Basics of Number Theory: Divisibility, Prime and Composite Numbers

1. CryptographyCS 555 Topic 6: Number Theory Basics Spring 2012/Topic 6

2. Outline and Readings • Outline • Divisibility, Prime and composite numbers, The Fundamental theorem of arithmetic, Greatest Common Divisor, Modular operation, Congruence relation • The Extended Euclidian Algorithm • Solving Linear Congruence • Readings: • Katz and Lindell: 7.1.1, 7.1.2 Spring 2012/Topic 6

3. Divisibility Definition Given integers a and b, with a  0, a divides b (denoted a|b) if  integer k, s.t. b = ak. a is called a divisor of b, and b a multiple of a. Proposition: (1) If a  0, then a|0 and a|a. Also, 1|b for every b (2) If a|b and b|c, then a | c. (3) If a|b and a|c, then a | (sb + tc) for all integers s and t. Spring 2012/Topic 6

4. Divisibility (cont.) Theorem (Division algorithm) Given integers a, b such that a>0, a<b then there exist two unique integers q and r, 0  r < a s.t. b = aq + r. Proof: Uniqueness of q and r: assume  q’ and r’ s.t b = aq’ + r’, 0  r’< a, q’ integer then aq + r=aq’ + r’  a(q-q’)=r’-r  q-q’ = (r’-r)/a as 0  r,r’ <a -a < (r’-r) < a -1 < (r’-r)/a < 1 So -1 < q-q’ < 1, but q-q’ is integer, therefore q = q’ and r = r’ Spring 2012/Topic 6

5. Prime and Composite Numbers Definition An integer n > 1 is called a prime number if its positive divisors are 1 and n. Definition Any integer number n > 1 that is not prime, is called a composite number. Example Prime numbers: 2, 3, 5, 7, 11, 13, 17 … Composite numbers: 4, 6, 25, 900, 17778, … Spring 2012/Topic 6

6. Decomposition in Product of Primes Theorem (Fundamental Theorem of Arithmetic) Any integer number n > 1 can be written as a product of prime numbers (>1), and the product is unique if the numbers are written in increasing order. Example: 84 = 2237 Spring 2012/Topic 6

7. Classroom Discussion Question (Not a Quiz) • Are the total number of prime numbers finite or infinite? Spring 2012/Topic 6

8. Greatest Common Divisor (GCD) Definition Given integers a > 0 and b > 0, we define gcd(a, b) = c, the greatest common divisor (GCD), as the greatest number that divides both a and b. Example gcd(256, 100)=4 Definition Two integers a > 0 and b > 0 are relatively prime if gcd(a, b) = 1. Example 25 and 128 are relatively prime. Spring 2012/Topic 6

9. GCD as a Linear Combination Theorem Given integers a, b > 0 and a > b, then d = gcd(a,b) is the least positive integer that can be represented as ax + by, x, y integer numbers. Proof: Let t be the smallest positive integer s.t. t = ax + by. We have d | a and d | b  d | ax + by, so d | t, so d  t. We now show t ≤ d. First t | a; otherwise, a = tu + r, 0 < r < t; r = a - ut = a - u(ax+by) = a(1-ux) + b(-uy), so we found another linear combination and r < t. Contradiction. Similarly t | b, so t is a common divisor of a and b, thus t ≤ gcd (a, b) = d. So t = d. Example gcd(100, 36) = 4 = 4  100 – 11  36 = 400 - 396 Spring 2012/Topic 6

10. GCD and Multiplication Theorem Given integers a, b, m >1. If gcd(a, m) = gcd(b, m) = 1, then gcd(ab, m) = 1 Proof idea: ax + ym = 1 = bz + tm Find u and v such that (ab)u + mv = 1 Spring 2012/Topic 6

11. GCD and Division Theorem Given integers a>0, b, q, r, such that b = aq + r, then gcd(b, a) = gcd(a, r). Proof: Let gcd(b, a) = d and gcd(a, r) = e, this means d | b and d | a, so d | b - aq , so d | r Since gcd(a, r) = e, we obtain d ≤ e. e | a and e | r, so e | aq + r , so e | b, Since gcd(b, a) = d, we obtain e ≤ d. Therefore d = e Spring 2012/Topic 6

12. Finding GCD Using the Theorem: Given integers a>0, b, q, r, such that b = aq + r, then gcd(b, a) = gcd(a, r). Euclidian Algorithm Find gcd (b, a) while a 0 do r  b mod a b  a a  r return b Spring 2012/Topic 6

