1. Number Theory

2. Integers and Division Definition 1:
Let a and b be integers where a0. We say a divides b, denoted by a|b, if there is an integer c such that b=ac.
If a does not divide b, we write ab.
When a|b, we say a is a factor of b or b is a multiple of of a.
Examples:
7 | 56, 1 | 56, 8 | , 7 and 8 are factors of 56.
7 | 53, 17 | 53.
Number Theory
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3. Properties of Divisibility
Theorem 1: Let a, b and c be any integers.
a|0, 1|a and a|a.
If a|b and a|c then a|(b+c).
If a|b then a|bc.
If a|b and b|c then a|c.
Example:
7|0, 1|7 and 7|7.
3|12 and 3|9. Then, 3|(12+9).
3|12. Then, 3|(127).
3|21 and 21|189. Then, 3|(189).
Number Theory
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4. Properties of Divisibility
Theorem 1: Let a, b and c be any integers.
a|0, 1|a and a|a.
If a|b and a|c then a|(b+c).
Proof:
Let a, b and c be any integers.
Since 0=a0, a|0. Since a=1a, 1|a. Since a=a1, a|a.
Let a|b and a|c. Then, there are integers k1 and k2 such that b=k1a and c=k2a. Thus, b+c=k1a+k2a = a(k1+k2). Therefore, a|(b+c).
Number Theory
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5. Properties of Divisibility
Theorem 1: Let a, b and c be any integers.
If a|b then a|bc.
If a|b and b|c then a|c.
Proof:
Let a, b and c be any integers.
Let a|b. Then, there is an integer k such that b=ka. Thus, b c=kac. Therefore, a|(bc).
Let a|b and b|c. Then, there are integers k1 and k2 such that b=k1a and c=k2b. Thus, c=k1k2a. Therefore, a|c.
Number Theory
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6. 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.
Proof:
Let a, b and c be integers, and a|b and a|c.
Since a|b, a|mb for any integer m. (from a|bc if a|b)
Since a|c, a|nc for any integer n.
Since a|mb and a|nc, a|mb+nc (from if a|b and a|c then a|b+c).
Q.E.D.
Number Theory
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7. Division Algorithm
Theorem 2: Let a be an integer and d be a positive integer. Then, there are unique q and r, with 0 r<d, such that a=dq+r.
Proof:
Let a be an integer and d be a positive integer.
Let S ={r | rZ, r>0, and r = a-dq where q is an integer}.
S is not empty because we can choose q as needed.
By the well-ordering property, there is the smallest element, say r0, in S. Then, there is q0 such that r0 = a-dq0.
If r0  d, there is a smaller integer a-dq0-d in S, which contradicts to the fact that r0 is the smallest element in S.
Thus, r < d.
Number Theory
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8. Division Algorithm
Now, we proved that there are q and r, with 0 r <d, such that a=dq+r. Next, we will prove that q and r are unique.
Assume there exist q, q', r and r' such that a = dq+r = dq'+r', with 0  r, r' <d.
Then, d(q - q') =r - r'.
That is, d | (r - r').
Since 0  r, r' <d, -d  r - r' < d.
From d | (r - r') and -d  r - r' < d, r - r' = 0, which means r = r'.
Then, q = q'.
Therefore, there are unique q and r such that a=dq+r.
Number Theory
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9. Division Algorithm Definition 2:
Let a be an integer and d be a positive integer, such that there exist integers q and 0  r < d where a=dq+r.
a is called the dividend, d is called the divisor, q is called the quotient, and r is called the remainder.
q = a div d r = a mod d
Number Theory
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10. Modular Arithmetic Definition 3:
If a and b are integers and m is a positive integer, then a is congruent to b modulo m (denoted by a  b (mod m) ) if m divides a-b.
If a is not congruent to b modulo m , we write a  b (mod m).
