127_Inside PK Cryptography:Math and Implementation

Inside PK
Cryptography:
Math and Implementation

Sriram Srinivasan
(“Ram”)



Agenda





Introduction to PK Cryptography
Essential Number Theory
 Fundamental Number Theorem
 GCD, Euclid’s algorithm
 Linear combinations

Modular Arithmetic

Euler’s Totient Function
Java implementation of RSA
Sriram Srinivasan

2/47


Security Issues






Authentication, Authorization, and
Encryption, Non-repudiation
Shared Secrets (e.g passwords, Enigma)
Something shared, something (else)
secret


Concept by Ellis, Cocks and Williams




Popularly attributed to Diffie and Hellman

Algorithm by Rivest, Shamir and Adelman


Used everywhere: https, SSL, email, certificates.

Sriram Srinivasan

3/47


Public Key Cryptography



Consider a pair of magic pens.





You want to send a message to me






Write with one, use the other to decode.
Symmetric: either can be used to encode
You borrow one of my pens and write with it.
I decode it with my other pen.
Avoids problems of shared secrets

Same tools for authentication, encryption
and non-repudiation.
Sriram Srinivasan

4/47


Mathematics

Fundamental Theorem of
Arithmetic




All numbers are expressible as a unique
product of primes

10 = 2 * 5, 60 = 2 * 2 * 3 * 5
Proof in two parts

1. All numbers are expressible as products
of primes

2. There is only one such product sequence
per number

Sriram Srinivasan

6/47


Fundamental Theorem proof


First part of proof
 All numbers are products of primes

Let S = {x | x is not expressible as a product of primes
Let c = min{S}.

c cannot be prime

Let c = c1 . c2
c1, c2 < c ⇒ c1, c2 ∉ S (because c is min{S})
∴c1, c2 are products of primes ⇒ c is too
∴S is an empty set
Sriram Srinivasan

7/47


Fundamental Theorem proof


Second part of proof
 The product of primes is unique
Let n = p1p2p3p4… = q1q2q3q4…

Cancel common primes. Now unique primes on both si
Now, p1 | p1p2p3p4

⇒ p1 | q1q2q3q4…
⇒ p1 | one of q1, q2, q3, q4…
⇒ p1 = qi which is a contradiction
Sriram Srinivasan

8/47

GCD (Greatest Common
Divisor)




gcd(a,b) = the greatest of the divisors of
a,b
Many ways to compute gcd
 Extract common prime factors







Express a, b as products of primes
Extract common prime factors
gcd(18, 66) = gcd(2*3*3, 2*3*11) = 2*3 = 6
Factoring is hard. Not practical

Euclid’s algorithm
Sriram Srinivasan

9/47

Euclid’s algorithm
a
1

b

r=a%b

b
2

r
r

3

r1

r1 = b % r

r % r1 = 0.

∴gcd (a,b) = r1

Sriram Srinivasan

10/47


Euclid’s algorithm proof



Proof that r1 divides a and b
r1 | r
b = r1 + r
a = qb + r
r1 | b
r1 | r

r1 | b

r1 | a

Sriram Srinivasan

11/47


Euclid’s algorithm proof


(contd)

Proof that r1 is the greatest divisor
Say, c | a and c | b
c | qb + r
c|r
c | q’b + r1
c | r1
Sriram Srinivasan

12/47


Linear Combination




ax + by = “linear combination” of a and b
 12x + 20y = {…, -12,-8,-4,0,4,8,12, … }
The minimum positive linear combination
of a & b = gcd(a,b)
 Proof in two steps:




1. If d = min(ax+by) and d > 0, then d | a, d
|b
2. d is the greatest divisor.
Sriram Srinivasan

13/47


GCD & Linear combination
(contd.)
Let S = {z = ax + by | z > 0 }
Let d = min{S} = ax1 + by1

Let a = qd + r. 0 <= r < d
r = a - qd = a - q(ax1 + by1)
r = a(1 - qx1) + (-qy1)b
If r > 0, r ∈ S

But r < d, which is a contradiction, because d = min{
∴r = 0

⇒ d|a
Sriram Srinivasan

14/47


GCD & Linear combination
(contd.)



