chapter 8 introduction to number theory. 2 contents prime numbers fermats and eulers theorems

Chapter 8 Introduction to Number Theory

2

Contents

Prime Numbers

Fermat’s and Euler’s Theorems

3

Prime Numbers

Primes numbers An integer p > 1 is a prime number if and only if it is divisible by only

1 and p.

4

Prime Numbers

Integer factorization

Any integer a > 1 can be factored in a unique way as

where p1 < p2 < … < pt are prime numbers and

each ai is a positive integer.

tat

aaa ppppa ...321321

91 = 7 × 13;11101 = 7 × 112 ×13

5

Prime Numbers

Another integer factorization If P is the set of all prime numbers, then any positive integer can be w

ritten uniquely in the following form:

The right side is the product over all possible prime numbers p. Most of the exponents ap will be 0.

0each whereP

pp

a apa p

3600 = 24×32×52×70×110×….

6

Prime Numbers

Another integer factorization The value of any given positive integer can be specified by listing all

the nonzero exponents.

The integer 12 =22×31 is represented by {a2=2, a3=1}.The integer 18 =21×32 is represented by {a2=1, a3=2}.The integer 91= 72×131 is represented by {a7= 2, a13= 1}.

7

Prime Numbers

Multiplication Multiplication of two numbers is adding the corresponding exponents.

k = 12 × 18 = 216

12 = 22 × 31

18 = 21 × 32

------------------216 = 23 × 33

8

Prime Numbers

Divisibility

a|b → ap ≤ bp for all p

a = 12; b= 36; 12|36

12 = 22×3;36 = 22×32

a2 = 2 = b2

a3 = 1 ≤ 2 = b3

9

Prime Numbers

GCD

k = gcd (a, b) → kp = min(ap, bp) for all p

300 = 22×31×52

18 = 21×32×50

gcd (18, 300) = 21×31×50 = 6

10

Fermat’s and Euler’s Theorems

Fermat’s theorem If p is prime and a is a positive integer not divisible by p,

then

ap-1 ≡ 1 (mod p)

11

Fermat’s and Euler’s Theorems

Proof of Fermat’s theorem. Outline

Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p}

Show .

Since is relatively prime to p, we multiply

to both sides to get .

papp p mod)!1()!1( 1

)!1( p -1)!1( p

pa p mod1 1

12

Fermat’s and Euler’s Theorems

Proof of Fermat’s theorem Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p}

Show ka mod p for any 1 ≤ k ≤ p-1 is in {1, 2, …, p-1}

by showing that ka mod p ≠ k’a mod p for k≠ k’.

Show ka mod p ≠ k’a mod p for 1 ≤ k≠ k’ ≤ p-1. Proof by contradiction Assume that ka ≡ k’a mod p for some 1 ≤ k≠ k’ ≤ p-1. Since a is relatively prime to p, we multiply a-1 to get k ≡ k’ mod p,

which contradiction the fact that k≠ k’.

13

Fermat’s and Euler’s Theorems

Proof of Fermat’s theorem Show .

{1, 2, …, p-1} = {a mod p, 2a mod p, …, (p-1)a mod p}

ppap p mod)!1()!1( 1

pa

papp

papaap...

ppappapap...

p

p

mod1

mod)!1()!1(

mod ])1(...2[)]1(21[

mod )]mod)1((...)mod2()mod[()]1(21[

1

1

14

Fermat’s and Euler’s Theorems

An alternative form of Fermat’s Theorem

ap ≡ a mod p

where p is prime and a is any positive integer.

Proof If a and p are relatively prime, we get ap ≡ a mod p by multi

plying a to each side of ap-1 ≡ 1 mod p.

If a and p are not relatively prime, a = cp for some positive integer c. So ap ≡ (cp)p ≡ 0 mod p and a ≡ 0 mod p, which means ap ≡ a mod p.

15

Fermat’s and Euler’s Theorems

An alternative form of Fermat’s Theorem

ap ≡ a mod p

where p is prime and a is any positive integer.

p = 5, a = 3 35 = 243 ≡ 3 mod 5

p = 5, a = 10 105 = 100000 ≡ 10 mod 5 ≡ 0 mod 5

16

Fermat’s and Euler’s Theorems

Euler’s Totient Function The number of positive integers less than n and relatively prime to n.

)(n

= 3637 is prime, so all the positive number from 1 to 36are relatively prime to 37.

