1. RSA
Preliminaries

2. Division Algorithm Given integers a and b≠0, there exist unique
Integers q and r such that
a = bq +r where 0 ≤ r < |b|
a is the dividend, b is the divisor, and r is the remainder.
b divides a or b is a factor of a if r=0.

3. Greatest Common Divisor
The greatest common divisor of two nonzero integers a and b is defined to be the largest positive integer that divides (i.e., is a factor of) both a and b.
The greatest common divisor of a and b is denoted by gcd(a,b).
We say that a and b are relatively prime if gcd(a,b)=1.

4. The Euclidean Algorithm
EUCLID(a,b)
if b=0
then return a
else return EUCLID(b, a mod b)

5. A useful relation
Fact: If d = gcd(a,b), then there exist unique integers x and y such that
d = ax + by
x and y can be found using the “extended Euclidean algorithm”

6. Euler’s phi function
For any positive integer n, φ(n) (Euler’s phi function) is defined to be the number of positive integers less than n that are relatively prime to n.
φ(n) = nπp|n(1-1/p), (where p runs through all prime factors of n)
Note that if p is prime, then φ(p) = p-1.

7. The Theorems of Euler and Fermat
Euler’s Theorem
aφ(n) = 1 mod n for all a in Zn* , where Zn*
is the set of all a in Zn such that gcd(a,n)=1
Fermat’s Theorem
If p is prime, then
ap = a mod p for all a in Zp and if gcd(a,p)=1,
ap-1 = 1 mod p for all a in Zp*

8. The Chinese Remainder Theorem
Suppose n1, n2, …, nk are positive integers which are pairwise relatively prime. Then, for any given integers a1,a2, …, ak, there exists an integer x solving the system of simultaneous congruences
x = a1 mod n1
x = a2 mod n All solutions are
… congruent modulo
x = ak mod nk N=n1n2 … nk

9. Public-key Cryptosystems
Each participant has a public key and a secret key.
Every public and secret key is a one-to-one function from the set D to D, where D is the set of permissible messages.
Alice: PA, SA
Bob: PB,SB

10. Sending a Message
Bob uses Alice’s public key to send an encrypted message M to Alice.
C = PA(M)
M -> PA > SA -> M
communication channel
Alice uses her secret key to decrypt M

11. Public and Secret keys are Matched Pairs
E.g.,
M=SA(PA(M)) = PA(SA(M))

12. Digital Signatures
Suppose Alice wished to send Bob a digitally signed response M’
1. Alice computes her digital signature for the message M’: σ = SA(M’)
2. Alice sends the pair (M’, σ) to Bob.
3. When Bob receives (M’, σ) , he can verify that it originated from Alice by verifying that M’ =PA (σ )

13. The RSA Cryptosystem
A participant creates his public and secret keys as follows:
Select an random two large primes, p and q
Compute n = pq.
Compute e that is relatively prime to φ(n)=(p-1)(q-1)
Compute the modulo φ(n) inverse d of e.
Publish the pair P=(e,n) as RSA public key.
Keep secret the pair S=(d,n) as RSA secret key.
The domain D of messages is Zn
The transformation of a message M associated with a public key P=(e,n) is P(M)=Me and the transformation of a “ciphertext” C associated with a secret key S=(d,n) is S(C)=Cd

14. The Correctness of RSA SA(PA(M)) = Med=Mde=PA(SA(M))
We have ed = 1 mod (p-1)(q-1). This means that when ed is divided by
(p-1)(q-1), the remainder is 1 and so by the division algorithm, ed = k(p-1)(q-1)+1 for some k.
Thus, Med = M1+k(p-1)(q-1)=M M(p-1)k(q-1)
=M(1)k(q-1) mod p (by Fermat’s theorem)
=M mod p if gcd(M,p)=1.

15. The Correctness of RSA 2 Also, Med = M1+k(p-1)(q-1)=M M(q-1)k(p-1)
=M(1)k(p-1) mod q (by Fermat’s theorem)
=M mod q if gcd(M,q)=1
Thus,
Med =M mod p
Med =M mod q
These two congruences hold even when
either p or q is not relatively prime to M and so
Med =M mod n (by the Chinese Remainder
Theorem)

16. Example (for an extremely simple, breakable code)
Suppose that Alice chooses p=7, q=11
and e=13. Then PA = (13,77) and d=37
since 13*37=1 mod 60 and so her secret key is SA = (37,n).
Suppose that Bob wants to send 52 to Alice. What is the encrypted message?

17. Encrypted message Bob uses Alice’s public key to encrypt his message:
5213 = 17 mod 77

18. Decrypted message Alice receives the encrypted message 17.
Alice uses her secret key (37,77) to decrypt Bob’s message
1737 = 52 mod 77

19. Verification
Suppose that Alice wants to verify to Bob that she received his message.
She sends encodes the message she received with her secret key and sends it to Bob:
5237 = 24 mod 77
Bob uses Alices’s public key to obtain
2413 = 52 mod 77

20. Breaking the code
Messages encrypted with RSA can be decrypted by determining primes p and q such that n=pq since in that case a d can be determined such that de = 1 mod φ(n)

21. Asignment
Write an openMP program such that given an integer n, the program determines two primes p and q such that n = pq, if such primes exist.