RSA算法实现在java中无法正常工作

Question

理论上我知道如果n=33、e(public key)=3 和d(private key)=7 我可以使用BigInteger 类和modPow(e, n) 加密plaintext，并使用modPow(d,n) 解密，但在解密plaintext 之后和第一个不一样。

这是我的代码：

  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);
}
}

输出是：

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

Answer 1

您的明文 (55) 大于模数 (33)，因此您实际上无法加密消息。考虑以下稍微不同的示例：

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

所以把你的代码改成：

  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