Calculate primes p and q from private exponent (d), public exponent (e) and the modulus (n)

How do I calculate the p and q parameters from e (publickey), d (privatekey) and modulus?

I have BigInteger keys at hand I can copy paste into code. One publickey, one privatekey and a modulus.

I need to calculate the RSA parameters p and q from this. But I suspect there is a library for that which I was unable to find with google. Any ideas? Thanks.

This does not have to be brute force, since I'm not after the private key. I just have a legacy system which stores a public, private key pair and a modulus and I need to get them into c# to use with RSACryptoServiceProvider.

So it comes down to calculating (p+q) by

public BigInteger _pPlusq()
    {
        int k = (this.getExponent() * this.getD() / this.getModulus()).IntValue();

        BigInteger phiN = (this.getExponent() * this.getD() - 1) / k;

        return phiN - this.getModul开发者_如何学Cus() - 1;

    }

but this doesn't seem to work. Can you spot the problem?

5 hours later... :)

Ok. How can I select a random number out of Zn* (http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n) in C#?

Let's assume that e is small (that's the common case; the Traditional public exponent is 65537). Let's also suppose that ed = 1 mod phi(n), where phi(n) = (p-1)(q-1) (this is not necessarily the case; the RSA requirements are that ed = 1 mod lcm(p-1,q-1) and phi(n) is only a multiple of lcm(p-1,q-1)).

Now you have ed = k*phi(n)+1 for some integer k. Since d is smaller than phi(n), you know that k < e. So you only have a small number of k to try. Actually, phi(n) is close to n (the difference being on the order of sqrt(n); in other words, when written out in bits, the upper half of phi(n) is identical to that of n) so you can compute k' with: k'=round(ed/n). k' is very close to k (i.e. |k'-k| <= 1) as long as the size of e is no more than half the size of n.

Given k, you easily get phi(n) = (ed-1)/k. It so happens that:

phi(n) = (p-1)(q-1) = pq - (p+q) + 1 = n + 1 - (p+q)

Thus, you get p+q = n + 1 - phi(n). You also have pq. It is time to remember that for all real numbers a and b, a and b are the two solutions of the quadratic equation X²-(a+b)X+ab. So, given p+q and pq, p and q are obtained by solving the quadratic equation:

p = ((p+q) + sqrt((p+q)² - 4*pq))/2

q = ((p+q) - sqrt((p+q)² - 4*pq))/2

In the general case, e and d may have arbitrary sizes (possibly greater than n), because all that RSA needs is that ed = 1 mod (p-1) and ed = 1 mod (q-1). There is a generic (and fast) method which looks a bit like the Miller-Rabin primality test. It is described in the Handbook of Applied Cryptography (chapter 8, section 8.2.2, page 287). That method is conceptually a bit more complex (it involves modular exponentiation) but may be simpler to implement (because there is no square root).

There is a procedure to recover p and q from n, e and d described in NIST Special Publication 800-56B R1 Recommendation for Pair-Wise Key Establishment Schemes Using Integer Factorization Cryptography in Appendix C.

The steps involved are:

1. Let k = de – 1. If k is odd, then go to Step 4.
2. Write k as k = 2^tr, where r is the largest odd integer dividing k, and t ≥ 1. Or in simpler terms, divide k repeatedly by 2 until you reach an odd number.
3. For i = 1 to 100 do:
   1. Generate a random integer g in the range [0, n−1].
   2. Let y = g^r mod n
   3. If y = 1 or y = n – 1, then go to Step 3.1 (i.e. repeat this loop).
   4. For j = 1 to t – 1 do:
      1. Let x = y² mod n
      2. If x = 1, go to (outer) Step 5.
      3. If x = n – 1, go to Step 3.1.
      4. Let y = x.
   5. Let x = y² mod n
   6. If x = 1, go to (outer) Step 5.
   7. Continue
4. Output “prime factors not found” and stop.
5. Let p = GCD(y – 1, n) and let q = n/p
6. Output (p, q) as the prime factors.