13. Euclidian Algorithm Example Find gcd(143, 110) 143 = 1  110 + 33 110 = 3  33 + 11 33 = 3  11 + 0 gcd (143, 110) = 11 Spring 2012/Topic 6

14. Modulo Operation Definition: Example: 7 mod 3 = 1 -7 mod 3 = 2 Spring 2012/Topic 6

15. Congruence Relation Definition: Let a, b, n be integers with n>0, we say that a  b (mod n), if a – b is a multiple of n. Properties:a  b (mod n) if and only if n | (a – b) if and only if n | (b – a) if and only if a = b+k·n for some integer k if and only if b = a+k·n for some integer k E.g., 327 (mod 5), -1237 (mod 7), 1717 (mod 13) Spring 2012/Topic 6

16. Properties of the Congruence Relation Proposition: Let a, b, c, n be integers with n>0 • a  0 (mod n) if and only if n | a • a  a (mod n) • a  b (mod n) if and only if b  a (mod n) • if a  b and b  c (mod n), then a  c (mod n) Corollary: Congruence modulo n is an equivalence relation. Every integer is congruent to exactly one number in {0, 1, 2, …, n–1} modulo n Spring 2012/Topic 6

17. Equivalence Relation • Definition • A binary relation R over a set Y is a subset of Y  Y. We denote a relation (a,b)  R as aRb. • example of relations over integers? • Definition • A relation is an equivalence relation on a set Y, if R is • Reflexive: aRa for all a  R • Symmetric: for all a, b  R, aRb  bRa . • Transitive: for all a,b,c  R, aRb and bRc  aRc • Example • “=“ is an equivalence relation on the set of integers Spring 2012/Topic 6

18. More Properties of the Congruence Relation Proposition: Let a, b, c, n be integers with n>0 If a  b (mod n) and c  d (mod n), then: a + c  b + d (mod n), a – c  b – d (mod n), a·c  b·d (mod n) E.g., 5  12 (mod 7) and 3  -4 (mod 7), then, … Spring 2012/Topic 6

19. Multiplicative Inverse Definition: Given integers n>0, a, b, we say that b is a multiplicative inverse of a modulo n if ab  1 (mod n). Proposition: Given integers n>0 and a, then a has a multiplicative inverse modulo n if and if only if a and n are relatively prime. Spring 2012/Topic 6

20. Towards Extended Euclidian Algorithm • Theorem: Given integers a, b > 0, then d = gcd(a,b) is the least positive integer that can be represented as ax + by, x, y integer numbers. • How to find such x and y? Spring 2012/Topic 6

21. First computes b = q1a + r1 a = q2r1 + r2 r1 = q3r2 + r3  rk-3 =qk-1rk-2+rk-1 rk-2 = qkrk-1 Then computes x0 = 0 x1 = 1 x2 = -q1x1+x0  xk = -qk-1xk-1+xk-2 The Extended Euclidian Algorithm And y0 = 1 y1 = 0 y2 = -q1y1+y0  yk = -qk-1yk-1+yk-2 We have axk + byk = rk-1 = gcd(a,b) Spring 2012/Topic 6

22. Extended Euclidian Algorithm Extended_Euclidian (a,b) x=1; y=0; d=a; r=0; s=1; t=b; while (t>0) { q = d/t; u=x-qr; v=y-qs; w=d-qt; x=r; y=s; d=t; r=u; s=v; t=w; } return (d, x, y) end Invariants: ax + by = d ar + bs = t Spring 2012/Topic 6

23. Another Way Find gcd(143, 111) 32 = 143  1  111 15 = 111  3  32 = 4111  3 143 2 = 32  2  15 = 7 143  9  111 1 = 15 - 7  2 = 67  111 – 52 143 • 143 = 1  111 + 32 • 111 = 3  32 + 15 • = 2  15 + 2 • 15 = 7  2 + 1 gcd (143, 111) = 1 Spring 2012/Topic 6

24. Linear Equation Modulo n If gcd(a, n) = 1, the equation has a unique solution, 0< x < n. This solution is often represented as a-1 mod n Proof: if ax1 1 (mod n) and ax2  1 (mod n), then a(x1-x2)  0 (mod n), then n | a(x1-x2), then n | (x1-x2), then x1-x2=0 How to compute a-1 mod n? Spring 2012/Topic 6

