Hacking the RSA cipher is possible with small prime numbers, but impossible with big numbers. The following aspects explain why it is difficult to break the RSA cipher −

• A brute force approach will fail because there are too many possible keys to sort over. Additionally, this takes a significant amount of time.
• The dictionary attack will not work with the RSA algorithm since the keys are numeric and include no characters.
• The frequency analysis of the characters is difficult to follow because a single encrypted block represents many characters.
• There are no unique mathematical strategies for hacking the RSA cipher.

The RSA decryption equation is −

M = C^d mod n

We can try hacking the RSA cipher using small prime numbers, and the sample code for doing so is provided below −

Hacking using Python

Example

def find_factors(n):
   factors = []
   for i in range(2, n):
      if n % i == 0:
         factors.append(i)
   return tuple(factors)

def calculate_euler_function(p, q):
   return (p - 1) * (q - 1)

def calculate_private_key(e, euler_value):
   for i in range(2, euler_value):
      if i * e % euler_value == 1:
         return i

def decrypt_message(private_key, modulus, ciphertext):
   return ciphertext ** private_key % modulus

def main():
   e = int(input("Enter value of e: "))
   n = int(input("Enter value of n: "))
   c = int(input("Enter ciphertext: "))

   factors = find_factors(n)
   euler_value = calculate_euler_function(factors[0], factors[1])

   d = calculate_private_key(e, euler_value)
   plain_text = decrypt_message(d, n, c)
   print("Decrypted plaintext: ", plain_text)

if __name__ == "__main__":
   main()

Run the Code

• Save your Python program.
• Open a terminal or command prompt.
• Navigate to the directory where your code is saved.

Run the script by typing −

python rsa_hacking.py

When you run the program, you will have to enter some values of e, n and ciphertext so after that you will get the plaintext.

Following is the output of the above example −

Input/Output

Enter value of e: 7
Enter value of n: 143
Enter ciphertext: 7
Decrypted plaintext:  123

Hacking using Java

So the above code can be written in java also. See the below code for hacking RSA with the help of Java −

Example

import java.util.Scanner;

public class RSAHacking {
   public static int[] findFactors(int n) {
      int[] factors = new int[2];
      for (int i = 2; i < n; i++) {
         if (n % i == 0) {
            factors[0] = i;
            factors[1] = n / i;
            break;
         }
      }
      return factors;
   }

   public static int calculateEulerFunction(int p, int q) {
      return (p - 1) * (q - 1);
   }

   public static int calculatePrivateKey(int e, int eulerValue) {
      for (int i = 2; i < eulerValue; i++) {
         if ((i * e) % eulerValue == 1) {
            return i;
         }
      }
      return -1; // No private key found
   }

   public static int modPow(int base, int exponent, int modulus) {
      int result = 1;
      base = base % modulus;
      while (exponent > 0) {
         if (exponent % 2 == 1) {
            result = (result * base) % modulus;
         }
         exponent = exponent >> 1;
         base = (base * base) % modulus;
      }
      return result;
   }

   public static void main(String[] args) {
      Scanner scanner = new Scanner(System.in);

      System.out.print("Enter value of e: ");
      int e = scanner.nextInt();
      System.out.print("Enter value of n: ");
      int n = scanner.nextInt();
      System.out.print("Enter ciphertext: ");
      int c = scanner.nextInt();

      scanner.close();

      int[] factors = findFactors(n);
      int eulerValue = calculateEulerFunction(factors[0], factors[1]);

      int d = calculatePrivateKey(e, eulerValue);
      int plainText = modPow(c, d, n);
      System.out.println("Decrypted plaintext: " + plainText);
   }
}

Following is the output of the above example −

Input/Output

Enter value of e: 7
Enter value of n: 143
Enter ciphertext: 7
Decrypted plaintext: 123

Hacking Methods

• Brute Force − Trying every possible key until you find the correct one. However, RSA keys are very big, making brute force impractical.
• Factoring − Attempting to divide the modulus (N) into prime factors in order to get the private key. This is challenging with large prime numbers.
• Timing Attacks − Using alterations to the time spent by encryption or decryption processes to access sensitive data.
• Side-channel attacks − These attacks target information leaked by the encryption algorithm”s physical implementation, like power consumption or electromagnetic emissions.
• Quantum Computing − In principle, quantum computers can crack RSA encryption more efficiently using algorithms like Shor”s algorithm. However, practical quantum computers able to break RSA are currently unavailable.

Summary

Overall, breaking RSA encryption is difficult due to the complexities of factoring large integers. However, researchers are always developing new strategies and technologies to improve encryption systems and secure against attacks in the future.