(a  b (mod m) means the residues of a/m and b/m are equal)
Example:
26  14 (mod 12), 26  14 (mod 4), 26  14 (mod 3)
Number Theory
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11. Theorem 3.1 Let a and b be integers and m be a positive integer.
a  b (mod m) if a mod m = b mod m.
Proof:
Let a and b be integers and m be a positive integer such that a mod m = b mod m.
Then, there exist integers q1, q2 and r such that a = q1m+r, and b = q2m+r (from division algorithm).
That is, a-b = (q1-q2)m. Then, m|a-b.
Thus, a  b (mod m).
Number Theory
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12. Theorem 3.2
Let a and b be integers and m be a positive integer. If a  b (mod m) then a mod m = b mod m.
Proof:
Let a and b be integers and m be a positive integer such that a  b (mod m).
Then, m|a-b.
That is, there exists an integer c such that a-b = cm.
There exist integers q1-q2=c, and a-b = m(q1-q2).
Then, there is an integer r such that r = a-q1m = b-q2m.
As a result, a = q1m+r, and b = q2m+r.
Thus, a mod m = b mod m.
Number Theory
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13. Theorem 4 Let a and b be integers, and m be a positive integer.
a  b (mod m) iff there is an integer k such that a = b + km.
Proof:
()
If a  b (mod m) then m | (a-b). This means there is an integer k such that a -b = km. Then, a = b + km.
()
If there is an integer k such that a = b + km, then a-b = km. Then, m | (a-b). That is, a  b (mod m).
Number Theory
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14. 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).
Proof:
Let a  b (mod m) and c  d (mod m).
Then, there are integers s and t such that b = a + sm and d = c + tm.
Then, b+d = a+c+(s+t)m and bd = ac+(sc+at+stm)m.
That is, a+c  b+d (mod m) and ac  bd (mod m).
Number Theory
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15. Corollary 2 Let m be a positive integer and let a and b be 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:
By the definitions of mod m and congruence,
a mod m(a mod m)(mod m), and b mod m(b mod m)(mod m).
From Theorem 5,
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16. Applications of Congruences
Hashing functions
h(k) = k mod m
Pseudorandom numbers
xn+1 = (axn + c) mod m
Caesar’s cipher
f(p) = (p + k) mod 26
Number Theory
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17. Primes

18. Primes
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 p greater than 1 is called composite if it is not prime.
Examples:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, … are prime.
Number Theory
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19. Sieve of Eratostheses 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
100
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20. Theorem 1 Fundamental Theorem of Arithmetic
Every positive integer greater than 1 can be written uniquely as a prime or as the product of two or more primes where the prime factors are written in the order of non-decreasing size.
Meaning
Any integer k = 2d2  3d3  5d5  7d7  11d11  13d13 …, where d2, d3, d5, d7, d11, d13, …  0.
Examples:
365 = 20  30  51  70  110  130  170  190  230  290  310  370  410  430  470  530  590  610  670  710  731
Number Theory
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21. Theorem 2
If n is a composite integer, then n has a prime divisor d n.
Proof:
If n is composite, then there is a factor 1 < a < n.
Then, there is integer b greater than 1 such that ab = n.
There are 5 possible cases of a and b.
If a <n and b <n, then ab <nn. But ab  n, which contradicts to our prior information.
If a >n and b >n, then ab >nn=n. But ab  n, which contradicts to our prior information.
If a > n and b <n , then ab is possibly equal to n.
If a < n and b >n , then ab is possibly equal to n.
If a = n and b =n, then ab=n.
Number Theory
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22. Theorem 2 That is, a  n or b n .
Then, if a (or b) is not a prime, a (or b) itself has a prime factor, say a', which is a divisor of n and a' n.
Therefore, n has a prime divisor d n
Number Theory
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23. Applications of Theorem 2
To test is a number n is prime, we need only to find prime divisors less than or equal to n.
Example:
Show that 271 is prime.
Since 271<17, we only need to find prime divisors which are  16.
