Institute of Information Science - University of Miskolc

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==== Breaking RSA ====

The weak point of the RSA algorithm lies in the key generation: specifically, the difficulty of factoring \( N \) into its prime components. This factorisation is only possible through trial and error, and the algorithm remains secure until someone discovers a heuristic method to do it efficiently.

You wouldn't believe how tricky it is to find \( p \) and \( q \) when \( N = 77 \). Once you've figured it out, think about how you did it. What was the process? How did you come up with the solution?

The RSA Labs used to offer $200,000 for the factorization of 2048-bit numbers. However, they removed the prize because more recent research has shown that the method can be broken if the primes \( p \) and \( q \) have a special relationship.

The RSA Labs previously offered a $100,000 prize to anyone who could submit the prime factors of the following integer \( N \), that is, the two primes whose product is \( N \):

<code>
N = p x q = 25195908475657893494027183240048398571429282126204
03202777713783604366202070759555626401852588078440
69182906412495150821892985591491761845028084891200
72844992687392807287776735971418347270261896375014
97182469116507761337985909570009733045974880842840
17974291006424586918171951187461215151726546322822
16869987549182422433637259085141865462043576798423
38718477444792073993423658482382428119816381501067
48104516603773060562016196762561338441436038339044
14952634432190114657544454178424020924616515723350
77870774981712577246796292638635637328991215483143
81678998850404453640235273819513786365643912120103
97122822120720357
</code>

You can sense how difficult this task is: we are dealing with 617 digits, and unfortunately, the last digit is not even.

==== Example: Breaking RSA if we find \( p \) and \( q \) ====

=== Step 1: Given Data ===
Suppose we know the **public key** \( (e, N) \):
  * Public exponent \( e = 17 \)
  * Modulus \( N = 55 \)

To break the RSA encryption, we need to find the **private key** \( d \). This requires us to factor \( N \) into its prime factors \( p \) and \( q \).

=== Step 2: Factor \( N \) ===
We need to factor \( N = 55 \):
  * \( p = 5 \)
  * \( q = 11 \)
  
These are the two prime factors of \( N \).

==== Step 3: Calculate \( \phi(N) \) (Euler’s Totient Function) ====
Once we have \( p \) and \( q \), we can compute \( \phi(N) \), which is required to compute the private key \( d \).

\[
\phi(N) = (p - 1) \times (q - 1)
\]
\[
\phi(55) = (5 - 1) \times (11 - 1) = 4 \times 10 = 40
\]

==== Step 4: Find the Private Key \( d \) ====
The private key \( d \) is the modular inverse of \( e \mod \phi(N) \), which satisfies the equation:

\[
d \times e \equiv 1 \mod \phi(N)
\]

We need to find \( d \) such that:

\[
d \times 17 \equiv 1 \mod 40
\]

Using the extended Euclidean algorithm, we find that:

\[
d = 33
\]

So the **private key** is \( d = 33 \).

==== Step 5: Decrypt the Ciphertext ====
Now that we have \( d \), we can decrypt the ciphertext. Suppose the ciphertext is \( C = 18 \).

To decrypt, we use the formula:

\[
m = C^d \mod N
\]
\[
m = 18^{33} \mod 55
\]

To compute this efficiently, we can use modular exponentiation:

\[
18^{33} \mod 55 = 2
\]

So, the original **message** \( m = 2 \).

=== Summary: ===
  * We started with the public key \( (e = 17, N = 55) \) and a ciphertext \( C = 18 \).
  * After factoring \( N \) into \( p = 5 \) and \( q = 11 \), we calculated \( \phi(N) = 40 \) and found the private key \( d = 33 \).
  * Using \( d \), we decrypted the ciphertext \( C = 18 \) to recover the original message \( m = 2 \).

This example illustrates how RSA encryption can be broken if someone manages to factor \( N \) into its prime factors \( p \) and \( q \).