Number Theory
이재원
School of Information Technology
Sungshin W. University
Division
• DEFINITION 1 If a and b are integers with a≠0,
we say that a divides b if there is an integer c
such that b=ac. When a divides b we say that
a is a factor of b and that b is a multiple of a.
The notation a | b denotes that a divides b.
We write a ł b when a does not divide b.
ex) 3 ł 7, 3|12
Division Cont.
• THEOREM 1 Let a, b, and c be integers. Then
(i) if a | b and a | c, then a | (b + c);
(ii) if a | b and a | bc, for all integers c;
(iii) if a | b and b | c, then a | c.
• COROLLARY 1 If a, b, and c are integers such
that a | b and a | c then a | mb + nc whenever
m and n are integers.
Division Cont.
• THE DIVISION ALGORITHM Let a be an
integer and d a positive integer. Then there
are unique integers q and r, with 0 ≤ r < d,
such that a = dq + r.
• THEOREM 2 In the equality given in the
division algorithm, d is called the divisor, a is
called the dividend, q is called the quotient,
and r is called the remainder.
• q = a div d, r = a mod d. (101/11?, -11/3?)
Modular Arithmetic
• DEFINITION 3 If a and b are integers and m is
a positive integer, then a is congruent to b
modulo m if m divides a – b.
a  b (mod m)
• THEOREM 3 Let a and b be and let m be a
positive integer, then a  b (mod m) if and
only if a mod m = b mod m.
• 17  5 (mod 6) ? Yes. 24  14 (mod 6) ? No.
Modular Arithmetic Cont.
• THEOREM 4 Let m be a positive integer. The
integer a and b are congruent modulo m if
and only if there is an integer k such that a = b
+ km
• THEOREM 5 Let m be a positive integer. If a 
b (mod m) and c  d (mod m), then
(a + c)  (b + d) (mod m) and ac  bd (mod m) .
ex) 7  2 (mod 5) and 11  1 (mod 5)
=> 18  3 (mod 5) and 77  2 (mod 5)
Modular Arithmetic Cont.
• COROLLARY 2 Let m be a positive integer and
let a and b are integers. Then,
(a + b) mod m = ((a mod m) + (b mod m)) mod m
and ab mod m = ((a mod m)(b mod m)) mod m
Proof)
a  (a mod m) (mod m) and b  (b mod m) (mod m)
-> a + b  (a mod m) + (b mod m) (mod m)
…
Applications of Congruences
• Hashing Functions h(k) = k mod m
• Pseudorandom Numbers
xn+1= (axn+ c) mod m
• Ex) if we choose m=9, a=7, c=4, and x0=3, we
get 3, 7, 8, 6, 1, 2, 0, 4, 5, 3, 7, 8, 6, …
Applications of Congruences
• Cryptology
Caesar’s encryption: f (p) = (p + 3) mod m
• EXAMPLE 9 secret message from “MEET YOU
IN THE PARK” using Caesar’s cipher?
12 4 4 19 24 14 20 8 13 19 7 4 15 0 17 10
15 7 7 22 1 17 23 11 16 22 10 7 18 3 21 13
“PHHW BRX LQ WKH SUDN”
• enhancing security: f (p) = (ap + b) mod 26
Primes and GCDs
• DEFINITION 1 A positive integer p greater than
1 is called prime if the only positive factors of
p are 1 and p. A positive integer that is greater
than 1 and is not prime is called composite.
• The primes less than 100 are 2, 3, 5, 7, 11, 13,
17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71, 73, 79, 83, 89, and 97.
Primes and GCDs
• THEOREM 2 If n is a composite integer, then n
has a prime divisor less than or equal to n.
• EXAMPLE 3 Show that 101 is prime.
Solution: The only primes not exceeding
are 2, 3, 5, and 7. …
101
• EXAMPLE 4 Find the prime factorization of
7007.
7007/7 = 1001, 1001/7 = 143, … 7007 = 72·11·13
Representation of Integers
• THEOREM 1 Let b be a positive integer greater
than 1. Then if n is a positive integer, it can be
expressed uniquely in the form
n= akbk + ak-1bk-1 + … + a1b1 + a0
Where k is a nonnegative integer, a0, a1, …, ak
are nonnegative integers less than b and ak ≠ 0.
Cont.
• The representation of n given in Theorem 1 is
called base b expansion of n. The base b
expansion of n is denoted by (akak-1…a1a0)b
• For instance, (245)8 represents 2·82 + 4·8 + 5 =
165.
Cont.
• Example 1 What is the decimal expansion of
the integer that has (1 0101 1111)2 as its
binary expansion?
• Solution:
(1 0101 1111)2 = 1·28 + 0·27 + 1·26 + 0·25 + 1·24
+ 1·23 + 1·22 + 1·21 + 1·20 = 351.
