如何计算d从P，Q和E RSA加密(How to calculate D for RSA encry

我试图找到D使用P ， Q和E （ Dp ， Dq和(p ^-1 mod q)可得）。

根据这个答案和这个答案并更新了这个问题用下面的方法，我应该得到D 。

为了测试这个我生成的密钥对，并试图从现有的计算组件，并与原件比较的结果。所有的结果是，除了良好D 。有什么毛病我的计算，我从上面的答案复制。这将是巨大的，如果有人能告诉我，我做错了什么。

测试代码

using System;
using System.Numerics;
using System.Security.Cryptography;
using System.Text;

class Program {

    static RSAParameters key = new RSAParameters() {
        P = new byte[]{
                0xDE, 0xA6, 0x35, 0x0B, 0x0A, 0xA5, 0xD7, 0xA0, 0x5C, 0x49, 0xEA, 0xD1, 0x3F, 0xA6, 0xF5, 0x12, 
                0x19, 0x06, 0x25, 0x8A, 0xD9, 0xA7, 0x07, 0xE7, 0x0D, 0x8A, 0x7C, 0xB1, 0xD4, 0x81, 0x64, 0xFD, 
                0x04, 0xEC, 0x47, 0x33, 0x42, 0x0B, 0x22, 0xF2, 0x60, 0xBB, 0x75, 0x62, 0x53, 0x3E, 0x1A, 0x97, 
                0x9D, 0xEF, 0x25, 0xA7, 0xE5, 0x24, 0x3A, 0x30, 0x36, 0xA5, 0xF9, 0x8A, 0xF5, 0xFF, 0x1D, 0x1B
            },

        Q = new byte[]{
                0xBE, 0xB9, 0x60, 0x12, 0x05, 0xB1, 0x61, 0xD9, 0x22, 0xD8, 0x84, 0x6E, 0x9A, 0x7B, 0xD1, 0x9B, 
                0x17, 0xA5, 0xDD, 0x02, 0x5E, 0x9D, 0xD8, 0x24, 0x06, 0x1B, 0xF3, 0xD8, 0x2F, 0x79, 0xFE, 0x78, 
                0x74, 0x3D, 0xC4, 0xE6, 0x17, 0xD2, 0xB7, 0x68, 0x78, 0x6F, 0x53, 0xE0, 0x38, 0x00, 0x86, 0xFB, 
                0x20, 0x2A, 0x1B, 0xBD, 0x91, 0x76, 0x3E, 0x33, 0x85, 0x9A, 0x31, 0xE6, 0x88, 0x60, 0x91, 0x81
            },

        DP = new byte[]{
                0xAC, 0x28, 0x92, 0x6D, 0x46, 0x3F, 0x74, 0x1A, 0xA0, 0x21, 0xDB, 0xBB, 0x0E, 0xDF, 0xD7, 0x31, 
                0xB6, 0x3D, 0xC5, 0x7B, 0xB6, 0xCE, 0x6B, 0xD2, 0xE1, 0xEA, 0x8A, 0x7E, 0xAA, 0xD5, 0x9E, 0xB3, 
                0xF2, 0x41, 0x8C, 0xD0, 0x7A, 0xA9, 0xC7, 0xCC, 0xE8, 0xB5, 0x2A, 0x8F, 0xEB, 0xD3, 0xE2, 0x96, 
                0x07, 0xDD, 0xEA, 0x1D, 0x07, 0x96, 0x5A, 0x93, 0xFB, 0x3D, 0x9D, 0x56, 0x30, 0xDE, 0xA1, 0xAF
            },

