Cracking short RSA keys

问题:

Given the following RSA keys, how does one go about determining what the values of p and q are?

Public Key: (10142789312725007, 5)
Private Key: (10142789312725007, 8114231289041741)

回答1:

Your teacher gave you:

Public Key: (10142789312725007, 5)

which means

n = 10142789312725007
e = 5 

where n is the modulus and e is the public exponent.

In addition, you're given

Private Key: (10142789312725007, 8114231289041741)

meaning that

 d = 8114231289041741

where d is the decryption exponent that should remain secret.

You can "break" RSA by knowing how to factor "n" into its "p" and "q" prime factors:

n = p * q

The easiest way is probably to check all odd numbers starting just below the square root of n:

Floor[Sqrt[10142789312725007]] = 100711415

You would get the first factor in 4 tries:

10142789312725007 mod 100711415 = 100711367
10142789312725007 mod 100711413 = 100711373
10142789312725007 mod 100711411 = 100711387
10142789312725007 mod 100711409 = 0 <-- Winner since it evenly divides n

So we have

 p = 100711409

Now,

 q = n / p 
   = 10142789312725007 / 100711409
   = 100711423

Why is this important? It's because d is a special number such that

d = e^-1 mod phi(n)
  = e^-1 mod (p-1)*(q-1)

We can verify this

d * e = 40571156445208705 = 1 mod 10142789111302176

This is important because if you have a plaintext message m then the ciphertext is

c = m^e mod n

and you decrypt it by

m = c^d = (m^e)^d = (m^(e*d)) = (m^(e*e^-1)) = m^1 (mod n)

For example, I can "encrypt" the message 123456789 using your teacher's public key:

m = 123456789

This will give me the following ciphertext:

c = m^e mod n 
  = 123456789^5 mod 10142789312725007
  = 7487844069764171

(Note that "e" should be much larger in practice because for small values of "m" you don't even exceed n)

Anyways, now we have "c" and can reverse it with "d"

m = c^d mod n
  = 7487844069764171^8114231289041741 mod 10142789312725007
  = 123456789

Obviously, you can't compute "7487844069764171^8114231289041741" directly because it has 128,808,202,574,088,302 digits, so you must use the modular exponentiation trick.

In the "Real World", n is obviously much larger. If you'd like to see a real example of how HTTPS uses RSA under the covers with a 617-digit n and an e of 65537, see my blog post "The First Few Milliseconds of an HTTPS Connection."

回答2:

Here's a relatively simple way to look at it (and one that is doable by hand). If you were to factor the number completely, then the highest factor you would need to consider is sqrt(N):

sqrt(10142789312725007) = 100711415.9999997567

The first prime below this is 100711409, just 6 below the sqrt(N).

10142789312725007 / 100711409 = 100711423 

therefore these are two factors of N. Your professor made it pretty easy - the trick is to recognize that no one would choose a small p or q so starting your check from the bottom (as in the python script someone posted) is a bad idea. If it's going to be practical by hand, the large p and q must lie near sqrt(N).

回答3:

There are various fast algorithms to solve the problem of factoring n given n, e, and d. You can find a good description of one such algorithm in the Handbook of Applied Cryptography, Chapter 8, section 8.2.2. You can find these chapters online for free download here.

回答4:

Wolframalpha tells me that the factors are 100711409 and 100711423

I just wrote a naive Python script to bruteforce it. As amdfan pointed out, starting from the top is a better approach:

p = 10142789312725007
for i in xrange(int(p**0.5+2), 3, -2):
    if p%i == 0:
        print i
        print p/i
        break

This could be heavily improved, but it still works without a problem. You could improve it by just testing primfactors, but for small values like yours this should be enough.

回答5:

The definition of RSA tells you that the modulus n = pq. You know n so you just need to find two numbers p and q that multiply to produce n. You know that p and q are prime, so this is the prime factorisation problem.

You can solve this by brute force for relatively small numbers but the overall security of RSA depends on the fact that this problem is intractable in general.

回答6:

Here is a Java implementation of the fast factoring method from the Handbook of Applied Cryptography chapter 8 section 8.2.2 (thanks to GregS for finding it):

/**
 * Computes the factors of n given d and e.
 * Given are the public RSA key (n,d)
 * and the corresponding private RSA key (n,e).
 */
public class ComputeRsaFactors
{
    /**
     * Executes the program.
     *
     * @param args  The command line arguments.
     */
    public static void main(String[] args)
    {
        final BigInteger n = BigInteger.valueOf(10142789312725007L);
        final BigInteger d = BigInteger.valueOf(5);
        final BigInteger e = BigInteger.valueOf(8114231289041741L);

        final long t0 = System.currentTimeMillis();

        final BigInteger kTheta = d.multiply(e).subtract(BigInteger.ONE);
        final int exponentOfTwo = kTheta.getLowestSetBit();

        final Random random = new Random();
        BigInteger factor = BigInteger.ONE;
        do
        {
            final BigInteger a = nextA(n, random);

            for (int i = 1; i <= exponentOfTwo; i++)
            {
                final BigInteger exponent = kTheta.shiftRight(i);
                final BigInteger power = a.modPow(exponent, n);

                final BigInteger gcd = n.gcd(power.subtract(BigInteger.ONE));
                if (!factor.equals(BigInteger.ONE))
                {
                    break;
                }
            }
        }
        while (factor.equals(BigInteger.ONE));

        final long t1 = System.currentTimeMillis();