Second part of proof


Any other divisor is smaller than d
Let c | a, c | b, c > 0
a = cm, b = cn
d = ax1 + by1 = c(mx1 + ny1)

⇒ c|d
⇒ d is the gcd
Sriram Srinivasan

15/47


Summary 1






All numbers are expressible as unique
products of prime numbers
GCD calculated using Euclid’s algorithm
gcd(a,b) = 1 ⇒ a & b are mutually prime
gcd(a,b) equals the minimum positive
ax+by linear combination

Sriram Srinivasan

16/47


Modular/Clock Arithmetic





1:00 and 13:00 hours are the same

 1:00 and 25:00 hours are the same
1 ≡ 13 (mod 12)
a ≡ b (mod n)
 n is the modulus
 a is “congruent” to b, modulo n
 a - b is divisible by n

a%n=b%n
Sriram Srinivasan

17/47


Modular Arithmetic



a ≡ b (mod n), c ≡ d (mod n)
Addition
a - b = jn
c - d = kn




aa++c c- (b
+ k) nn)
≡ +
b d)
+=

d (j(mod

Multiplication
 ac ≡ bd (mod n)
Sriram Srinivasan

18/47


Modular Arithmetic (contd.)


Power

a ≡ b (mod n) ⇒ ak ≡ bk (mod n)
Using induction,
If ak ≡ bk (mod n),
a . ak ≡ b . bk (mod n), by multiplication rule



∴ ak+1 ≡ bk+1 (mod n)
Going n times around the clock
 a + kn ≡ b (mod n)
Sriram Srinivasan

19/47


Chinese Remainder Theorem



m ≡ a (mod p), m ≡ a (mod q)
⇒ m ≡ a (mod pq) (p,q are primes)
m-a = cp.
Now, m-a is expressible as p1. p2 .p3 . . .
If m - a is divisible by both p and q,
p and q must be one of p1 , p2 , p3
⇒ m - a is divisible by pq
Sriram Srinivasan

20/47


GCD and modulus


If gcd(a,n) = 1, and a = b (mod n),
then gcd(b,n) = 1
a ≡ b (mod n) ⇒ a = b + kn
gcd(a,n) = 1
ax1 + ny1 = 1, for some x1 and y1
(b + kn)x1 + ny1 = 1
bx1 + n(kx1 + y1) = bx1 + ny2 = 1
gcd(b,n) = 1
Sriram Srinivasan

21/47

Multiplicative Inverse


If a, b have no common factors, there
exists ai such that a.ai ≡ 1 (mod b)
 ai is called the “multiplicative inverse”
gcd(a,b) = 1 = ax1+ by1, for some x1 and y1
ax1 = 1 – by1
ax1 = 1 + by2

(making y2 = -y1)

ax1 - 1 = by2
ax1 ≡ 1 (mod b) (x1 is the multiplicative inverse)
Sriram Srinivasan
22/47


Summary 2






Modular arithmetic
 Addition, multiplication, power, inverse
Chinese Remainder Theorem
 If m ≡ a (mod p) and m ≡ a (mod q),
then m ≡ a (mod pq)

Relationship between gcd and modular
arithmetic
 gcd(a,b) = 1
⇒ aai ≡ 1 (mod b)
Sriram Srinivasan

23/47


Euler’s Totient function




φ(n) = Totient(n)
= Count of integers ≤ n coprime to n
 φ (10) = 4
(1, 3, 7, 9 are coprime to 10)
 φ (7) = 6 (1, 2, 3, 4, 5, 6 coprime to 10)
φ(p) = p - 1, if p is a prime

Sriram Srinivasan

24/47


Totient lemma #2: product


φ(pq) = (p - 1)(q - 1) = φ(p) . φ(q)

 if p and q are prime
Which numbers ≤ pq share factors with pq?
1.p, 2.p, 3.p, … (q-1)p and
1.q, 2.q, 3.q, … (p-1)q and
pq
The rest are coprime to pq. Count them.
φ(pq) = pq - (p - 1) - (q - 1) - 1 = (p - 1)(q - 1)
Sriram Srinivasan

25/47