That is, we need to consider 2, 3, 5, 7, 11 and 13.
All of them do not divides 271.
Thus, 271 is prime.
Number Theory
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24. Applications of Theorem 2
To find the prime factorization of a composite n, we need only to consider prime divisors less than or equal to n, as shown in the following example.
Example:
Find the prime factorization of 3003.
3003 < 55. We only need to try 2, 3, 5, 7, 11, 13, … , 47 and 53.
3003/3=1001.
1001 < 32. We only need to try 2, 3, 5, 7, 11, 13, … and 31.
1001/7=143.
143 < 12. We only need to try 2, 3, 5, 7, and 11.
None divides Then, 143 is prime.
That is, the prime factorization of 3003 is 3  7  143.
Number Theory
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25. Theorem 3 There are infinitely many primes. Proof:
Assume there are finite primes p1, p2, …, pn.
Let Q = p1 p2  …  pn +1.
Assume Q is not prime.
Then, there is a prime pi, for some 1  i  n, such that pi divides Q (Q = c  pi).
Then, pi divides Q - p1 p2  …  pn, which is 1 (from the way we set Q). That is a contradiction.
Thus, Q is prime.
Since Q is prime and Q > pn , it contradicts to our assumption.
That is, there are infinitely many primes.
Number Theory
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26. Prime Number Theorem
The ratio of the number of primes not exceeding x and x/ln x approaches 1 as x grows without bound.
In other words,
Let (x) be the number of
primes not exceeding x.
lim (x) = 1
x   x/ln x
That is, (x)  x/ln x. x
x/ln(x)
10.00
4.34
144.76
Number Theory
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27. Greatest Common Divisors
Definition 2: Let a and b be integers, not both zero.
The largest integer d such that d | a and d | b is called the greatest common divisor of a and b, denoted by gcd(a, b).
Examples:
Find gcd(125, 75) = 25. (125 = 555, 75 = 355)
Find gcd(23, 161) = 23. (161 = 23  7)
Find gcd(23, 127) = 1. (23 and 127 are primes.)
Find gcd(69, 194) = 1. (69 = 323, 194 = 297)
Number Theory
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28. Relatively Prime
Definition 3: The integers a and b are relatively prime if gcd(a, b) = 1.
Example: gcd(69, 194) = 1. (69=323, 194=297) Then, 69 and 194 are relatively prime.
Definition 4: The integers a1, a2, …, an are pairwise relatively prime if gcd(ai, aj) = 1, for all 1 i, j  n.
Example: gcd(21, 25)=1, gcd(25, 32)=1, gcd(21,32)=1.
Then, 21, 25, and 32 are relatively prime.
Number Theory
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29. Least Common Multiples
Definition 5: Let a and b be positive integers.
The smallest integer d such that a | d and b | d is called the least common multiple of a and b denoted by lcm(a, b).
Example:
lcm(23 52 11, 22 33 53 132) = 23  33  53 11 132.
Number Theory
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30. gcd and lcm Theorem 5: Let a and b be positive integers. Then,
a b = gcd(a, b)lcm(a, b).
Number Theory
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31. Integer Representation & Algorithms

32. 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 + … + a1b + a0, where k is a nonnegative integer, ak, ak-1 , …, a1 , a0 are nonnegative integers less than b, and ak  0.
This is called base b expansion of n.
Examples:
Let b = = 3   
Let b = = 1    21 + 1
Number Theory
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33. 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 + … + a1b + a0, where k is a nonnegative integer, ak, ak-1, …, a1, a0 are nonnegative integers less than b, and ak  0.
Proof:
Basis: Consider 0 < k < b. n can be expressed as k.
Induction hypothesis: For n < bk, n can be expressed as ak-1bk-1 + … + a1b + a0.
Induction Step: For bk  n < bk+1, let m = n – abk, for largest possible a which makes m positive.
m < bk, and m can be expressed as ak-1bk-1 + … + a1b + a0 (from the induction hypothesis.)