        DQ = new byte[]{
                0xA6, 0x9C, 0x44, 0x1B, 0x9A, 0x53, 0x89, 0xD9, 0xE8, 0xC1, 0xE2, 0x76, 0xC8, 0x87, 0x6F, 0xE5, 
                0x1F, 0x74, 0x6A, 0xAC, 0x5E, 0x41, 0x5F, 0x86, 0xA0, 0xBB, 0x9C, 0x79, 0xF7, 0x87, 0x87, 0xD0, 
                0x6C, 0x23, 0x65, 0xB5, 0x67, 0x8C, 0x51, 0x62, 0x77, 0x0B, 0x31, 0xE7, 0x86, 0xA4, 0x97, 0x46, 
                0x1B, 0xA4, 0x0D, 0x55, 0xBE, 0x13, 0xE0, 0x64, 0x9B, 0xCA, 0xC6, 0xDA, 0xCF, 0xBA, 0x24, 0x81
            },

        InverseQ = new byte[]{
                0x02, 0x42, 0x90, 0xAE, 0xFF, 0xFE, 0xB6, 0xCB, 0x53, 0xFF, 0x96, 0x17, 0xC6, 0xE4, 0x3F, 0xE6, 
                0xC7, 0xBC, 0xB2, 0xEB, 0x53, 0xA9, 0x47, 0xEE, 0x10, 0x36, 0x98, 0xEF, 0xA8, 0x3E, 0x9C, 0xF7, 
                0xF9, 0xCF, 0x24, 0xE5, 0xD7, 0x9A, 0xAF, 0x09, 0xCF, 0x28, 0xAA, 0x5D, 0x2A, 0xB7, 0x27, 0x73, 
                0x47, 0x2D, 0x54, 0x54, 0x61, 0xC5, 0xCE, 0x3E, 0xA4, 0x91, 0xF6, 0x9D, 0xF4, 0x65, 0x08, 0xDD
            },

        Exponent = new byte[]{
                0x00, 0x01, 0x00, 0x01, 
            },

        Modulus = new byte[]{
                0xA5, 0xE0, 0x95, 0x08, 0x87, 0x69, 0x2B, 0xB4, 0x7F, 0x08, 0xFB, 0x4F, 0x66, 0x85, 0xD9, 0x95, 
                0x53, 0x0F, 0x7C, 0x99, 0x95, 0x16, 0xF4, 0x0D, 0xAD, 0x9E, 0x31, 0xD8, 0x20, 0xF4, 0x88, 0x63, 
                0xAE, 0x51, 0x04, 0xC2, 0xE9, 0x92, 0x3C, 0x1C, 0x90, 0xF8, 0xF4, 0x38, 0x6A, 0x86, 0xFD, 0x8F, 
                0xDE, 0x85, 0x22, 0xDD, 0xE8, 0x7E, 0x8D, 0xF2, 0xC5, 0xC9, 0x4E, 0x71, 0x2B, 0x56, 0x25, 0x1A, 
                0xEA, 0x66, 0x15, 0x19, 0x63, 0x70, 0x53, 0x79, 0xDF, 0x38, 0x49, 0x30, 0x74, 0x45, 0xBE, 0xA3, 
                0x28, 0x0D, 0x0E, 0x7A, 0x7D, 0xB6, 0x8B, 0xCA, 0x09, 0x56, 0x21, 0xE7, 0x98, 0x3E, 0x4B, 0x8B, 
                0xD0, 0x31, 0x27, 0x8E, 0x6F, 0x10, 0xA6, 0x6C, 0x1C, 0x48, 0xB5, 0x5E, 0x89, 0x7B, 0x74, 0x74, 
                0xB2, 0x57, 0x72, 0x6D, 0x18, 0xEB, 0xF3, 0xF5, 0x53, 0xCA, 0x8C, 0xBE, 0xB7, 0x29, 0xF5, 0x9B
            },