= 2435 = 5×71, 2, 3, 4, 6, 8, 9,11, 12, 13, 16, 17, 18, 19, 22, 23, 24, 26, 27, 29, 31, 32, 33, 34

)37(

)35(

17

Fermat’s and Euler’s Theorems

How to compute In general,

For a prime n, (Zn = {1,2,…, n-1})

For n = pq, p and q are prime numbers and p≠ q

)(n

1)( nn

)1()1()( qpn

npp

nn dividing primes theallover runs where,)1

1()(

18

Fermat’s and Euler’s Theorems

Proof of

is the number of positive integers less than pq that are relatively prime to pq.

can be computed by subtract from pq – 1 the number of positive integers in {1, …, pq – 1} that are not relatively prime to pq.

The positive integers that are not relatively prime to pq are a multiple of either p or q. { p, 2p,…,(q – 1)p}, {q, 2q, …,(p – 1)q} There is no same elements in the two sets. So, there are p + q – 2 elements that are not relatively prime to pq.

Hence, = pq – 1– (p + q – 2) = pq – p – q +1 = (p – 1)(q – 1)

)(n

)1()1()( qpn

)(n

)(n

19

Fermat’s and Euler’s Theorems

Φ(21) = Φ(3)×Φ(7) = (3-1)×(7-1) = 2 ×6 = 12

Z21={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}Φ(3)={3,6,9,12,15,18}Φ(7)={7,14}

where the 12 integers are {1,2,4,5,8,10,11,13,16,17,19,20}

20

Fermat’s and Euler’s Theorems

Euler’s theorem For every a and n that are relatively prime:

na n mod1)(

a = 3; n = 10; Φ(10) = 4; 34 = 81 ≡ 1 mod 10a = 2; n = 11; Φ(11) = 10; 210 = 1024 ≡ 1 mod 11

21

Fermat’s and Euler’s Theorems

Proof of Euler’s theorem

If n is prime, it holds due to Fermat’s theorem.

Otherwise (If n is not prime), define two sets R and S. show the sets R and S are the same. then, show

na n mod1)(

nan mod11

na n mod1)(

22

Fermat’s and Euler’s Theorems

Proof of Euler’s theorem

Set R The elements are positive integers less than n and relatively prime to n. The number of elements is

R={x1, x2,…, xΦ(n)} where x1< x2<…< xΦ(n)

Set S Multiplying each element of R by a∈R modulo n S ={(ax1 mod n), (ax2 mod n),…(axΦ(n) mod n)}.

)(n

23

Fermat’s and Euler’s Theorems

Proof of Euler’s theorem

The sets R and S are the same. We show S has all integers less than n and relatively prime to n.

S ={(ax1 mod n), (ax2 mod n),…(axΦ(n) mod n)}

1. All the elements of S are integers less than n that are relatively prime to n because a is relatively prime to n and xi is relatively prime to n, axi must also be relatively prime to n.

2. There are no duplicates in S.If axi mod n = axj mod n, then xi = xj. by cancellation law.

24

Fermat’s and Euler’s Theorems

Proof of Euler’s theorem Since R and S are the same sets,

)(mod1

)(mod

)(mod

)mod(

)(

)(

1

)(

1

)(

)(

1

)(

1

)(

1

)(

1

na

nxxa

nxax

xnax

n

n

ii

n

ii

n

n

i

n

iii

n

i

n

iii

25

Fermat’s and Euler’s Theorems

Alternative form of the theorem

If a and n are relatively prime, it is true due to Euler’s theorem.

Otherwise, ….

)(mod1)( naa n

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Fermat’s and Euler’s Theorem

The validity of RSA algorithm Given 2 prime numbers p and q, and integers n = pq and m,

with 0<m<n, the following relationship holds.

If m and n are re relatively prime, it holds by Euler’s theorem. If m and n are not relatively prime, m is a multiple of either p or q.

nmmm qpn mod1)1)(1(1)(

27

Fermat’s and Euler’s Theorem

Case 1: m is a multiple of p m=cp for some positive integer c. gcd(m, q)=1, otherwise, m is a multiple of p and q and yet m<pq because gcd(m, q)=1, Euler’s theorem holds

by the rules of modular arithmetic,

Multiplying each side by m=cp

qmq mod11

kkqm

qm

qm

n

n

pq

integer somefor ,1

mod1

mod1][

)(

)(

11

nmm

kcnmkcpqmmn

n

mod1)(

1)(

28

Fermat’s and Euler’s Theorem

Case 2: m is a multiple of q prove similarly.

Thus, the following equation is proved.

nmmm qpn mod1)1)(1(1)(

29

Fermat’s and Euler’s Theorem

An alternative form of this corollary is directly relevant to RSA.

nm

nm

nmm

m

k

kn

nk

mod

theoremsEuler'by ,mod])1[(

mod][( 1)(

1)(