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Códigos y Criptografía Francisco Rodríguez Henríquez CINVESTAV e-mail: [email protected] Códigos y Criptografía Francisco Rodríguez Henríquez Number Theory: Some definitions and Theorems Códigos y Criptografía Francisco Rodríguez Henríquez Definitions The set of integers {…, -3, -2, -1, 0, 1, 2, 3, …} is denoted by the symbol Z. Let a, b be integers. Then a divides b if there exists an integer c such that b = ac. If a divides b, then it is denoted by a|b. Examples: -3|18, since 18 = (-3)(-6); any integer a divides 0, a|0, since 0 = (a)(0). Códigos y Criptografía Francisco Rodríguez Henríquez Definitions: integers The following are some elementary properties of divisibility: Fact: (properties of divisibility) For all a, b, c, Z, the following are true: i. a|a ii. If a|b and b|c, then a|c iii. If a|b and a|c, then a|(bx+cy) for all x, y Z. iv. If a|b and b|a, then a = ±b Códigos y Criptografía Francisco Rodríguez Henríquez Definitions: division for integers Definition (division algorithm for integers) If a and b are integers with b≥1, then ordinary long division of a by b yields integers q (the quotient) and r (the remainder) such that a = qb+r, where 0 ≤ r <b Moreover, q and r are unique. The remainder of the division is denoted a mod b, and the quotient is denoted a div b. Definition An integer c is a common divisor of a and b if c|a and c|b. Códigos y Criptografía Francisco Rodríguez Henríquez Definitions: gcd Definition A non-negative integer d is the greatest common divisor of integers a and b, namely d = gcd(a, b), if i. d is a common divisor of a and b; and ii. Whenever c|a and c|b, then c|d. Equivalently, gcd(a, b) is the largest positive integer that divides both a and b, with the exception that gcd(0,0) = 0. Definition Two integers a and b are said to be relatively prime or coprime if gcd(a, b)=1 Definition An integer p≥2 is said to be prime if its only positive divisor are 1 and p. Otherwise, p is called composite. Códigos y Criptografía Francisco Rodríguez Henríquez Definitions: lcm Definition A non-negative integer d is the least common multiple of integers a and b, namely d = lcm(a, b), if i. a|d is and b|d; and ii. Whenever a|c and b|c, then d|c. Equivalently, lcm(a, b) is the smallest positive integer divisible by both a and b. Fact If a and b are positive integers, then lcm(a, b)=a*b/gcd(a, b). Códigos y Criptografía Francisco Rodríguez Henríquez Definitions: Prime Numbers Definition An integer p≥2 is said to be prime if its only positive divisor are 1 and p. Otherwise, p is called composite. Fact If p is prime and p|ab, then either p|a or p|b or both. (is it true if p is composite?). Fact There are an infinite number of prime numbers (how can we prove it?) Fact (prime number theorem) Let (x) denote the number of prime numbers ≤ x. Then lim x Códigos y Criptografía x x / ln x 1 Francisco Rodríguez Henríquez Definitions: Prime Numbers Fact (upper and lower bounds for (x)). Let (x) denote the number of prime numbers ≤ x. Then for x≥17 and for x > 1, x x ln x x x 1.25506 ln x Códigos y Criptografía Francisco Rodríguez Henríquez Fundamental Theorem of Arithmetic • Every integer n ≥ 2 has a factorization as a product of prime powers: n p p p , e1 1 e2 2 ek k • Where the pi are distinct primes, and the ei are positive integers. Furthermore, the factorization is unique up to the rearrangement of factors. Códigos y Criptografía Francisco Rodríguez Henríquez Fundamental Theorem of Arithmetic • Proof: existence [sketch] Suppose there exist positive integers that are not product of primes. Let n be the smallest such integer. Then n cannot be 1 or a prime, so n must be composite. Therefore n = ab with 1 < a, b < n. Since n is the smallest positive integer that is not a product of primes, both a and b are product of primes. But a product of primes times a product of primes is a product of primes, so n = ab is a product of primes. Therefore, every positive integer is a product of primes. Códigos y Criptografía Francisco Rodríguez Henríquez Fundamental Theorem of Arithmetic • Proof: uniqueness [sketch] If p is a prime and p divides a product of integers ab, then either p|a or p|b (or both!), (is this statement true for composite numbers?). Suppose that an integer n can be written as a product of primes in two different ways: as at a1 a2 a1 a2 n p1 p2 ps q1 q2 qt , • If a prime occurs in both factorizations divide both sides by it to obtain a shorter relation. Now take a prime that occurs on the left side, say p1. Since p1 divides n then it must divide one of the factors of the right side, say qj. But since p1 is prime, we are forced to write p1= qj, which is a contradiction with the original hyphotesis. Códigos y Criptografía Francisco Rodríguez Henríquez Prime Numbers: How many? Fact There are an infinite number of prime numbers (how can we prove it?) Euclid did it! But how? Should we have a quizz???? Hint: Follow the same line of reasoning used for FTA… Any idea??? Códigos y Criptografía Francisco Rodríguez Henríquez Fundamental Theorem of Arithmetic • Fact If a p1e1 p2e2 pkek , b p1f1 p2f 2 pkf k , where each ei ≥ 0 and fi ≥ 0, then gcd a, b p min e1 , f1 1 p min e2 , f 2 2 p min ek , f k k i k p min ei , f i i i 1 and i k lcm a, b p1max e1 , f1 p2max e2 , f 2 pkmax ek , f k pimax ei , f i i 1 Códigos y Criptografía Francisco Rodríguez Henríquez Fundamental Theorem of Arithmetic Example: Let a = 4864 = 2819, b = 3458 = 2 7 13 19. Then gcd(4864, 3458) = 2 19 = 38 and, lcm(4864, 3458)= 287 13 19 = 442624 Códigos y Criptografía Francisco Rodríguez Henríquez Definitions: Euler phi Function Definition For n ≥ 1, let (n) denote the number on integers in the interval [1, n], which are relatively prime to n. The function is called the Euler phi function (or the Euler totient function). Fact (properties of Euler phi function) i. If p is a prime, then (p) = p-1. ii. The Euler phi function is multiplicative. That is, if gcd(m, n) = 1, then (mn) = (m)(n). Códigos y Criptografía Francisco Rodríguez Henríquez Definitions: Euler phi Function iii. If n p1e p2e pke , then n n1 1 is the prime factorization of n, k 2 1 1 1 1 1 p1 p2 pk iv. For all integers n ≥ 5, n n 6 ln ln n Códigos y Criptografía Francisco Rodríguez Henríquez Euclidean algorithm m,n Euclidean Algorithm gcd(m,n) Fact If a and b are positive integers with a>b, then gcd(a,b)=gcd(b, a mod b); gcd(m, n) x = m, y = n while(y > 0) r = x mod y x=y y=r return x Códigos y Criptografía Francisco Rodríguez Henríquez Euclidean algorithm Example The following are the division steps for computing gcd(4864, 3458) = 38: 4864 = 1*3458 + 1406 3458 = 2*1406 + 646 1406 = 2*646 + 114 646 = 5*114 + 76 114 = 1*76 + 38 76 = 2*38 + 0 (Which method is more efficient and why??) Códigos y Criptografía Francisco Rodríguez Henríquez gcd: Computational Complexity Assuming mod operation complexity is K: integer euclid(m, n) x = m, y = n while( y > 0) r = x mod y x=y y=r return x K+ ¿? ( O (1) + K + O (1) + O (1) ) + O (1) = ¿? K O(1) Where “¿?” is the number of while-loop iterations. Códigos y Criptografía Francisco Rodríguez Henríquez gcd: Computational Complexity Facts: (x’ = next value of x, etc. ) 1. x can only be less than y at very beginning of algorithm –once x > y, x’ = y > y’ = x mod y 2. When x > y, two iterations of while loop guarantee that new x is < ½ original x –because x’’ = y’ = x mod y. Two cases: I. II. y > ½ x x mod y = x – y < ½ x y ≤ ½ x x mod y < y ≤ ½ x Códigos y Criptografía Francisco Rodríguez Henríquez gcd: Computational Complexity (1&2) After first iteration, size of x decreases by factor > 2 every two iterations. i.e. after 2i+1 iterations, x < original_x / 2i Q: When –in terms of number of iterations i– does this process terminate? Códigos y Criptografía Francisco Rodríguez Henríquez gcd: Computational Complexity After 2i+1 steps, x < original_x / 2i A: While-loop exits when y is 0, which is right before “would have” gotten x = 0. Exiting while-loop happens when 2i > original_x, (why??) so definitely by: i = log2 ( original_x ) Therefore running time of algorithm is: O(2i+1) = O(i) = O (log2 (max (a, b)) ) Códigos y Criptografía Francisco Rodríguez Henríquez gcd: Computational Complexity Measuring input size in terms of n = number of digits of max(a,b): n = (log10 (max(a,b)) ) = (log2 (max(a,b)) ) Therefore running time of algorithm is: O(log2 (max(a,b)) ) = O(n) (Except fot the mod operation complexity K, which in general is operand-size dependant) A more formal derivation of the complexity of Euclidean gcd can be found in section 4.5.3, Volume II of Knuth’s “The Art of Computing Programming” Códigos y Criptografía Francisco Rodríguez Henríquez Euclidean gcd: Revisited Properties: i. By definition gcd(0, 0) = 