        D = new byte[]{
                0x9F, 0x86, 0xE1, 0x4D, 0x96, 0x8C, 0xFA, 0xCF, 0x57, 0xED, 0x17, 0x64, 0x41, 0x41, 0x31, 0x04, 
                0x7F, 0x21, 0x41, 0xBF, 0xA2, 0xB6, 0xB4, 0x78, 0x03, 0x25, 0x44, 0xE2, 0x8A, 0xAF, 0x22, 0x0C, 
                0x5B, 0xB4, 0xE7, 0x53, 0x5C, 0xB6, 0x9A, 0xC1, 0x0E, 0x5B, 0x9E, 0xE4, 0x32, 0xEF, 0x28, 0x24, 
                0x98, 0xE8, 0x89, 0xA3, 0xC8, 0xD9, 0x0D, 0x43, 0x12, 0x1C, 0x8C, 0x28, 0x22, 0x79, 0x72, 0xAC, 
                0x66, 0x7B, 0x7D, 0xD2, 0xF9, 0x48, 0x06, 0xCD, 0x9D, 0x9A, 0xE6, 0x42, 0x92, 0xBA, 0x56, 0xA6, 
                0x63, 0x07, 0x1E, 0x25, 0x4E, 0xC8, 0x07, 0x58, 0x5B, 0x88, 0x60, 0x97, 0x92, 0xE2, 0xD5, 0xB9, 
                0xC6, 0x70, 0xBB, 0x63, 0x5A, 0xC3, 0xC3, 0xA6, 0x46, 0x5A, 0x1C, 0x9C, 0xBF, 0x61, 0x57, 0x9E, 
                0x9E, 0xFA, 0xC0, 0xC4, 0x8A, 0xC2, 0xBA, 0x88, 0x46, 0xA9, 0x7A, 0xF2, 0x7D, 0x4F, 0x6C, 0x01
            }
    };

    public static BigInteger FromBigEndian(byte[] p) {
        Array.Reverse(p);
        if (p[p.Length - 1] > 127) {
            Array.Resize(ref p, p.Length + 1);
            p[p.Length - 1] = 0;
        }
        return new BigInteger(p);
    }

    static void Main(string[] args) {

        using (RSACryptoServiceProvider rsa = new RSACryptoServiceProvider() { PersistKeyInCsp = false }) {
            rsa.ImportParameters(key);

            Console.Write("Testing Encrypt/Decrypt ... ");
            string message = "Testing Some Data to Encrypt";
            byte[] buffer = Encoding.ASCII.GetBytes(message);
            byte[] encoded = rsa.Encrypt(buffer, true);
            byte[] decoded = rsa.Decrypt(encoded, true);
            string message1 = ASCIIEncoding.ASCII.GetString(decoded);

            if (message == message1) {
                Console.WriteLine("Ok :)");
            } else {
                Console.WriteLine("Bad Encryption :(");
                Console.ReadKey();
                return;
            }
        }

        //Convert Key to BigIntegers
        BigInteger P = FromBigEndian(key.P);
        BigInteger Q = FromBigEndian(key.Q);
        BigInteger DP = FromBigEndian(key.DP);
        BigInteger DQ = FromBigEndian(key.DQ);
        BigInteger InverseQ = FromBigEndian(key.InverseQ);
        BigInteger E = FromBigEndian(key.Exponent);
        BigInteger M = FromBigEndian(key.Modulus);
        BigInteger D = FromBigEndian(key.D);


        Console.WriteLine("Testing Numbers ... ");
        BigInteger M1 = BigInteger.Multiply(P, Q); // M = P*Q
        if (M1.CompareTo(M) == 0) {
            Console.WriteLine("  M Ok :)");
        } else {
            Console.WriteLine("  Bad M:(");
            Console.ReadKey();
            return;
        }

        BigInteger PMinus1 = BigInteger.Subtract(P, BigInteger.One); // M = P*Q
        BigInteger DP1 = BigInteger.Remainder(D, PMinus1); // M = P*Q
        if (DP1.CompareTo(DP) == 0) {
            Console.WriteLine("  DP Ok :)");
        } else {
            Console.WriteLine("  Bad DP :(");
            Console.ReadKey();
            return;
        }

        BigInteger QMinus1 = BigInteger.Subtract(Q, BigInteger.One); // M = P*Q
        BigInteger DQ1 = BigInteger.Remainder(D, QMinus1); // M = P*Q
        if (DQ1.CompareTo(DQ) == 0) {
            Console.WriteLine("  DQ Ok :)");
        } else {
            Console.WriteLine("  Bad DQ :(");
            Console.ReadKey();
            return;
        }