Since m = n – abk, we have n = abk + ak-1bk-1 + … + a1b + a 
Number Theory
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34. Constructing Base b Expansion
procedure expand(b, n: positive integers)
q := n
k := 0
while q  0
begin
ak := q mod b
q := q/b
k := k+1
end
Number Theory
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35. Base b Expansion Base 16 (Hexadecimal) digits
A B C D E F
(5A)16 = (516 +10)10 = (90)10
Base 8 (Octal) digits
(403)8=(4 82+08+3)10=(259)10
Number Theory
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36. Base 2 Addition
Let a = (an-1 an-2 … a1 a0)2 and b = (bn-1 bn-2 … b1 b0)2.
s = a + b.
a0 + b0 = 2c0 + s0
a1 + b1 + c0= 2c1 + s1 …
an + bn + cn-1= 2cn + sn a b s
Number Theory
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37. Base 2 Addition
Let a = (an-1 an-2 … a1 a0)2 and b = (bn-1 bn-2 … b1 b0)2.
procedure add (a, b: positive integers)
c := 0
for j := 0 to n-1
begin
d := (aj + bj + c) / 2
si := aj + bj + c – 2d
c := d
end
sn := c
Number Theory
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38. Base 2 Multiplication
Let a = (an-1 an-2 … a1 a0)2 and b = (bn-1 bn-2 … b1 b0)2.
ab = a (bn-1 bn-2 … b1 b0)2
= a (2n-1bn-1 + 2n-2bn-2 + … + 21b1 + 20b0)
= a(2n-1bn-1) + a(2n-2bn-2) + … + a(21b1)+ a(20b0) 
Number Theory
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39. Base 2 Multiplication
Let a = (an-1 an-2 … a1 a0)2 and b = (bn-1 bn-2 … b1 b0)2.
procedure multiply (a, b: positive integers)
for j := 0 to n-1
begin
if bj = 1 then cj := a << j {<< means shift}
else cj := 0
end
p := 0
p := p + cj
Number Theory
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40. Base 2 Multiplication a = (1001)2 = (9)10 b = (1011)2 = (11)10
Let a = (an-1 an-2 … a1 a0)2 and b = (bn-1 bn-2 … b1 b0)2.
procedure multiply (a, b: positive integers)
p := 0
for j := 0 to n-1
begin
if bj = 1 then p := p + a
a := a << 1
end bj p a 1
(1001)2 = 9 9
(10010)2 = 18 27
(100100)2 = 36
( )2 = 72 99
( )2 = 144
a = (1001)2 = (9)10
b = (1011)2 = (11)10
Number Theory
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41. Division d = 3 procedure division (a, d: positive integers) q := 0
r := a
while r  d
begin
r := r - d
q := q + 1
end
{r is a div d, q is a mod d}
d = 3 q r 19 1 16 2 13 3 10 4 7 5 6
Number Theory
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42. Modular Exponentiation
Let a = (an-1 an-2 … a1 a0)2
= (2n-1an-1 + 2n-2an-2 + … + 21a1 + a0)
ba = b2n-1an-1  b2n-2an-2  …  b23a3  b22a2  b2a1  ba0
From (ab) mod m  ((a mod m)(b mod m)) (mod m) ,
ba mod m
= (b2n-1an-1 mod m)(b2n-2an-2 mod m)…(b23a3 mod m)(b22a2 mod m) (b2a1 mod m)(ba0 mod m)
0 or 1
square
square
square
square
Number Theory
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43. Modular Exponentiation
procedure exp (b, n , m : positive integers)
x := 1
power := b
for j := 0 to k-1 (k-bit binary a)
begin
if ai = 1 then x := (x  power) mod m
power := (power  power) mod m
end
{x is bn mod m}
Number Theory
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44. LEMMA 1: Euclidean Algorithm
Let a = bq+r, where a, b, q and
r are integers.
