Cracking short RSA keys

Question

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)

Answer 1

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.

Answer 2

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."

Answer 3

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.

Answer 4

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

Answer 5

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)

Answer 6

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).