it back into base 27, giving digits $(13, 9)$, and then computing their corresponding position in the alphabet ("M", "I").