Then, gcd(a,b) = gcd(b,r).
Proof:
Let a, b, q and r be integers such that a = bq+r.
Suppose d is a common divisor of a and b.
Then, d divides both a and b.
Then, d divides r = a – bq.
Thus, d is also a common divisor of b and r.
Suppose d is a common divisor of b and r.
Then, d divides both b and r.
Then, d divides a = bq + r.
Thus, d is a common divisor of a and b.
Therefore, d is a common divisor of a and b iff it is a common divisor of b and r.
That is, gcd(a,b) = gcd(b,r). 
Number Theory
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45. 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
{x is gcd(a, b)} x y
165 70 15 10 5
mod
mod
mod
mod
Number Theory
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46. THEOREM
If a and b are positive integers, then there exist integers s and t such that gcd(a, b) = sa + tb.
y3= x2-y2 x3= y2
y2= x1-4y1 x2= y1
y1= x0-2y0 x1= y0
y3 = x2-y2
= y1 -(x1 -4y1)
= 5y1 - x1
= 5(x0-2y0) -y0
= 5x0-11y0 (x0 = a, y0 = b)
 gcd(a, b) = 5a – 11b  i x y
165 70 1
y0= 70
x0-2y0 = 15 2
y1= 15
x1-4y1 = 10 3
y2= 10
x2- y2 = 5 4
y3= 5
x3-2y3 = 0
Number Theory
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47. LEMMA 1
If a, b and c are positive integers such that gcd(a,b)=1 and a | bc, then a | c.
Proof:
Let a, b and c be positive integers such that gcd(a,b)=1 .
By Theorem 1, there are integers s and t such that sa +tb = gcd(a,b) = 1.
Then, sac + tbc = c. s = (c – tbc)/a
Therefore, a|c 
Number Theory
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48. LEMMA 2
If p is a prime and p | a1 a2 … an, where each ai is an integer, then p | ai for some i.
Number Theory
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49. THEOREM
Let m be a positive integer and let a, b, and c be integers. If ac  bc (mod m) and gcd(c,m) = 1, then a  b (mod m).
Proof:
Let m be a positive integer and a, b, and c be integers such that ac  bc (mod m) and gcd(c,m) = 1.
Since ac  bc (mod m) , m | ac – bc.
From gcd(c,m) = 1, m does not divide c.
Then, m | a – b.
That is, a  b (mod m) 
Number Theory
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50. Linear Congruence
Let m be a positive integer, a and b be integers and x be a variable.
ax  b (mod m) is called a linear congruence.
Example:
3x  4 (mod 7)
x  6 (mod 7) x 3x
3x mod 7 1 3 2 6 9 4 12 5 15 18 7 21 8 24
Number Theory
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51. Inverse of a modulo m
Let m be a positive integer, a and b be integers and x be a variable.
If ax  1 (mod m), a is an inverse of x modulo m.
Example:
From 3x  1 (mod 7) , x  5 (mod 7)
Then, 3 is an inverse of 5 modulo 7.
Number Theory
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52. Theorem 3
If a and m are relatively prime integers, and m > 1, then an inverse of a modulo m exists.
Proof:
Let a and m are relatively prime integers, and m > 1.
Then, gcd(a, m) =1.
From Theorem 1, there exist integers s and t such that sa + tm = 1.
Therefore, sa + tm  1 (mod m).
Since tm  0 (mod m), sa  1 (mod m).
Thus, s is an inverse of a modulo m.  
Number Theory
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53. Chinese Remainder Theorem
Let m1, m2, …, mn be pairwise relatively prime positive integers and a1, a2, …, an be arbitrary integers. Then, the system
x  a1 (mod m1),
x  a2 (mod m2), …
x  an (mod mn)
has a unique solution modulo m = m1 m2…  mn. 
Number Theory
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54. Chinese Remainder Theorem: Proof
Let m1, m2, …, mn be pairwise relatively prime positive integers, m = m1 m2…  mn, and a1, a2, …, an be arbitrary integers.