I recently wrote an implementation in Java. Not directly useful for C# I realise, but perhaps it can be easily ported:

// Step 1: Let k = de – 1. If k is odd, then go to Step 4
BigInteger k = d.multiply(e).subtract(ONE);
if (isEven(k)) {

    // Step 2 (express k as (2^t)r, where r is the largest odd integer
    // dividing k and t >= 1)
    BigInteger r = k;
    BigInteger t = ZERO;

    do {
        r = r.divide(TWO);
        t = t.add(ONE);
    } while (isEven(r));

    // Step 3
    Random random = new Random();
    boolean success = false;
    BigInteger y = null;

    step3loop: for (int i = 1; i <= 100; i++) {

        // 3a
        BigInteger g = getRandomBi(n, random);

        // 3b
        y = g.modPow(r, n);

        // 3c
        if (y.equals(ONE) || y.equals(n.subtract(ONE))) {
            // 3g
            continue step3loop;
        }

        // 3d
        for (BigInteger j = ONE; j.compareTo(t) <= 0; j = j.add(ONE)) {
            // 3d1
            BigInteger x = y.modPow(TWO, n);

            // 3d2
            if (x.equals(ONE)) {
                success = true;
                break step3loop;
            }

            // 3d3
            if (x.equals(n.subtract(ONE))) {
                // 3g
                continue step3loop;
            }

            // 3d4
            y = x;
        }

        // 3e
        BigInteger x = y.modPow(TWO, n);
        if (x.equals(ONE)) {

            success = true;
            break step3loop;

        }

        // 3g
        // (loop again)
    }

    if (success) {
        // Step 5
        p = y.subtract(ONE).gcd(n);
        q = n.divide(p);
        return;
    }
}

// Step 4
throw new RuntimeException("Prime factors not found");

This code uses a few helper definitions/methods:

private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = BigInteger.valueOf(2);
private static final BigInteger ZERO = BigInteger.ZERO;

private static boolean isEven(BigInteger bi) {
    return bi.mod(TWO).equals(ZERO);
}

private static BigInteger getRandomBi(BigInteger n, Random rnd) {
    // From http://stackoverflow.com/a/2290089
    BigInteger r;
    do {
        r = new BigInteger(n.bitLength(), rnd);
    } while (r.compareTo(n) >= 0);
    return r;
}

I've adapted the Java code provided by Duncan in C#, if anyone is interested:

    public static void RecoverPQ(
        BigInteger n,
        BigInteger e,
        BigInteger d,
        out BigInteger p,
        out BigInteger q
        )
    {
        int nBitCount = (int)(BigInteger.Log(n, 2)+1);

        // Step 1: Let k = de – 1. If k is odd, then go to Step 4
        BigInteger k = d * e - 1;
        if (k.IsEven)
        {
            // Step 2 (express k as (2^t)r, where r is the largest odd integer
            // dividing k and t >= 1)
            BigInteger r = k;
            BigInteger t = 0;

            do
            {
                r = r / 2;
                t = t + 1;
            } while (r.IsEven);

            // Step 3
            var rng = new RNGCryptoServiceProvider();
            bool success = false;
            BigInteger y = 0;

            for (int i = 1; i <= 100; i++) {

                // 3a
                BigInteger g;
                do
                {
                    byte[] randomBytes = new byte[nBitCount / 8 + 1]; // +1 to force a positive number
                    rng.GetBytes(randomBytes);
                    randomBytes[randomBytes.Length - 1] = 0;
                    g = new BigInteger(randomBytes);
                } while (g >= n);

                // 3b
                y = BigInteger.ModPow(g, r, n);

                // 3c
                if (y == 1 || y == n-1) {
                    // 3g
                    continue;
                }

                // 3d
                BigInteger x;
                for (BigInteger j = 1; j < t; j = j + 1) {
                    // 3d1
                    x = BigInteger.ModPow(y, 2, n);

                    // 3d2
                    if (x == 1) {
                        success = true;
                        break;
                    }

                    // 3d3
                    if (x == n-1) {
                        // 3g
                        continue;
                    }

                    // 3d4
                    y = x;
                }

                // 3e
                x = BigInteger.ModPow(y, 2, n);
                if (x == 1) {

                    success = true;
                    break;

                }

                // 3g
                // (loop again)
            }

            if (success) {
                // Step 5
                p = BigInteger.GreatestCommonDivisor((y - 1), n);
                q = n / p;
                return;
            }
        }
        throw new Exception("Cannot compute P and Q");
    }

This uses the standard System.Numerics.BigInteger class.

This was tested by the following unit test:

BigInteger n = BigInteger.Parse("9086945041514605868879747720094842530294507677354717409873592895614408619688608144774037743497197616416703125668941380866493349088794356554895149433555027");
BigInteger e = 65537;
BigInteger d = BigInteger.Parse("8936505818327042395303988587447591295947962354408444794561435666999402846577625762582824202269399672579058991442587406384754958587400493169361356902030209");
BigInteger p;
BigInteger q;
RecoverPQ(n, e, d, out p, out q);
Assert.AreEqual(n, p * q);

I implemented the method described by Thomas Pornin.

