1. Integers Number Theory = Properties of Integers
(For this part, assume all values are integers.)
“a|b” = “a divides b” =  nZ (b=na)
“b is a multiple of a.” “a is a factor of b.”
“Multiple” always means “integer multiple”
Thrm: If a|b and a|c, then a|(b+c).
Thrm: If a|b, then m a|mb.
Thrm: If a|b and b|c, then a|c.
UCI ICS/Math 6D

2. Division Algorithm Thrm:
There is a unique.
There is one and only one.
Thrm:
If a,dZ  d>0, then ! q,rZ (0≤r<d  a=qd+r)
d is the “divisor” (a is the “dividend”)
q is the “quotient,” q = a div d
(quotient = # of multiples of d which fit into a, if a≥0)
r is the “remainder,” r = a mod d (“a modulo d”)
Functions on pairs (a,d) a d
q = a div d
r = a mod d 17 5 3 2 51
-17 -4
UCI ICS/Math 6D

3. Congruent ... Modulo
For a, b, m integers with m>0, we say “a is congruent to b modulo m,” written a  b (mod m), iff m | (a-b)
Thrm: For a, b, m integers with m>0, a  b (mod m) iff kZ a=b+km
Thrm: For a, b, m integers with m>0, a  b (mod m) iff (a mod m) = (b mod m)
Thrm: For a, b, c, d, m integers with m>0, if a  b (mod m) and c  d (mod m), then a+c  b+d (mod m) and ac  bd (mod m).
UCI ICS/Math 6D

4. Applications of Congruences
Hashing Functions: hm(n) = (n mod m)
Range(hm) = {n | 0≤n<m}
Not injective (not one-to-one)  Collisions
{0,1,2,3,...,m-1} = “Zm”
Pseudorandom Number Generator:
nk+1 =(ank+c) mod m
Example: (a,c,m)=(3,4,7), i.e. nk+1 =(3nk+4) mod 7
n1=0 ; n2=4 ; n3=2 ; n4=3 ; n5=6 ; n6=1 ; n7=0 ; …
UCI ICS/Math 6D

5. Applications of Congruences (cont)
Example: (a,c,m)=(3,4,7), i.e. nk+1 =(3nk+4) mod 7
n1=0 ; n2=4 ; n3=2 ; n4=3 ; n5=6 ; n6=1 ; n7=0 ; …
Ceasar’s Cipher (“Shift Cipher”):
p = plaintext, encoded as integer in Z26
c = ciphertext, encoded as integer in Z26
Encrypt each letter using a fixed offset k from the alphabet’s start, e.g.:
c = Ek (p) = (p+k) mod 26
Actually, any bijection, f:Z26Z26, provides an encryption algorithm:
Examples: E(p) = (3n+13) mod 26
E(p) = (15n+7) mod 26
UCI ICS/Math 6D

6. Primes
n>1 is “prime” iff the only positive divisors of n are 1 and n itself.
n is “composite” = n is not prime.
We say “d is a factor of n” iff d is positive and d is a divisor of n.
We call d a trivial factor of n if d = 1 or n.
=> n is prime if it has no non-trivial factors.
Fundamental Theorem of Arithmetic:
Every integer n>1 is either a prime or can be written uniquely as the product of prime factors.
(“Uniquely” means “in exactly one way ignoring differences in ordering”.
e.g. 30=2·3·5 and 30=5·3·2 are same factorizations.)
UCI ICS/Math 6D

7. Factorization into Primes
e.g.
420 = 42·10 = 6·7·2·5 = 2·3·7·2·5 = 2·2·3·5·7
17 is prime
Sieve of Eratosthenes
Thrm: If n is composite, n has a prime factor whose square is at most n.
289 is not prime:
just test for i=1 to 20, if i2 | 289
(can do it only for i = 11,13,17,19…)
UCI ICS/Math 6D

8. Prime Facts Thrm: There are infinitely many primes.
Equivalently: There is no largest prime.
Prime Number Theorem:
If H(n)=|{kN | k<n  k is prime}|, then
loge(n)·H(n) / n gets arbitrarily close to as n grows large.
Consequently, H(n) ≈ n / log(n)
Proportion of numbers in [0,n] which are prime is about 1/log(n)
How to pick a 100-bit prime (e.g. for hash or a cryptosystem)?
Answer: Try random 100-bit number, test for primality.
Probability of success ≈ 1/100
=> Expected number of attempts before success ≈ 100
Thrm: If f is a (non-constant) polynomial with integer coefficients, there is an integer n s.t. f(n) is composite.
UCI ICS/Math 6D

9. Prime Conjectures Goldbach’s Conjecture:
Every even integer greater than 2 can be written as the sum of two primes.
The Twin Prime Conjecture:
There are infinitely many primes p such that p+2 is also prime.
UCI ICS/Math 6D

10. Greatest Common Divisor (gcd)
When a and b are integers, not both 0,
the “greatest common divisor” of a and b, denoted gcd(a,b),
is the largest integer d such that d|a and d|b.
Note: If a≠0, gcd(a,0)=|a|
Thrm: When a and b are integers, not both 0, if d|a and d|b, then d|gcd(a,b).
Thrm: If a and b are integers, not both 0, gcd(a,b)=gcd(b,a)
Thrm: If a and b are integers, not both 0, gcd( a , b ) = gcd( a , b mod a ) = gcd( a mod b , b )
Ref:
UCI ICS/Math 6D

11. Least Common Multiple (lcm)
If a,b>0, the “least common multiple” of a and b, denoted lcm(a,b), is the smallest m>0 such that a|m and b|m.
Thrm: If a,b>0, then a · b = gcd(a,b) · lcm(a.b)
Integers a and b are said to be “relatively prime” iff gcd(a,b)=1.
Set S of integers is said to be “pairwise relatively prime” iff each pair of (different) elements in S is relatively prime.
UCI ICS/Math 6D

