09-FermatEuler.ppt

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:
)
(mod
)
(mod
n
b
x
m
a
x



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
Q

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






















or
)
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
ap


)
(mod
1
1
p
ap


)
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 f(n)?
13
)
(mod
1
)
(
n
a n

f
4

f(n) is the number of integers a,
such that 1 ≤ a ≤ n and gcd(a,n) = 1.
Examples:
1. f(10) = 4.
2. When p is prime, f(p) = ____
3. When n =pq (product of 2 primes), f(n) = ____
14
5

The general formula for f(n)
Example: f(12)4
[Bill Waite, RHIT 2007]
 






 

n
p p
p
n
n
|
1
)
(
f
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 f(n).
)
(mod
1
)
(
n
a n

f
7-10

09-FermatEuler.ppt

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