RSA实现在C＃ [英] RSA Implementation in C#

问题描述

我想实现RSA算法在C＃。 。试图复制的RSA-100或更大，其中p和q是非常大时，下面的代码工作当p和q是小，但不

例如当：

  p = 61，q = 53，N = 3233，披（N）= 3120，E = 17，D = 2753 
  

一旦被解密，我得到了正确的原始一封邮件。我从 RSA维基百科页面得到这些值。代码也适用于p和q的其他小的值。

然而，使用 RSA-100 或更高，我不回来我的原始邮件。我一直在使用的不同值指数（e）尝试并确保它是互质与皮皮（n）的，但我不能得到正确的结果。我失去了一些东西简单/显而易见的？

感谢您提前为您的帮助！

  // p和q的RSA-100 
 //串p =37975227936943673922808872755445627854565536638199; 
 //串Q =40094690950920881030683735292761468389214899724061; 
 
串P =61; 
串Q =53; 
 
 //字符串转换为BigInteger的
的BigInteger rsa_p = BigInteger.Parse（P）; 
的BigInteger rsa_q = BigInteger.Parse（Q）; 
 
 // N = P * Q 
的BigInteger rsa_n = BigInteger.Multiply（rsa_p，rsa_q）; 
 
 //披（N）=（P-1）*（Q-1）
的BigInteger rsa_fn = BigInteger.Multiply（（rsa_p  -  1），（rsa_q  -  1））; 
 
的BigInteger rsa_e = 17; 
 
 //计算ð
的BigInteger rsa_d = BigInteger.ModPow（rsa_e，（rsa_fn  -  1），rsa_fn）; 
 
 //加密该消息，在这种情况下，3007 
 // C =（3007 ^ rsa_e）模rsa_n 
的BigInteger C = BigInteger.ModPow（3007，rsa_e，rsa_n） ; 
 
 //解密消息，男应该等于3007 
 // M =（3007 ^ rsa_d）模rsa_n 
的BigInteger M = BigInteger.ModPow（C，rsa_d，rsa_n） ; 
  

解决方案

D = E ^（PHI（N）-1 ）MOD岛（N）看起来我错了。您也需要D = E ^（PHI（PHI（N）） - 1）。MOD岛（N），或者你可以使用扩展欧几里德

I am trying to implement the RSA Algorithm in C#. The code below works when p and q are small, but not when trying to replicate RSA-100 or greater where p and q are very large.

For example when:

p = 61, q = 53, n = 3233, phi(n) = 3120, e = 17, d = 2753

Once decrypted, I get the correct original messsage. I got these values from the RSA Wikipedia page. The code also works for other small values of p and q.

However, when using RSA-100 or greater, I do not get back my original message. I have tried using different values for the exponent (e) and made sure it is coprime with phi(n) but I cannot get the correct result. Am I missing something simple/obvious?

Thank you in advance for your help!

//p and q for RSA-100
//string p = "37975227936943673922808872755445627854565536638199";
//string q = "40094690950920881030683735292761468389214899724061";

string p = "61";
string q = "53";

//Convert string to BigInteger
BigInteger rsa_p = BigInteger.Parse(p);
BigInteger rsa_q = BigInteger.Parse(q);

//n = p * q
BigInteger rsa_n = BigInteger.Multiply(rsa_p, rsa_q);

//phi(n) = (p-1)*(q-1)
BigInteger rsa_fn = BigInteger.Multiply((rsa_p - 1), (rsa_q - 1));

BigInteger rsa_e = 17;

//Compute d
BigInteger rsa_d = BigInteger.ModPow(rsa_e, (rsa_fn - 1), rsa_fn);

//Encrypt the message, in this case 3007
//C = (3007^rsa_e) mod rsa_n
BigInteger C = BigInteger.ModPow(3007, rsa_e, rsa_n);

//Decrypt the message, M should equal 3007
//M = (3007^rsa_d) mod rsa_n
BigInteger M = BigInteger.ModPow(C, rsa_d, rsa_n);

解决方案

d=e^(phi(n)-1) mod phi(n) looks wrong to me. You either need d=e^(phi(phi(n))-1) mod phi(n), or you could use extended euclid.

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