12. Finding gcd’s and lcm’s
Method 1: Factor each number into primes
a=p1j1·p2j2·...·pnjn, b=p1k1·p2k2·...·pnkn.
Then
gcd(a,b)=p1min(j1,k1)·p2min(j2,k2)·...·pnmin(jn,kn).
lcm(a,b)=p1max(j1,k1)·p2max(j2,k2)·...·pnmax(jn,kn).
Method 2: Euclidean Algorithm: Find gcd(a,b) [using gcd(a,b)=gcd(a mod b,b)=gcd(b,a mod b)]
Can then compute lcm(a,b)=a·b/gcd(a,b).
Ref:
UCI ICS/Math 6D

13. Euclidean Algorithm procedure gcd(a,b: positive integers)
x := a; y := b;
repeat
r := x mod y;
x := y;
y := r
until y=0;
{gcd(a,b) is x}
(x,y) := (a,b);
(x,y) := (y, x mod y);
gcd := x
UCI ICS/Math 6D

14. Euclidean Algorithm Example
gcd(309,171)
= gcd(171,138)
= gcd(138,33)
= gcd(33,6)
= gcd(6,3)
= gcd(3,0) = 3
309=1·
171=1·138+33
138=4·33+6
33=5·6+3
6=2·3+0
UCI ICS/Math 6D

15. Greatest Common Divisor Represented as Linear Combination of a & b:
Thrm: If a and b are integers, not both 0, then  s,tZ sa + tb = gcd(a,b)
(s,t) can be found by an Extended (version of the) Euclidean Algorithm.
Ref:
UCI ICS/Math 6D

16. Extended Euclidean Algorithm: Example
gcd(309,171)
= gcd(171,138)
= gcd(138,33)
= gcd(33,6)
= gcd(6,3)
= gcd(3,0) = 3
309=1·
171=1·138+33
138=4·33+6
33=5·6+3
6=2·3+0
You can represent the final gcd (= 3) as a linear combination of value (a,b) at each step, going bottom up, i.e.
(a,b) = (33,6), (138,33), (171,138), (309,171), and finally (309,171)
3 = 33-5·6 = 33-5·(138-4·33)
= -5·138+21·33= -5·138+21·(171-1·138)
= 21·171-26·138 = 21·171-26·(309-1·171)
= -26·309+47·171 [= =3]
UCI ICS/Math 6D

17. Representations of Integers
Thrm: If b is an integer greater than 1, then any positive integer n can be written uniquely as n=akbk+ak-1bk a1b+a0, where ak≠0, 0≤ai<b for all i
(akak-1...a1a0) is a “base b expansion of n”,
(or “base b representation of n”)
Notation: (akak-1...a1a0)b
Example: (5739)10=5·103+7·102+3·101+9·100
Ref:
UCI ICS/Math 6D

18. Representations of Integers
Commonly used bases: 2, 4, 8, 10, 12, 16.
For 10<b≤36, the letters “A” to “Z” are used to designate the decimal values 10 to 35.
In particular, for base 16 (“hexadecimal”) A=10, B=11, C=12, D=13, E=14, F=15
Examples:
(231)4=2·42+3·4+1=(45)10
(276)8=2·82+7·8+6=(190)10
(2D)16=2·16+13=(45)10
(AB)16=10·16+11=(171)10
(1AB)16=1·162+10·16+11=(427) 10
UCI ICS/Math 6D

19. Computing Base Expansions
Converting from base b to base 10:
Using the powers of the base b (5134)b = 5·b3+1·b2+3·b1+4·b0
Avoiding using the powers of the base b (5134)b = b·(b·(b·5 + 1) + 3) + 4
Why? To perform fewer multiplications
[also for the “square and multiply” exponentiation algorithm on slide 21]
Converting between bases where one base is a power of the other is very easy (e.g., 2 and 8, 2 and 16), because we can do it block-by-block. For example:
( )2 = (6 9 1 D)16
General procedure for computing base b expansion of integer n:
procedure base-b-expansion (n: positive integer)
q:=n; k:=0;
repeat
ak := q mod b; q:= q div b; k := k+1;
until q=0;
{ the base b expansion of n is (akak-1...a1a0)b }
UCI ICS/Math 6D

20. Arithmetic with Base Expansions
( )2+( )2 =?
(421)8+(75)8 =?
(A1)16+(3D)16 =?
( )2 ·( )2 =?
(342)8-(173)8=?
References
9*16+12*16+3
UCI ICS/Math 6D

21. Modular Exponentiation: “Square and Multiply” Algorithm
modular exponentiation (b: integer; a,m: positive integers)
{computes ba (mod m)}
Let a = (anan-1...a1a0)2; Let x := 1 mod m; Let k := n;
repeat
if ak = 1 then x := x·b (mod m) (1)
x := x2 (mod m);
k := k-1
until k<0; {x equals ba mod m when the loop terminates}
Why does it work? First do the (base-2)→(base-10) conversion on exponent a.
Example: n=3, a=(a3a2a1a0)2 = 2·(2·(2·a3+a2)+a1)+a0
Note that we can replace the whole line (1) by the following: x := x·bak
Note also that if x=be then x·bak = be+ak. Also, if x=be then x2= b2·e.
Now look at the values of x computed in the above loop:
(k,x) = initially (3,1), then (2,b2·a3), then (1,b2·(2·a3+a2)), then (0,b2·(2·(2·a3+a2)+a1), and finally (-1,b2·(2·(2·(2·a3+a2)+a1)+a0), so the output is correct!
UCI ICS/Math 6D