        BigInteger Phi = BigInteger.Multiply(PMinus1, QMinus1);
        BigInteger PhiMinus1 = BigInteger.Subtract(Phi, BigInteger.One);
        BigInteger D1 = BigInteger.ModPow(E, PhiMinus1, Phi);
        if (D1.CompareTo(D) == 0) {
            Console.WriteLine("  D Ok :)");
        } else {
            Console.WriteLine("  Bad D :(");
            Console.ReadKey();
            return;
        }

        Console.ReadKey();
    }
}

测试结果

Testing Encrypt/Decrypt ... Ok :)
Testing Numbers ...
  M Ok :)
  DP Ok :)
  DQ Ok :)
  Bad D :(

Answer 1:

首先，你需要确认GCD(e, φ) = 1 ，因为d只存在如果财产持有。然后，计算出的模反元素e模phi ，我在介绍我的回答“1 / BigInteger的在C＃” 。

你的代码似乎假设e^(φ(n)-1) mod φ(n)是逆，但是这是不正确。我认为正确的公式将是e^(φ(φ(n))-1) mod φ(n) ，但是这是使用不方便，因为你只知道φ(n)但不φ(φ(n))

我建议通过移植维基百科伪代码C＃中使用扩展欧几里德算法。

作为一个侧面说明：对于常为多发的相当值d ，因为你不需要e*d mod φ(n)=1 ，但只是e*d mod λ(n)=1其中λ是卡迈克尔功能看“为什么RSA加密密钥是基于模（PHI（N）），而不是crypto.SE n为模”

Answer 2:

扩展欧几里德算法可以用来计算模逆，使用此链接： http://www.di-mgt.com.au/euclidean.html#extendedeuclidean获得细节，我在C＃测试的源代码如下，其结果是匹配，

public static BigInteger modinv(BigInteger u, BigInteger v)
{
   BigInteger inv, u1, u3, v1, v3, t1, t3, q;
   BigInteger iter;
   /* Step X1. Initialise */
   u1 = 1;
   u3 = u;
   v1 = 0;
   v3 = v;
   /* Remember odd/even iterations */
   iter = 1;
   /* Step X2. Loop while v3 != 0 */
   while (v3 != 0)
   {
       /* Step X3. Divide and "Subtract" */
       q = u3 / v3;
       t3 = u3 % v3;
       t1 = u1 + q * v1;
       /* Swap */
       u1 = v1; v1 = t1; u3 = v3; v3 = t3;
       iter = -iter;
   }
   /* Make sure u3 = gcd(u,v) == 1 */
   if (u3 != 1)
       return 0;   /* Error: No inverse exists */
       /* Ensure a positive result */
       if (iter < 0)
           inv = v - u1;
       else
           inv = u1;
       return inv;
}

Answer 3:

d可由下式计算：

    var qq = BigInteger.Multiply(totient, n);
    var qw = BigInteger.Multiply(totient, qq);
    BigInteger d = BigInteger.ModPow(e, (qw - 1), totient);

Answer 4:

 Console.Write("Testing Encrypt/Decrypt using BigInteger ");
        string message2 = "Testing Some Data to Encrypt";
        byte[] buffer2 = Encoding.ASCII.GetBytes(message2);
        BigInteger m = new BigInteger(buffer2);
        BigInteger c = BigInteger.ModPow(m, E, M); //encrypt
        BigInteger m2 = BigInteger.ModPow(c, D, M); //decrypt, m2 also equals m
        byte[] decoded2 = m2.ToByteArray();

        if (decoded2[0] == 0) 
        {
            decoded2 = decoded2.Where(b => b != 0).ToArray();
        }
        string message3 = ASCIIEncoding.ASCII.GetString(decoded2);

        if (message2 == message3)
        {
            Console.WriteLine("Ok :)");
        }
        else
        {
            Console.WriteLine("Bad Encryption :(");
            Console.ReadKey();
            return;
        }

我与你parametrs尝试它和它的作品，所以E，d和M必须是有效的。