While implementing RSA encryption/decryption (using python), the plaintext doesn't match with the decrypted ciphertext for large values of plaintext. Works fine for smaller values for plaintext (numeric value).

Input: p=53 q=59 e=3 plaintext = 1000 (private key computed as 2011)

Here, the decryption gives 1000 as the plaintext, which is correct. Now, if

Input: p=53 q=59 e=3 plaintext = 10000 (private key computed as 2011)

Here, after the decryption, the computed plaintext is 619 (which should be 10000)

The code for the same is here,

    def encrypt(plaintext):
            ciphertext = (plaintext**publicKey) % (self.n)
            return ciphertext
   def decrypt(ciphertext):
            plaintext = (ciphertext**privateKey) % (self.n)
            return plaintext

Considering the algorithm will be used to encrypt/decrypt alphanumeric text, which will produce large numeric values, what modifications are needed or am I missing something?

ps: Checking for multiple values of plaintext,it may be happening because n=3127 and any plaintext greater than 3127 will not produce the original plaintext upon decryption. I might be wrong.

How to make it work for plaintext greater than n, here 3127?

  • 1
    $\begingroup$ RSA is a trapdoor permutation! How do you expect to get a plaintext that is larger than the modulus size encrypted and decrypted correctly? The message $m$ must be $m < N$ where $N$ is the modulus and $m$ is the message. Before implementing read at least Wikipedia RSA? $\endgroup$
    – kelalaka
    Oct 9 at 17:48
  • $\begingroup$ may be a duplicate of How should I address message size limits in RSA encryption? $\endgroup$
    – kelalaka
    Oct 9 at 17:52
  • $\begingroup$ $10000 \pmod{53\cdot 51} = 619$ as expected! $\endgroup$
    – kelalaka
    Oct 9 at 17:57
  • 1
    $\begingroup$ The better way is to use hybrid encryption. Pick a random number between 0 and N. Call that number X. Use a KDF to derive a key from X. Encrypt using an AEAD like ChaCha20-Poly1305. Encrypt X with RSA. Send the encrypted X and the ciphertext of the message (and IV & tag). Recipient can decrypt X, use the same KDF, verify the tag, and decrypt the message. This is less complicated than using RSA-OAEP and treating RSA like a block cipher. $\endgroup$ Oct 9 at 18:48
  • 1
    $\begingroup$ After reading your suggestion @MaartenBodewes, I split up the numeric plaintext into digits and encrypted it, and upon decrypting, clubbed the decrypted digits. ALso, thank you for your responses SAI and Maarten $\endgroup$
    – ayush7ad6
    Oct 11 at 4:55