1. Public-key encryption

2. Symmetric-key encryption
Invertible function
Security depends on the shared secret – a particular key.
Fast, highly secure
Fine for repeated communication
Poor fit for one-shot communication, signatures

3. Asymmetric-key (public key) encryption
The basic idea:
A user has two keys: a public key and a private key.
A message can be encrypted with the public key and decrypted with the private key to provide security.
A message can be encrypted with the private key and decrypted with the public key to provide signatures.

4. One-way functions
Most common functions are invertible; for any F(x) = y, there is an F-1(y) = x.
Multiplication and division
DES
A function which is easy to compute in one direction, but hard to compute in the other, is known as a one-way function.
Hashing, modular arithmetic.
A one-way function that can be easily inverted with an additional piece of knowledge is called a trapdoor one-way function.

5. One-way functions
Public key encryption is based on the existence of trapdoor one-way functions.
Encryption with the public key is easy.
Decryption is computationally hard.
Knowledge of the private key opens the trapdoor, making inversion easy.
Password systems also use one-way functions.

6. Overview of RSA
RSA is the most common and well-known public key cryptosystem
Basic notation: a key pair (e,d) contains two keys:
e is the public key (used to encrypt documents)
d is the private key (used to decrypt documents)
M is the plaintext message.
Let R be the encryption function.
R(e,M) = C. R(d,C) = M. - encryption
R(d,M) = C’ R(e,C’) = M - signing
R(e,R(d,M)) = M = R(d,R(e,M))
Same function is used for both operations.

7. The RSA Algorithm Pick two large (100 digit) primes p and q.
Let n = pq
Select a relatively small integer d that is prime to (p-1)(q-1)
Find e, the multiplicative inverse of d mod (p-1)(q-1)
(d,n) is the public key. To encrypt M, compute
En(M) = Me(mod n)
(e,n) is the private key. To decrypt C, compute
De(C) = Cd(mod n)

8. RSA example Let p = 11, q = 13 n = pq = 143
(p-1)(q-1) = 120 = 3 x 23 x 5
Possible d: 7, 11, 13, 17, … (let’s use 7)
Find e: e*7 = 1(mod 120) = 103
Public key: (7, 143)
Private key: (103, 143)
En(42) = 427 (mod 143) = 81
De(81) = 81103(mod 143) = 42

9. Correctness of RSA
To show RSA is correct, we must show that encryption and decryption are inverse functions:
En(De(M)) = De(En(M)) = M = Med (mod n)
Since d and e are multiplicative inverses, there is a k such that:
ed=1+ kn = 1 + k(p-1)(q-1)
Med = M1+k(p-1)(q-1) = M*(Mp-1)k(q-1)
By Fermat: Mp-1=1(mod p)
Med = M(1)k(q-1)(mod p) = M(mod p)

10. Correctness of RSA Med = M(1)k(q-1)(mod p) = M(mod p)
Med = M(1)k(q-1)(mod q) = M(mod q)
By Chinese Remainder Thm, we get:
M^{ed} = M (mod p) M (mod q) =
M (mod pq) = M (mod n)
Therefore, RSA reproduces the original message and is correct.

11. Strengths of RSA No prior communication needed
Highly secure (for large enough keys)
Well-understood
Allows both encryption and signing

12. Weaknesses of RSA Large keys needed (1024 bits is current standard)
Relatively slow
Not suitable for very large messages
Public keys must still be distributed safely.

13. Security of RSA
The security of RSA is dependent on the assumption that it’s difficult to generate the private key d from the public key e and the modulus n.
Equivalent to integer factorization problem.
This is how we got e and d in the first place.
Factoring is thought to be computationally hard.
No proof, though!