RSA算法在Java中的实现不正常

Question

理论上我知道，如果n=33、e（公钥）=3和d（私钥）=7我可以使用biginteger类用modpow(e,n)加密明文并用modpow(d,n)解密，但解密后明文与第一次不一样。

下面是我的代码：

  public class KeyTest {
private BigInteger n = new BigInteger("33");
private BigInteger e = new BigInteger("3");
private BigInteger d = new BigInteger("7");

public static void main(String[] args) {
    KeyTest test = new KeyTest();

    BigInteger plaintext = new BigInteger("55");
    System.out.println("Plain text: " + plaintext);

    BigInteger ciphertext = test.encrypt(plaintext);
    System.out.println("Ciphertext: " + ciphertext);

    BigInteger decrypted = test.decrypt(ciphertext);
    System.out.println("Plain text after decryption: " + decrypted);
}

public BigInteger encrypt(BigInteger plaintext) {

    return plaintext.modPow(e, n);
}

public BigInteger decrypt(BigInteger ciphertext) {

    return ciphertext.modPow(d, n);
}
}

输出为：

Plain text: 55 Ciphertext: 22 Plain text after decryption: 22

Answer 1

您的明文(55)大于模数(33)，因此实际上无法加密消息。考虑以下略有不同的示例：

• p=11
• Q=17
• n=187
• phi(n)=160
• 选择E=3
• 如果d=107则e*d=321=1 mod phi(n)

因此将代码更改为：

  private BigInteger n = new BigInteger("187");
  private BigInteger e = new BigInteger("3");
  private BigInteger d = new BigInteger("107");

  public static void main(String[] args) {
    KeyTest test = new KeyTest();

    BigInteger plaintext = new BigInteger("55");
    System.out.println("Plain text: " + plaintext);

    BigInteger ciphertext = test.encrypt(plaintext);
    System.out.println("Ciphertext: " + ciphertext);

    BigInteger decrypted = test.decrypt(ciphertext);
    System.out.println("Plain text after decryption: " + decrypted);
  }

  public BigInteger encrypt(BigInteger plaintext) {

    return plaintext.modPow(e, n);
  }

  public BigInteger decrypt(BigInteger ciphertext) {

    return ciphertext.modPow(d, n);
  }
}

输出：

Plain text: 55
Ciphertext: 132
Plain text after decryption: 55