(Note that "e" should be much larger in practice because for small values of "m" you don't even exceed n)Īnyways, now we have "c" and can reverse it with "d" m = c^d mod n This will give me the following ciphertext: c = m^e mod n This is important because if you have a plaintext message m then the ciphertext is c = m^e mod nĪnd you decrypt it by m = c^d = (m^e)^d = (m^(e*d)) = (m^(e*e^-1)) = m^1 (mod n)įor example, I can "encrypt" the message 123456789 using your teacher's public key: m = 123456789 Why is this important? It's because d is a special number such that d = e^-1 mod phi(n) The easiest way is probably to check all odd numbers starting just below the square root of n: Floor] = 100711415 You can "break" RSA by knowing how to factor "n" into its "p" and "q" prime factors: n = p * q Where d is the decryption exponent that should remain secret. Where n is the modulus and e is the public exponent.
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