仅已知n、e、c的RSA解密替代方案技术咨询

Hey, I feel your pain—spending 24 hours brute-forcing an RSA modulus and getting nowhere is super frustrating, especially since this n looks like a pretty large one (around 150 digits, by my count). Brute force is basically a dead end here, so let's go through some practical alternative approaches you can try right now:

1. Use Efficient Factorization Algorithms Instead of Brute Force #

Forget naive brute search—Pollard's Rho Algorithm is the industry standard for factoring large integers efficiently, and it’s orders of magnitude faster than trying every possible prime. You can test it easily with Python’s sympy library, or use dedicated tools like YAFU. Here’s a quick snippet to run:

from sympy import factorint

n = 58900433780152059829684181006276669633073820320761216330291745734792546625247
factors = factorint(n)
print(f"Found factors: {factors}")

Also, quickly rule out trivial cases: check if n is a perfect square (meaning p=q, though this is rare) by computing the integer square root and squaring it to see if it matches n. Pollard's Rho will also catch if n is a smooth number (all prime factors are relatively small) automatically.

2. Test for Key-Specific Attacks #

Your e is 65537, the standard secure public exponent, but there are still a few angles to check:

• Wiener's Attack: This targets cases where the private exponent d is small relative to n. While 65537 is a large e, it’s still worth verifying with tools designed for this attack (like Python’s rsa-wiener-attack package) to rule it out entirely.
• Common Modulus Attack: If you can get another ciphertext encrypted with the same n but a different e, you could combine them to solve for the plaintext—but this requires additional data you don’t have right now.

3. Enumerate Likely Plaintexts (If You Have Context) #

If you have any idea what the plaintext might look like (e.g., it’s a short number, an English phrase, a formatted token), dictionary or targeted brute force can work:

• For example, if the plaintext is a 10-digit number or a common password, you can generate possible candidates, encrypt them with (e, n), and compare to c. Here’s a simple snippet for numeric plaintexts:

  n = 58900433780152059829684181006276669633073820320761216330291745734792546625247
  e = 65537
  c = 56191946659070299323432594589209132754159316947267240359739328886944131258862
  
  # Adjust the range based on your plaintext guess
  for m in range(0, 10**10):
      if pow(m, e, n) == c:
          print(f"Found plaintext: {m}")
          break
  

  This only works if the plaintext has a small, predictable search space—but it’s worth trying if you have any context about what the plaintext might be.

4. Check Public Factor Databases #

Many RSA moduli (especially those generated with weak randomness) end up in public factorization databases. You can search for your n in databases like FactorDB—if someone has already factored this n, you’ll get p and q instantly, which lets you compute φ(n) and then the private exponent d.

5. Side Channel Attacks (If Physical Access is Available) #

If this encryption was performed on a physical device (like a smart card or microcontroller), you could try side channel attacks—things like timing attacks (measuring how long decryption takes for different inputs) or power analysis (tracking power consumption during operations). But this requires physical access to the device, so it’s only relevant if you’re not just working with the numeric n/e/c values.

内容的提问来源于stack exchange，提问作者Teodor Vecerdi