25. Examples Example 1: • Observe that 3·5  1 (mod 7). • Let us try to solve 3·x+4  3 (mod 7). • Subtracts 4 from both side, 3·x  -1 (mod 7). • We know that -1  6 (mod 7). • Thus 3·x  6 (mod 7). • Multiply both side by 5, 3·5·x  5·6 (mod 7). • Thus, x  1·x  3·5·x  5·6  30  2 (mod 7). • Thus, any x that satisfies 3·x+4  3 (mod 7) must satisfy x  2 (mod 7) and vice versa. Question: To solve that 2x  2 (mod 4). Is the solution x1 (mod 4)? Spring 2012/Topic 6

26. Linear Equation Modulo (cont.) To solve the equation When gcd(a,n)=1, compute x = a-1 b mod n. When gcd(a,n) = d >1, do the following • If d does not divide b, there is no solution. • Assume d|b. Solve the new congruence, get x0 • The solutions of the original congruence are x0, x0+(n/d), x0+2(n/d), …, x0+(d-1)(n/d) (mod n). Spring 2012/Topic 6

27. Solving Linear Congruences Theorem: • Let a, n, z, z’ be integers with n>0. If gcd(a,n)=1, then azaz’ (mod n) if and only if zz’ (mod n). • More generally, if d:=gcd(a,n), then azaz’ (mod n) if and only if zz’ (mod n/d). Example: • 5·2  5·-4 (mod 6) • 3·5  3·3 (mod 6) Spring 2012/Topic 6

28. Basic Number Theory Divisibility Let a,b be integerswith a≠0. if there exists an integer k such that b=ka, we say a divides b which is denoted by a|b 11|143, 1993|3980021 ◇ if a≠0, then a|0 and a|a; 1|b for each b a|b and b|c → a|c a|b and a|c → a|sb+tc for all s, t

29. Prime Numbers An integer p>1 that is divisible only by 1 and itself is called a prime number, otherwise it iscalled composite (P.64) primegen.c generates prime numbers Let π(x) be the number of primes less than x, then π(x) ≈x/ln(x) as x→∞ Exercise Plot π(x) vs. x for x=216 to 232

30. A Plot of π(x)≈x/ln(x) vs. x

31. Prime Factorization Theorem Every positive integer is a product of primes. This factorization into primes is unique, up to reordering the factors 49500=22 32 5311 If a prime p|ab, then either p|a or p|b Moreover, p|x1 x2 … xn →p|xj for some j 7|14•30,

32. Greatest Common Divisor gcd gcd(343, 63)=7, gcd(12345,11111)=1 gcd(1993,3980021)=1993 Euclidean Algorithm to compute gcd(a,b) does not require the factorization of the numbers and is fast. gcd(482,1180)=2

33. Solving ax+by=1 when gcd(a,b)=1 Let a,b be integers with a2 +b2 ≠0, and gcd(a,b)=1, then ax+by=1 has an integer solution (x,y) ♪ Euclidean Algorithm Example 7(-2) + 5(3) =1 Solving ax+by=d with gcd(a,b)=d can be reduced as solving a0x + b0y = 1 where a=a0d, b=b0d

34. Congruences Let a,b,n be integers with n≠0. We say that a≡b (mod n) {read as a is congruent to b mod n} if n|(a-b) a=b+nk for an integer k is another description Example 32≡7 (mod 5)

35. Simple Properties Let a,b,c,n be integers with n≠0 (1) a≡0 (mod n) iff n|a (2) a≡a (mod n) (3) a≡b (mod n) iff b≡a (mod n) (4) a≡b and b≡c (mod n) → a≡c (mod n) (5) a≡b and c≡d (mod n) → a+c≡b+d, a−c≡b−d, ac≡bd (mod n) (6) ab≡ac (mod n) with n≠0, and gcd(a,n)=1, then b≡c (mod n)

36. Computational Properties Finding a-1 (mod n) Solving ax≡c (mod n) when gcd(a,n)=1 What if gcd(a,n)>1 ☺Solve 11111x≡4 (mod 12345) ☻Solve 12x≡21 (mod 39) ♫ How to solve x2 ≡a (mod n)? □ Working with fractions (inverse ?)

