DTTF/NB479: Dszquphsbqiz
Announcements:
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Homework 2 due now
Computer quiz Thursday on chapter 2
Questions?
Today:
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Finish congruences
Fermat’s little theorem
Euler’s theorem
Important for RSA public key crypto
– pay careful attention!
Day 9
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:
x  a(modm)
x  b(modn)
CRT Equivalences let us use systems of
congruences to solve problems
Solve the system:
How many solutions?

Find them.
x  3(mod7)
x  5(mod15)
x 2  1(mod35)
The Chinese Remainder Theorem also applies
generally to n factors

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:
x  a1 (modm1 )
x  a2 (modm2 )
...
x  ak (modmk )
Modular Exponentiation is extremely efficient since
the partial results are always small
Compute the last digit of 32000
Compute 32000 (mod 19)
Idea:
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Get the powers of 3 by repeatedly squaring 3, BUT
taking mod at each step.
Q
Modular Exponentiation Technique and Example
(All congruences are mod 19)
Compute
(mod 19)
32000
32  9
34  9 2  81  5
38  52  25  6
Technique:
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316  6 2  36  17(or  2)
Repeatedly square
3, but take mod at
each step.
332  172  289  4
Then multiply the
terms you need to
get the desired
power.
3256  5
Book’s
powermod()
364  4 2  16
3128  162  256  9
3512  6
31024  17
32000  (31024 )(3512 )(3256 )(3128 )(364 )(316 )
32000  (17)(6)(5)(9)(16)(17)
32000  (1248480)
32000  9(mod19)
Modular Exponentiation Example
32000
Compute
(mod 152)
32  9
34  9 2  8 1
38  8 12  6 5 6 1 2 5
316  2 52  6 2 5  1 7
332  1 72  2 8 9  1 3 7
364  1 3 72  1 8 7 6 9 7 3
3128  9
3256  8 1
3512  2 5
31024  1 7
32000  (31024 )(3512 )(3256 )(3128 )(364 )(316 )
32000  (17)(25)(81)(9)(73)(17)
32000  (384492875
)
32000  9(mod152)
Fermat’s Little Theorem:
If p is prime and gcd(a,p)=1, then a(p-1)≡1(mod p)
1-2
Fermat’s Little Theorem:
If p is prime and gcd(a,p)=1, then a(p-1)≡1(mod p)
S=
1
2
3
4
5
6
Examples:
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22=1(mod 3)
64 =1(mod ???)
(32000)(mod 19)
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:
p 1
a
 1(mod p)
If p is prime and doesn’t divide a,
Converse:
If
a
p 1
 1(mod p) , then p is prime and doesn’t divide a.
This is almost always true when a = 2. Rare
counterexamples:
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n = 561 =3*11*17, but
560
2
 1(mod561)
n = 1729 = 7*13*19
Can do first one by hand if use Fermat and combine results with
Chinese Remainder Theorem
Primality testing schemes typically use the
contrapositive of Fermat
n
Even?
no
div by other small primes?
no
Prime by Factoring/
advanced techn.?
yes
prime
3
Primality testing schemes typically use the
contrapositive of Fermat
n
Even?
no
Use Fermat as a filter since it’s
faster than factoring (if
calculated using the powermod
method).
Fermat: p prime
Contrapositive?
2p-1
≡ 1 (mod p)
Why can’t we just compute 2n-1(mod n)
using Fermat if it’s so much faster?
div by other small primes?
no
nn11
?
?
n ) n1)
22 (mod
1(mod
yes
Prime by Factoring/
advanced techn.?
yes
prime
Euler’s Theorem is like Fermat’s, but for composite
moduli
If gcd(a,n)=1, then
f ( n)
a
So what’s f(n)?
 1(modn)
4
f(n) is the number of integers a,
such that 1 ≤ a ≤ n and gcd(a,n) = 1.
Examples:
14
1.
f(10) = 4.
2.
When p is prime, f(p) = ____
3.
When n =pq (product of 2 primes), f(n) = ____
5
The general formula for f(n)
p are distinct primes
6
 p 1 

f (n)  n 
p 
p|n 
Example: f(12)4
[Intuition from Bill Waite, RHIT 2007]
7-10
Euler’s Theorem can also lead to computations
that are more efficient than modular exponentiation
f ( n)
a
 1(modn)
as long as gcd(a,n) = 1
Basic
Principle: when working mod n, view the exponents mod f(n).
Examples:
1.
2.
3.
4.
Find last 3 digits of 7803
Find 32007 (mod 12)
Find 26004 (mod 99)
Find 26004 (mod 101)