Let Mk = m/mk, for k = 1, 2, …, n.
gcd(mk, Mk) = 1 because m1, m2, …, mn are pairwise relatively prime.
From Theorem 3, there is an integer yk, an inverse of Mk modulo mk. That is, Mk yk  1 (mod mk).
Let x = a1 M1 y1 + a2 M2 y2 + … + an Mn yn.
Since Mk yk  1 (mod mk), x  akMk yk  ak (mod mk).
Then, x is a simultaneous solution to the n congruences.
The rest is to prove the uniqueness.
Number Theory
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55. Chinese Remainder Theorem : Example
Find x such that x  2 (mod 3),
x  3 (mod 5),
x  2 (mod 7).
Since 3,5 and 7 are pairwise relatively prime, from Chinese Remainder Theorem
x = a1 M1 y1 + a2 M2 y/ + … + a3 M3 y3, where
a1=2, a2=3, a3=2, m1=3, m2=5, m3=7. Then, m = m1 m2 m3= 357 = 105.
M1= m/m1= 357/3 = 35, M2= m/m2= 357/5 = 21, M3= m/m3= 357/7 = 15.
Then, we need to solve the following linear congruence Mk yk  1 (mod mk), for k = 1,2,3.
35 y1  1 (mod 3)
21 y2  1 (mod 5)
15 y3  1 (mod 7)
We have y1 = 2, y2 = 1, y3 = 1.
Thus, x = 2352 + 3211 + 2151 = 233  23 (mod 105).
Number Theory
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56. Large Integer Representation
Let m1, m2, …, mn be pairwise relatively prime positive integers greater than 1, and m = m1 m2…  mn. An integer a with 0  a  m can be uniquely represented by the n-tuple (a mod m1, a mod m2, …, a mod mn).
Example:
3 and 4 are pairwise relatively prime. Any integer not greater than 34 = 12 can be represented uniquely by an order pair.
0 = (0 mod 3, 0 mod 4) = (0, 0) = ( 6 mod 3, 6 mod 4) = (0, 2)
1 = (1 mod 3, 1 mod 4) = (1, 1) = ( 7 mod 3, 7 mod 4) = (1, 3)
2 = (2 mod 3, 2 mod 4) = (2, 2) = ( 8 mod 3, 8 mod 4) = (2, 0)
3 = (3 mod 3, 3 mod 4) = (0, 3) = ( 9 mod 3, 9 mod 4) = (0, 1)
4 = (4 mod 3, 4 mod 4) = (1, 0) 10 = (10 mod 3, 10 mod 4) = (1, 2)
5 = (5 mod 3, 5 mod 4) = (2, 1) 11 = (11 mod 3, 11 mod 4) = (2, 3)
Number Theory
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57. Computer Arithmetic with Large Integers
Example:
99, 98, 97, and 95 are pairwise relatively prime, and and are less than 99989795.
can be represented by ( mod 99, mod 98, mod 97 , mod 95) = (33,8,9,89).
can be represented by ( mod 99, mod 98, mod 97 , mod 95) = (32,92,42,16).
Number Theory
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58. Computer Arithmetic with Large Integers
= (33,8,9,89) + (32,92,42,16)
= (65 mod 99, 100 mod 98, 51 mod 97 , 105 mod 95)
= (65, 2, 51, 10)
x  65 (mod 99)
x  2 (mod 98)
x  51 (mod 97)
x  10 (mod 95)
From the system of linear congruences, x =
Number Theory
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59. Fermat’s Little Theorem
If p is prime and a is an integer not divisible by p, then
ap-1  1 (mod p).
For every integer a, ap  a (mod p).
Example:
Since 2 is prime, and 341 is not divisible by 2,
2340  1 (mod 341). 
Number Theory
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60. Psuedoprime Definition
Let b be a positive integer. If n is a composite positive integer, and bn-1  1 (mod n), then n is called a pseudoprime to the base b.
