PollardRsaCracker

This is my implementation of RSA cracking algorithm based on Pollard factorization. It will work if the numbers p - 1 and q - 1 that form n = p * q (as a part of the public key, p and q are prime), have no large maximum prime power factors. Of course, all modern RSA implementations don't have this security issue so this is more of an educational project.

Usage

In order to start the script, simply run python cracker.py

You will need 4 inputs:

1. n - the modulus in the public key so that m = c ^ d mod n where m is the message, c is the encrypted message, d is the private key.
2. e - the second public key component (so that gcd(e, phi(n)) = 1).
3. c - the encrypted message.
4. b - the maximum prime factor of n. This field is optional. However, if you leave it blank, the brute-force algorithm will be used for factorizing n.

Demo

Enter n (as part of the public key): 186444745729857899758373984272541398503249351266417000699738642133172271283265124803102459
Enter e (as part of the public key): 65537
Enter c (the encrypted message to decrypt): 159178142916077677757648147687519523540045276157456113470673097514775229976995968698190914
Enter b (the maximum prime factor in n, optional): 200000
Factorizing n with the Pollard algorithm...
k has 86871 digits
Trying to solve with a = 2
(a ^ k mod n) - 1 = 95072790833932492612013226959843266212858324985456485206430737027466196168851937286400160
p = 2962244945158622279601750283735698362054297
q = 62940354083336669282335053470277744418281466547
Factorizing complete.
phi(n) = 186444745729857899758373984272541398503249288323100672417910737518517050721785008159581616
Solving e * d = 1 mod phi(n) with the euklidian algorithm...
d (private key) = 144519783985790947216463957780263103956010556583510294319695217448779562333745493606767873
m (decrypted message) = 2020

Demo with small numbers

Brute-force

Enter n (as part of the public key): 493
Enter e (as part of the public key): 45
Enter c (the encrypted message to decrypt): 56
Enter b (the maximum prime factor in n, optional):
Factorizing n with the Bruteforce algorithm...
p = 17
q = 29
Factorizing complete.
phi(n) = 448
Solving e * d = 1 mod phi(n) with the euklidian algorithm...
d (private key) = 229
m (decrypted message) = 490

Pollard

Enter n (as part of the public key): 493
Enter e (as part of the public key): 45
Enter c (the encrypted message to decrypt): 56
Enter b (the maximum prime factor in n, optional): 493
Factorizing n with the Pollard algorithm...
k has 216 digits
Trying to solve with a = 2
(a ^ k mod n) - 1 = 0
Trying to solve with a = 3
(a ^ k mod n) - 1 = 0
Trying to solve with a = 4
(a ^ k mod n) - 1 = 0

...

Trying to solve with a = 509
(a ^ k mod n) - 1 = 0
Trying to solve with a = 510
(a ^ k mod n) - 1 = 203
p = 29
q = 17
Factorizing complete.
phi(n) = 448
Solving e * d = 1 mod phi(n) with the euklidian algorithm...
d (private key) = 229
m (decrypted message) = 490

Credits

• This awesome lecture where I learnt about RSA
• Mirko Höhn for helping me understand the mathematics behind RSA.

License

Copyright (c) 2020 Alexander Mayorov.

This project is licenced under the MIT Licence.

Please leave a copyright notice if you use/modify this software or parts of it.

For more information see the LICENCE file.