Cont.
• Example 3 Find the base 8, or octal, expansion
of (12345)10.
Solution: 12345 = 8·1543 + 1.
1543 = 8·192 + 7.
192 = 8·24 + 0.
24 = 8·3 + 0.
3 = 8·0 + 3.
(12345)10 = (30071)8
Modular Exponentiation
• Example 11 Find 3644 mod 645.
Solution:
- 644를 2진수로 표현하면 (1010000100)2이다. 따
라서 644 = 29 + 27 + 22 이며, 3644 = (3의 29승) · (3의
27승) · (3의 22승) 이다.
- 세 수의 곱을 645로 나눈 나머지는 각각의 수를
644로 나눈 나머지들을 모두 곱한 후 다시 645로
나눈 나머지와 같다(교재 205 페이지의 따름정리
(corollary) 2 참고).
- 이 원리를 이용하여 만들어진 알고리즘 5에 의해,
Modular Exponentiation Cont.
• i=0: a0=0, x=1, p=32 mod 645 = 9 mod 645 = 9;
• i=1: a1=0, x=1, p=92 mod 645 = 81 mod 645 = 81 (3의 22승);
• i=2: a2=1, x=(1·81) mod 645 = 81, p=812 mod 645 = 6561 mod
645 = 111; ( (6561 · 6561) mod 645 = (111 · 111) mod 645 )
• i=3: a3=0, x=81, p=1112 mod 645 = 12,321 mod 645 = 66;
• i=4: a4=0, x=81, p=662 mod 645 = 4356 mod 645 = 486;
• i=5: a5=0, x=81, p=4862 mod 645 = 236,196 mod 645 = 126;
• i=6: a6=0, x=81, p=1262 mod 645 = 15,876 mod 645 = 396;
• i=7: a7=1, x=(81·396) mod 645 = 471, p=3962 mod 645 =
156,816 mod 645 = 81;
• i=8: a8=0, x=471, p=812 mod 645 = 6561 mod 645 = 111;
• i=9: a9=1, x=(471·111) mod 645 = 52,281 mod 645 = 36.
Modular Exponentiation Cont.
ALGORITHM 5 Modular Exponentiation
Procedure modular exponentiation(b: integer,
n = (ak-1ak-2, …, a1a0)2, m: positive integer)
x := 1
power := b mod m
for i = 0 to k -1
begin
if ai = 1 then x := (x · power) mod m
power := (power · power) mod m
end (x equals bn mod m)
The Euclidean Algorithm
• More efficient method of finding the greatest
common divisor
• Finding gcd(91, 287)
- 287 = 91 · 3 + 14
- any divisor of 91 and 287 must also be a divisor of
287 - 91 · 3 = 14. Also, any divisor of 91 and 14 must
also be a divisor of 287 = 91 · 3 + 14.
- Hence, the greatest common divisor of 91 and 287 is
the same as the greatest common divisor of 91 and
14.
The Euclidean Algorithm Cont.
- This means that the problem of finding gcd(91, 287)
has been reduced to the problem of finding gcd(91,
14).
- Next, divide 91 by 14 to obtain
- 91 = 14 · 6 + 7
- Continue by dividing 14 by 7(gcd(91, 14) = gcd(14, 7))
- 14 = 7 · 2
- Because 7 divides 14, it follows that gcd(14, 7) = 7.
Furthermore, because gcd(287, 91) = gcd (91, 14) …,
the original problem has been solved.
The Euclidean Algorithm Cont.
• LEMMA 1 Let a = bq + r, where a, b, q, and r
are integers. Then gcd(a, b) = gcd(b, r).
• Proof: Suppose that d divides both a and b. Then d
also divides a – bq = r (from Theorem 1 of section
3.4). Hence, any common divisor of a and b is also a
common divisor of b and r.
Likewise, suppose that d divides both b and r. Then d
also divides bq + r = a. Hence, any common divisor of
b and r is also a common divisor of a and b. => gcd(a,
b) = gcd(b, r)
The Euclidean Algorithm Cont.
• EXAMPLE 12 Find the greatest common
divisor of 414 and 662.
• Solution: Successive uses of the division
algorithm give:
662 = 414 · 1 + 448
414 = 248 · 1 + 166
248 = 166 · 1 + 82
166 = 82 · 2 + 2
82 = 2 · 41.
Hence, gcd(414, 662) = 2, because 2 is the last
nonzero remainder.
The Euclidean Algorithm Cont.
ALGORITHM 6 The Euclidean Algorithm
Procedure gcd(a, b: positive integers)
x := a
y := b
while y ≠ 0
begin
r := x mod y
x := y
y := r
end (gcd(a, b) is x)