Number Theory
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61. Cryptography
An Introduction

62. Cryptography receiver sender eavesdropper My password is 3791.
Number Theory
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63. Cryptography plaintext plaintext decryption encryption ciphertext
AOD4BNU6DRTU7O
TYTBPTJODE9AOF
My password is 3791.
decryption
encryption
ciphertext
AOD4BNU6DRTU7O
TYTBPTJODE9AOF
My password is 3791.
receiver
sender
eavesdropper
Number Theory
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64. Keys plaintext ciphertext plaintext Decryption key Encryption key
Number Theory
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65. Cryptography plaintext plaintext encryption decryption ciphertext
receiver
sender
eavesdropper
Number Theory
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66. Private Key Encryption
plaintext
plaintext
KEY = a
KEY = a
Easy to find decryption,
when encryption key is
known.
encryption
decryption
ciphertext
KEY = a
receiver
sender
The key a must be a secret
kept between the sender
and the receiver.
What if the eavesdropper
gets the key?
eavesdropper
Number Theory
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67. Public Key Encryption key b key a key a plaintext plaintext KEY = a, b
public KEY = a
encryption
decryption
key b
secret
ciphertext
key a
key a
receiver
sender
Eavesdroppers can only
encrypt messages,
but cannot decrypt
any message.
eavesdropper
Number Theory
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68. RSA Cryptosystem Let C denote a ciphertext and M denote a plaintext.
Let p and q be large primes, and n=pq.
Let e be an integer that is relatively prime to (p-1)(q-1).
Let d be an inverse of e modulo (p-1)(q-1).
Encryption:
C = Me mod n.
Decryption:
M  Cd (mod n).
Number Theory
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69. RSA Cryptosystem
Let p and q be large primes, and n=pq. Let e be an integer which is relatively prime to (p-1)(q-1), and d be an inverse of e modulo (p-1)(q-1).
Prove that if C = Me mod n, then Cd  M (mod n)
Proof:
Since d is an inverse of e modulo (p-1)(q-1),
de  1 (mod (p-1)(q-1))
Cd  (Me)d = Mde = M1+k(p-1)(q-1) (mod n)
Thus, Cd  M1+k(p-1)(q-1) = M (Mq-1)k(p-1) (mod p), and
Cd  M1+k(p-1)(q-1) = M (Mp-1)k(q-1) (mod q)
Number Theory
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70. RSA Cryptosystem
From Cd  M1+k(p-1)(q-1) = M (Mq-1)k(p-1) (mod p), and
Cd  M1+k(p-1)(q-1) = M (Mp-1)k(q-1) (mod q)
By Fermat’s Little Theorem, if gcd(M, p) = gcd(M,q) =1 then Mp-11 (mod p) and Mq-11 (mod q).
Then, Cd  M (Mq-1)k(p-1)  M1  M (mod p)
Cd  M (Mq-1)k(p-1)  M1  M (mod q)
By the Chinese Remainder Theorem,
Cd  M (mod pq)
Number Theory
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71. Example: RSA Cryptosystem
Let p = 3, and q = 19, n = 319 = 57.
Let e = 23, which is relatively prime to 218=36.
Since an inverse of 23 mod 36 = 11, d = 11.
Encryption: C = M 23 mod 57.
Given a plaintext M = 2,
C = mod 57 = 32.
Decryption: B = C 11.
B = 3211 mod 57 = 2.
Number Theory
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72. Example: RSA Cryptosystem
Let p = 43, and q = 59, n = 4359 = 2537.
Let e = 13, which is relatively prime to 4258=2436.
Since an inverse of 13 mod 2436 = 937, d = 937.
Encryption: C = M13 mod 2537.
Given a plaintext M = 1819,
C = mod 2537 = 2081.
Decryption: B = C 937.
B = mod 2537 =1819.
Number Theory
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