tanszek:oktatas:techcomm:breaking_rsa
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| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| tanszek:oktatas:techcomm:breaking_rsa [2024/10/07 15:12] – knehez | tanszek:oktatas:techcomm:breaking_rsa [2024/11/26 08:23] (current) – [Example: Breaking RSA if we find \( p \) and \( q \)] knehez | ||
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| To break the RSA encryption, we need to find the **private key** \( d \). This requires us to factor \( N \) into its prime factors \( p \) and \( q \). | To break the RSA encryption, we need to find the **private key** \( d \). This requires us to factor \( N \) into its prime factors \( p \) and \( q \). | ||
| - | #### Step 2: Factor \( N \) | + | === Step 2: Factor \( N \) === |
| We need to factor \( N = 55 \): | We need to factor \( N = 55 \): | ||
| * \( p = 5 \) | * \( p = 5 \) | ||
| Line 94: | Line 94: | ||
| So, the original **message** \( m = 2 \). | So, the original **message** \( m = 2 \). | ||
| - | ### Summary: | + | === Summary: |
| * We started with the public key \( (e = 17, N = 55) \) and a ciphertext \( C = 18 \). | * We started with the public key \( (e = 17, N = 55) \) and a ciphertext \( C = 18 \). | ||
| * After factoring \( N \) into \( p = 5 \) and \( q = 11 \), we calculated \( \phi(N) = 40 \) and found the private key \( d = 33 \). | * After factoring \( N \) into \( p = 5 \) and \( q = 11 \), we calculated \( \phi(N) = 40 \) and found the private key \( d = 33 \). | ||
tanszek/oktatas/techcomm/breaking_rsa.1728313930.txt.gz · Last modified: 2024/10/07 15:12 by knehez