37. The Chinese Remainder Theorem Let m1,m2, …, mk be integers with gcd(mi,mj) = 1, there exists only one solution x (mod m1 m2…mk) to the simultaneous congruences [P.76-78] x≡a1 (mod m1) x≡a2 (mod m2) : : x≡ak (mod mk)

38. Fermat's Little Theorem How to fast evaluate 21234 (mod 789)? How to fast evaluate Xa (mod n)? If p is a prime and gcd(p,a)=1, then ap-1 ≡ 1 (mod p)

39. Euler’s φ-Function and Theorem φ(n)= #{a | 1 ≤ a ≤ n, gcd(a,n)=1}, that is, the number of positive integers which are relatively prime to n Examples: φ(15)=8, φ(16)=8, φ(17)=16 φ(pq)=(p-1)(q-1) if p and q are primes φ(p)=p-1 if p is a prime number φ(pr)=pr-pr-1=pr(1- 1/p) If gcd(a,n)=1, then aφ(n)≡ 1 (mod n)

40. Examples and Basic Principle [Page 82] What are the last three digits 7803 ? Compute 243210 (mod 101) Let a,n,x,y be integers with n≥1 and gcd(a,n)=1. If x≡y (mod φ(n)), then ax ≡ ay (mod n) (Hint) x=y+kφ(n); by Euclidean Theorem

41. Primitive Roots If p is a prime, a primitive root mod p is a number g whose power yield every nonzero class mod p. {gk|0<k<p}={1,2,…,p-1} Proposition: Let g be a primitive root mod p gn≡1 (mod p) iff (p-1)|n or n≡0 (mod p-1) gj≡gk (mod p) iff j≡k (mod p-1) ♪ 3 is a primitive root mod 7 but not for mod 13

42. Inverting Matrices (mod n) A matrix M is invertible under (mod n) if gcd(det(M), n)=1 The inverse of A=[1 2;3 4] (mod 11) is A-1 =[9 1 ; 7 5] and det(A)= -2≡9 (mod 11) The inverse of M=[1 1 1; 1 2 3; 1 4 9] under (mod 11) is [3 3 6; 8 4 10; 1 4 6], where det(M)= ½ ≡ 6 (mod 11)

43. Square Roots mod n (1/9) X2 ≡71 (mod 77) has solutions ±15, ±29 How to (efficiently) solve X2 ≡b (mod pq), where p,q are (very close) primes? Every prime p (except 2) must satisfy p≡1 (mod 4) or p≡3 (mod 4) The square roots of 5 mod 11 are ±4

44. Square Roots mod n (2/9) Let p≡3 (mod 4) be prime and y is an integer such that x≡y(p+1)/4 (mod p). ♪ If y has a square root mod p, then the square roots of y mod p are x and –x ♪ If y has no square roots mod p, then –y has a square root mod p, and the square roots of –y are x and –x.

45. Square Roots mod n (3/9) Proof: x4 ≡yp+1≡y2 . yp-1 ≡y2 (mod p) → (x2 + y) (x2 - y) ≡ 0 (mod p) Suppose both y and –y are squares mod p This is impossible.

46. Square Roots mod n (4/9) Lemma: Let p ≡ 3 (mod 4) be prime, then X2 ≡ -1 (mod p) has no solutions. Proof: Let p = 4q+3 X2 ≡ -1→ Xp-1 ≡ -1(p-1)/2≡ -12q+1 ≡-1 But Xp-1 ≡ 1 (Fermat’s theorem)

47. Square Roots mod n (5/9) Suppose both y and –y are squares mod p, say y ≡ a2 and -y ≡ b2. Then (a/b)2 ≡ -1 (mod p) But according to the previous lemma, (a/b)2 ≡ -1 (mod p) is impossible

48. Square Roots mod n (6/9) y ≡ x2 (mod p), the square roots of y are ± x. -y ≡ x2 (mod p), the square roots of -y are ± x.

49. Examples for Square Roots (7/9) x2 ≡ 5 (mod 11) (p+1)/4 = 3 x≡ 53 ≡ 4(mod 11) Since 43 ≡ 5 (mod 11), the square root of 5 mod 11 are ±4

50. Examples for Square Roots (8/9) ◎ To solve x2≡ 71 (mod 77) x2≡ 1 (mod 7) → x ≡±1 (mod 7) x2≡ 5 (mod 11) → x ≡±4 (mod 11) By Chinese remainder theorem x ≡±15 , x ≡±29 (mod 77)