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

nbxmax

≡≡

CRT Equivalences let us use systems of congruences to solve problems

Solve the system:

How many solutions? Find them.

)15(mod5)7(mod3

≡≡

xx

)35(mod12 ≡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

22

11

kk max

maxmax

≡

≡≡

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()

1736353

92561631643

4289173)2(173663

6255358193

93

1024

512

256

2128

264

232

216

28

24

2

≡

≡

≡

≡≡≡

≡=

≡≡=

−≡≡=

≡≡=

≡≡=

≡

or

)19(mod93)1248480(3

)17)(16)(9)(5)(6)(17(3)3)(3)(3)(3)(3)(3(3

2000

2000

2000

166412825651210242000

≡

≡

≡

≡

(All congruences are mod 19)

Modular Exponentiation Example

Compute 32000 (mod 152)

17325381393

7318769137313728917317625253256561813

819393

1024

512

256

128

264

232

216

28

24

2

≡

≡

≡

≡

≡≡=

≡≡=

≡≡=

≡≡=

≡=

≡

)152(mod93)384492875(3

)17)(73)(9)(81)(25)(17(3)3)(3)(3)(3)(3)(3(3

2000

2000

2000

166412825651210242000

≡

≡

≡

≡

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

)(mod11 pa p ≡−

)(mod11 pa p ≡−

)561(mod12560 ≡

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)(mod2

?1 ≡− nn

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?

)(mod12?

1 nn ≡−

3

Euler’s Theorem is like Fermat’s, but for composite moduli

If gcd(a,n)=1, then So what’s φ(n)?

13

)(mod1)( na n ≡φ

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]

∏

−=

np ppnn

|

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).

)(mod1)( na n ≡φ

7-10