DTTF/NB479: Dszquphsbqiz Day 9 Announcements: Homework 2 due now - - PowerPoint PPT Presentation
DTTF/NB479: Dszquphsbqiz Day 9 Announcements: Homework 2 due now Computer quiz Thursday on chapter 2 Questions? Today: Wrap up congruences Fermats little theorem Eulers theorem Both really important for RSA pay
DTTF/NB479: Dszquphsbqiz Day 9 Announcements:
Homework 2 due now Computer quiz Thursday on chapter 2
Questions? Today:
Wrap up congruences Fermat’s little theorem Euler’s theorem Both really important for RSA – pay careful attention!
The Chinese Remainder Theorem establishes an equivalence A single congruence mod a composite number is equivalent to a system of congruences mod its factors Two-factor form
Given gcd(m,n)=1. For integers a and b, there exists
exactly 1 solution (mod mn) to the system:
CRT Equivalences let us use systems of congruences to solve problems Solve the system: How many solutions?
Find them.
) 15 (mod 5 ) 7 (mod 3 ≡ ≡ x x
) 35 (mod 1
2 ≡
x
Chinese Remainder Theorem n-factor form
Let m1, m2,… mk be integers such that gcd(mi, mj)=1
when i ≠ j. For integers a1, … ak, there exists exactly 1 solution (mod m1m2…mk) to the system:
) (mod ... ) (mod ) (mod
2 2 1 1 k k
m a x m a x m a x ≡ ≡ ≡
Modular Exponentiation is extremely efficient since the partial results are always small Compute the last digit of 32000 Compute 32000 (mod 19) Idea:
Get the powers of 3 by repeatedly squaring 3, BUT
taking mod at each step. Q
Modular Exponentiation Technique and Example
Compute 32000 (mod 19) Technique:
Repeatedly square
3, but take mod at each step.
Then multiply the
terms you need to get the desired power.
Book’s powermod()
17 3 6 3 5 3 9 256 16 3 16 4 3 4 289 17 3 ) 2 ( 17 36 6 3 6 25 5 3 5 81 9 3 9 3
1024 512 256 2 128 2 64 2 32 2 16 2 8 2 4 2
≡ ≡ ≡ ≡ ≡ ≡ ≡ = ≡ ≡ = − ≡ ≡ = ≡ ≡ = ≡ ≡ = ≡
) 19 (mod 9 3 ) 1248480 ( 3 ) 17 )( 16 )( 9 )( 5 )( 6 )( 17 ( 3 ) 3 )( 3 )( 3 )( 3 )( 3 )( 3 ( 3
2000 2000 2000 16 64 128 256 512 1024 2000
≡ ≡ ≡ ≡
(All congruences are mod 19)
Modular Exponentiation Example
Compute 32000 (mod 152) 17 3 25 3 81 3 9 3 73 18769 137 3 137 289 17 3 17 625 25 3 25 6561 81 3 81 9 3 9 3
1024 512 256 128 2 64 2 32 2 16 2 8 2 4 2
≡ ≡ ≡ ≡ ≡ ≡ = ≡ ≡ = ≡ ≡ = ≡ ≡ = ≡ = ≡
) 152 (mod 9 3 ) 384492875 ( 3 ) 17 )( 73 )( 9 )( 81 )( 25 )( 17 ( 3 ) 3 )( 3 )( 3 )( 3 )( 3 )( 3 ( 3
2000 2000 2000 16 64 128 256 512 1024 2000
≡ ≡ ≡ ≡
Fermat’s Little Theorem: If p is prime and gcd(a,p)=1, then a(p-1)≡1(mod p)
8
1-2
Fermat’s Little Theorem: If p is prime and gcd(a,p)=1, then a(p-1)≡1(mod p) Examples:
22=1(mod 3) 64 =1(mod ???) (32000)(mod 19)
9
1 2 3 4 5 6 S= f(1)=2 f(2)=4 f(3)=6 f(4)=1 f(5)=3 f(6)=5 Example: a=2, p=7
1-2
The converse when a=2 usually holds Fermat: If p is prime and doesn’t divide a, Converse:
If , then p is prime and doesn’t divide a. This is almost always true when a = 2. Rare counterexamples:
n = 561 =3*11*17, but n = 1729 = 7*13*19 Can do first one by hand if use Fermat and combine results with
Chinese Remainder Theorem
) (mod 1
1
p a p ≡
−
) (mod 1
1
p a p ≡
−
) 561 (mod 1 2560 ≡
Primality testing schemes typically use the contrapositive of Fermat
Even? div by other small primes? Prime by Factoring/ advanced techn.? n no no yes prime
Primality testing schemes typically use the contrapositive of Fermat
Use Fermat as a filter since it’s faster than factoring (if calculated using the powermod method).
1 ) (mod 2
? 1
≡
−
n
n
Even? div by other small primes? Prime by Factoring/ advanced techn.? n no no yes yes prime
Fermat: p prime 2p-1 ≡ 1 (mod p) Contrapositive?
Why can’t we just compute 2n-1(mod n) using Fermat if it’s so much faster?
) (mod 1 2
? 1
n
n
≡
−
3
Euler’s Theorem is like Fermat’s, but for composite moduli If gcd(a,n)=1, then So what’s φ(n)?
13
4
φ(n) is the number of integers a, such that 1 ≤ a ≤ n and gcd(a,n) = 1. Examples:
1.
φ(10) = 4.
2.
When p is prime, φ(p) = ____
3.
When n =pq (product of 2 primes), φ(n) = ____
14
5
The general formula for φ(n)
Example: φ(12)=4
[Bill Waite, RHIT 2007]
− =
n p
p p n n
|
1 ) ( φ
6
p are distinct primes
Euler’s Theorem can also lead to computations that are more efficient than modular exponentiation as long as gcd(a,n) = 1 Examples:
1.
Find last 3 digits of 7803
2.
Find 32007 (mod 12)
3.
Find 26004 (mod 99)
4.
Find 26004 (mod 101) Basic Principle: when working mod n, view the exponents mod φ(n).
) (
n ≡ φ
7-10