The BigInteger class is Chew Keong TAN's C# version (check codeproject comments for bug fixes)

    /// EXAMPLE (Hex Strings)
    /// N(MODULUS) = "DB2CB41E112BACFA2BD7C3D3D7967E84FB9434FC261F9D090A8983947DAF8488D3DF8FBDCC1F92493585E134A1B42DE519F463244D7ED384E26D516CC7A4FF7895B1992140043AACADFC12E856B202346AF8226B1A882137DC3C5A57F0D2815C1FCD4BB46FA9157FDFFD79EC3A10A824CCC1EB3CE0B6B4396AE236590016BA69"
    /// D(PRIVATE EXPONENT) = "18B44A3D155C61EBF4E3261C8BB157E36F63FE30E9AF28892B59E2ADEB18CC8C8BAD284B9165819CA4DEC94AA06B69BCE81706D1C1B668EB128695E5F7FEDE18A908A3011A646A481D3EA71D8A387D474609BD57A882B182E047DE80E04B4221416BD39DFA1FAC0300641962ADB109E28CAF50061B68C9CABD9B00313C0F46ED"
    /// E(PUBLIC EXPONENT) = "010001"
    /// RESULTS: 
    /// DP = "899324E9A8B70CA05612D8BAE70844BBF239D43E2E9CCADFA11EBD43D0603FE70A63963FE3FFA38550B5FEB3DA870D2677927B91542D148FA4BEA6DCD6B2FF57"
    /// DQ = "E43C98265BF97066FC078FD464BFAC089628765A0CE18904F8C15318A6850174F1A4596D3E8663440115D0EEB9157481E40DCA5EE569B1F7F4EE30AC0439C637"
    /// INVERSEQ = "395B8CF3240C325B0F5F86A05ABCF0006695FAB9235589A56759ECBF2CD3D3DFDE0D6F16F0BE5C70CEF22348D2D09FA093C01D909D25BC1DB11DF8A4F0CE552"
    /// P = "ED6CF6699EAC99667E0AFAEF8416F902C00B42D6FFA2C3C18C7BE4CF36013A91F6CF23047529047660DE14A77D13B74FF31DF900541ED37A8EF89340C623759B"
    /// Q = "EC52382046AA660794CC1A907F8031FDE1A554CDE17E8AA216AEDC92DB2E58B0529C76BD0498E00BAA792058B2766C40FD7A9CC2F6782942D91471905561324B"

    public static RSACryptoServiceProvider CreateRSAPrivateKey(string mod, string privExponent, string pubExponent)
    {
        var rsa = new RSACryptoServiceProvider
        {
            PersistKeyInCsp = false
        };
        var n = new BigInteger(mod, 16);
        var d = new BigInteger(privExponent, 16);
        var e = new BigInteger(pubExponent, 16);

        var zero = new BigInteger(0);
        var one = new BigInteger(1);
        var two = new BigInteger(2);
        var four = new BigInteger(4);


        BigInteger de = e*d;
        BigInteger modulusplus1 = n + one;
        BigInteger deminus1 = de - one;
        BigInteger p = zero;
        BigInteger q = zero;

        BigInteger kprima = de/n;

        var ks = new[] {kprima, kprima - one, kprima + one};

        bool bfound = false;
        foreach (BigInteger k in ks)
        {
            BigInteger fi = deminus1/k;
            BigInteger pplusq = modulusplus1 - fi;
            BigInteger delta = pplusq*pplusq - n*four;

            BigInteger sqrt = delta.sqrt();
            p = (pplusq + sqrt)/two;
            if (n%p != zero) continue;
            q = (pplusq - sqrt)/two;
            bfound = true;
            break;
        }

        if (bfound)
        {
            BigInteger dp = d%(p - one);
            BigInteger dq = d%(q - one);

            BigInteger inverseq = q.modInverse(p);

            var pars = new RSAParameters
            {
                D = d.getBytes(),
                DP = dp.getBytes(),
                DQ = dq.getBytes(),
                Exponent = e.getBytes(),
                Modulus = n.getBytes(),
                P = p.getBytes(),
                Q = q.getBytes(),
                InverseQ = inverseq.getBytes()
            };
            rsa.ImportParameters(pars);
            return rsa;
        }

        throw new CryptographicException("Error generating the private key");
    }