        System.out.printf("%s %s (%dms)\n", factor, n.divide(factor), t1 - t0);
    }


    private static BigInteger nextA(final BigInteger n, final Random random)
    {
        BigInteger r;
        do
        {
            r = new BigInteger(n.bitLength(), random);
        }
        while (r.signum() == 0 || r.compareTo(n) >= 0);
        return r;
    }
}

A typical output is

100711423 100711409 (3ms)

回答7:

These two papers could possibly come in useful

• Andrej Dujella - A Variant of Wiener's Attack on RSA
• Andrej Dujella - Continued Fractions and RSA with small secret exponent

Came across them when I was doing some basic research on continued fractions.

回答8:

The algorithm to do this is (and this will work for any example, not only this small one that can be factored easily by any computer):

ed - 1 is a multiple of phi(n) = (p-1)(q-1), so is at least a multiple of 4.
ed - 1 can be computed as 40571156445208704 which equals 2^7 * 316962159728193, and we call s=7 and t = 316962159728193. (in general: any even number is a power of 2 times an odd number). Now pick a in [2,n-1) at random, and compute (by successive squaring modulo n) the sequence a^t (mod n), a^(2t) (mod n), a^(4t) (mod n).. until at most a^((2^7)*t) (mod n), where the last one is guaranteed to be 1, by the construction of e and d.

We now look for the first 1 in that sequence. The one before it will either be +1 or -1 (a trivial root of 1, mod n) and we redo with a different a, or some number x which does not equal +1 or -1 mod n. In the latter case gcd(x-1, n) is a non-trivial divisor of n, and so p or q, and we are done. One can show that a random a will work with probability about 0.5, so we need a few tries, but not very many in general.

回答9:

Sorry for the necromancy, but a friend asked me about this, and after pointing him here, I realized that I didn't really like any of the answers. After factoring the modulus and getting the primes (p and q), you need to find the totient, which is (p-1)*(q-1).

Now, to find the private exponent, you find the inverse of the public exponent mod the totient.

public_exponent * private_exponent = 1 mod totient

And now you have your private key, that easy. All of this except for the factorization can be done almost instantly for huge integers.

I wrote some code:

// tinyrsa.c
//
// apt-get install libgmp-dev
// yum install gmp-devel
//
// gcc tinyrsa.c -o tinyrsa -lm -lgmp

#include<stdio.h>
#include<gmp.h>

int main()
{
  // declare some multi-precision integers
  mpz_t pub_exp, priv_exp, modulus, totient, fac_p, fac_q, next_prime;

  mpz_init_set_str(pub_exp,"5",10);
  mpz_init_set_str(modulus,"10142789312725007",10);

  mpz_init(priv_exp);
  mpz_init(totient);
  mpz_init(fac_p);
  mpz_init(fac_q);

  // now we factor the modulus (the hard part)
  mpz_init(next_prime);
  mpz_sqrt(next_prime,modulus);
  unsigned long removed=0;
  while(!removed)
  {
    mpz_nextprime(next_prime,next_prime);
    removed=mpz_remove(fac_p,modulus,next_prime);
  }

  mpz_remove(fac_q,modulus,fac_p);
  // we now have p and q

  // the totient is (p-1)*(q-1)  
  mpz_t psub, qsub;
  mpz_init(psub);
  mpz_init(qsub);

  mpz_sub_ui(psub,fac_p,1);
  mpz_sub_ui(qsub,fac_q,1);
  mpz_mul(totient,psub,qsub);

  // inverse of the public key, mod the totient..
  mpz_invert(priv_exp,pub_exp,totient);

  gmp_printf("private exponent:\n%Zd\n",priv_exp);

}

The factorization algorithm I used is stupid, but concise, so grain of salt there. In this particular example the code runs almost instantly, but that is largely because the instructor in question provided an example that uses two primes in a row, which isn't really realistic for RSA.

If you wanted to cut out my stupid iterative search, you could put in some real factorization algorithm, and factor keys likely up to around 256 bits in a reasonable amount of time.

回答10:

I suggest you read about the Quadratic Sieve. If you implement one yourself, this is surely worth the credit. If you understand the principles, you already gained something.

回答11:

You need to factorize the modulus, that's the first parameter of the public key, 10142789312725007. Brute force will do (check every odd number from 3 to sqrt(n) if it's a factor), although more sophisticated/fast algorithms exist.

Since the number is too big to fit into a conventional integer (even 64-bit), you might want a numeric library that supports arbitrary-lenth integers. For C, there's GMP and MPIR (more Windows-friendly). For PHP, there's Bignum. Python comes with a built-in one - the built-in integer datatype is already arbitrary-length.

回答12:

There is a lot of bad speculation about factorization of large semi primes which go into brute force or sieving neither of which is required to factorise the semi prime. 64 bit takes 1 - 2 seconds on my pc, and 256 bit generally less than 2 days