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tanszek:oktatas:techcomm:rsa_encryption [2024/10/07 13:28] – [RSA] kneheztanszek:oktatas:techcomm:rsa_encryption [2024/11/26 08:25] (current) – [Example] knehez
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 ==== The Basic Model of Public Key Systems ==== ==== The Basic Model of Public Key Systems ====
- 
-=== Basic communication model === 
  
 {{:tanszek:oktatas:techcomm:pasted:20241007-131716.png}} {{:tanszek:oktatas:techcomm:pasted:20241007-131716.png}}
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 2.) **Compute \( N \)** (the modulus): 2.) **Compute \( N \)** (the modulus):
-   \[+ 
 +\[
    N = p \times q = 61 \times 53 = 3233    N = p \times q = 61 \times 53 = 3233
-   \]+\]
 So, \( N = 3233 \). So, \( N = 3233 \).
  
 3.) **Compute Euler’s totient function** \( \phi(N) \): 3.) **Compute Euler’s totient function** \( \phi(N) \):
-   \[+ 
 +\[
    \phi(N) = (p - 1) \times (q - 1) = (61 - 1) \times (53 - 1) = 60 \times 52 = 3120    \phi(N) = (p - 1) \times (q - 1) = (61 - 1) \times (53 - 1) = 60 \times 52 = 3120
-   \]+\]
 So, \( \phi(N) = 3120 \). So, \( \phi(N) = 3120 \).
  
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 5. **Calculate the private exponent** \( d \), which is the modular multiplicative inverse of \( e \mod \phi(N) \): 5. **Calculate the private exponent** \( d \), which is the modular multiplicative inverse of \( e \mod \phi(N) \):
-   \[+\[
    d \times e \equiv 1 \mod \phi(N)    d \times e \equiv 1 \mod \phi(N)
-   \]+\]
 Using the extended Euclidean algorithm, we find that \( d = 2753 \). Using the extended Euclidean algorithm, we find that \( d = 2753 \).
  
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 2.) **Encryption formula**:  2.) **Encryption formula**: 
-   \[+\[
    c = m^e \mod N    c = m^e \mod N
-   \]+\]
    * \( m = 65 \), \( e = 17 \), \( N = 3233 \).    * \( m = 65 \), \( e = 17 \), \( N = 3233 \).
    * Compute \( 65^{17} \mod 3233 \):    * Compute \( 65^{17} \mod 3233 \):
  
 Using modular exponentiation: Using modular exponentiation:
-   \[+\[
    65^{17} \mod 3233 = 2790    65^{17} \mod 3233 = 2790
-   \]+\]
  
 So, the **ciphertext** \( c = 2790 \). So, the **ciphertext** \( c = 2790 \).
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 1.) **Private key** is \( (d, N) = (2753, 3233) \). 1.) **Private key** is \( (d, N) = (2753, 3233) \).
 +
 2.) **Decryption formula**: 2.) **Decryption formula**:
-   \[+\[
    m = c^d \mod N    m = c^d \mod N
-   \]+\]
    * \( c = 2790 \), \( d = 2753 \), \( N = 3233 \).    * \( c = 2790 \), \( d = 2753 \), \( N = 3233 \).
    * Compute \( 2790^{2753} \mod 3233 \):    * Compute \( 2790^{2753} \mod 3233 \):
  
 Again, using modular exponentiation: Again, using modular exponentiation:
-   \[+\[
    2790^{2753} \mod 3233 = 65    2790^{2753} \mod 3233 = 65
-   \]+\]
  
 Thus, Alice successfully decrypts the message and retrieves the original plaintext **65**. Thus, Alice successfully decrypts the message and retrieves the original plaintext **65**.
  
-### Summary of the RSA Example:+=== Summary of the RSA Example: ===
  
   * Public key: \( (e = 17, N = 3233) \)   * Public key: \( (e = 17, N = 3233) \)
tanszek/oktatas/techcomm/rsa_encryption.1728307735.txt.gz · Last modified: 2024/10/07 13:28 by knehez