1. Discrete Math for CS CMPSC 360 LECTURE 14 Last time:
Chinese Remainder Theorem
Today:
Public Key Cryptography: RSA
CMPSC
360
2/27/2019

2. Cryptography Secure communications between two parties. Applications:
Military, Intelligence (NSA)
E-Commerce
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3. Cryptography Alice encrypts message 𝑚 to 𝐸(𝑚) Alice send 𝐸(𝑚) to Bob
Bob decrypts secret message 𝐸(𝑚) to 𝐷(𝐸(𝑚)) = 𝑚.
Eve can intercept 𝐸(𝑚) but can’t read m.
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4. Private Key Cryptography
Example: Julius Caesar (shifting 3 letters)
Today is Friday.
Wrgdb lv Iulgdb.
Requires a secret encryption/decryption key between two parties.
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5. Public Key Cryptography
Bob announces a public key 𝑒 to everyone.
Everyone can encrypt a message 𝑚 to 𝑒(𝑚).
Only Bob can decrypt 𝑒(𝑚) using his private decryption key 𝑑 for 𝑒.
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6. The RSA algorithm Need the following numbers:
Two large primes (i.e., 512 bits) 𝒑 and 𝒒
𝑵 = 𝒑𝒒
A public encryption key 𝒆 , which is relatively prime to (𝒑−𝟏)(𝒒−𝟏).
A private decryption key 𝒅, which is 𝒆’s inverse modulo (𝒑−𝟏)(𝒒−𝟏)
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7. 𝐸 𝑥 = 𝑥 𝑒 mod 𝑁 𝐷 𝑥 = 𝑥 𝑑 mod 𝑁 𝑒𝑑≡1 mod (𝑝−1)(𝑞−1) The RSA algorithm
Encrypting a message 𝑥:
𝐸 𝑥 = 𝑥 𝑒 mod 𝑁
Decrypting a message 𝑥:
𝐷 𝑥 = 𝑥 𝑑 mod 𝑁
Numbers 𝑝, 𝑞 are prime, and 𝑒, 𝑑 satisfy:
𝑒𝑑≡1 mod (𝑝−1)(𝑞−1)
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8. RSA is correct Fermat’s Little Theorem 𝐷(𝐸(𝑥)) = 𝑥 mod 𝑁 in RSA
(Blackboard)
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9. Running time of RSA
Efficient algorithm (repeated squaring) for exponentiation
To create the key pairs, need to find large primes:
Primes are relatively densely distributed.
Testing if a number is prime is fast.
Randomly generate enough large numbers and test which ones are prime.
Finding 𝑒 relatively prime to (𝑝−1)(𝑞−1) and its inverse is fast.
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10. RSA is secure Need 𝑝 and 𝑞 to find 𝑑 from 𝑒.
Factoring a large number 𝑁 is hard.
2/27/2019