0. ii. gcd(u, v) = gcd(v, u) iii. gcd(u, v) = gcd(-u, v) iv. gcd(u, 0) = |u| v. gcd(u, v)w = gcd(uw, vw) if w ≥0 vi. lcm(u, v)w = lcm(uw, vw) if w ≥0 vii. uv = gcd(u, v) lcm(u, v) if u, v ≥0 viii. gcd(lcm(u, v), lcm(u, w)) = lcm(u, gcd(v, w)); ix. lcm(gcd(u, v), gcd(u, w)) = gcd(u, lcm(v, w)) Códigos y Criptografía Francisco Rodríguez Henríquez Euclidean gcd Revisited Binary Properties: i. If u and v are both even, then gcd(u, v) = 2 gcd(u/2, v/2); i. If u is even and v is odd, then gcd(u, v) = gcd(u/2, v); i. gcd(u, v) = gcd(u-v, v). ii. If u and v are both odd, then u-v is even and |u-v| < max(u, v). Códigos y Criptografía Francisco Rodríguez Henríquez Binary gcd algorithm Input: u, v positive integers, such that u > v. Output: w = gcd(u, v). 1. for (k = 0; u, v both even; k++) { u /= 2; v /= 2; }; /* [Find power of 2] */ 2. [Initialize] if (u is odd) t =-v else t = u; 3. [halve t] while (t is even) t /= 2; 4. if (t > 0) u = t else v = -t; 5. [Subtract] t = u-v. If t ≠ 0 go back to 3, otherwise output w = u2k. Códigos y Criptografía Francisco Rodríguez Henríquez Binary gcd algorithm: Example Example find the gcd of u =40902, v = 24140. t u v -12070, -6035 +14416, +901 -5134, -2567 40902 20451 20451 901 24140 6035 6035 6035 -1666, -833 +68, +34, +17 -816, -51 901 901 17 2567 833 833 -34, -17 17 51 0 17 17 Códigos y Criptografía w=17*21=34 Francisco Rodríguez Henríquez Extended Euclidean Algorithm The Euclidean algorithm can be extended so that it not only yields the greatest common divisor d of two integers a and b, but also generates x and y satisfying ax +by = d. Códigos y Criptografía Francisco Rodríguez Henríquez Modular Inverses THM1: e has an inverse modulo N if and only if e and N are relatively prime. This will follow from the following useful fact. THM2: If a and b are positive integers, the gcd of a and b can be expressed as an integer combination of a and b. I.e., there are integers s, t for which gcd(a,b) = sa + tb Códigos y Criptografía Francisco Rodríguez Henríquez Modular Inverses Proof of THM1 using THM2: If an inverse d exists for e modulo N, we have de 1 (mod N) so that for some k, de = 1 +kN, so 1 = de – kN. This equation implies that any number dividing both e and N must divide 1, so must be 1, so e,N are relatively prime. Códigos y Criptografía Francisco Rodríguez Henríquez Modular Inverses On the other hand, suppose that e,N are relatively prime. Using THM2, write 1 = se + tN. Rewrite this as se = 1-tN. Evaluating both sides mod N gives se 1 (mod N) . Therefore s is seemingly the inverse e except that it may be in the wrong range so set d = s mod N. • Códigos y Criptografía Francisco Rodríguez Henríquez Extended Euclidean Algorithm A constructive version of THM2 which gives s and t will give explicit inverses. This is what the extended Euclidean algorithm does. The extended Euclidean algorithm works the same as the regular Euclidean algorithm except that we keep track of more details –namely the quotient q = x/y in addition to the remainder r = x mod y. This allows us to backtrack and write the gcd(a,b) as a linear combination of a and b. Códigos y Criptografía Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step 0 x = qy + r - Códigos y Criptografía x y 244 117 gcd = ax+by Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 244=2·117+10 Códigos y Criptografía x y gcd = ax+by 244 117 117 10 Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 2 244=2·117+10 117=11·10+7 Códigos y Criptografía x y gcd = ax+by 244 117 117 10 10 7 Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 2 3 244=2·117+10 117=11·10+7 10=7+3 Códigos y Criptografía x y gcd = ax+by 244 117 117 10 10 7 7 3 Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 2 3 4 244=2·117+10 117=11·10+7 10=7+3 7=2·3+1 Códigos y Criptografía x y gcd = ax+by 244 117 117 10 10 7 7 3 3 1 Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 2 3 4 5 244=2·117+10 117=11·10+7 10=7+3 7=2·3+1 3=3·1+0 Códigos y Criptografía x y gcd = ax+by 244 117 117 10 10 7 7 3 3 1 1 0 Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 2 3 4 5 244=2·117+10 117=11·10+7 10=7+3 7=2·3+1 3=3·1+0 Códigos y Criptografía x y 244 117 117 10 10 7 7 3 3 1 1 0 gcd = ax+by 1=7-2·3 Solve for r. Plug it in. Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 2 244=2·117+10 117=11·10+7 x y 244 117 117 10 10 7 3 10=7+3 7 3 4 5 7=2·3+1 3 1 1 0 3=3·1+0 Códigos y Criptografía gcd = ax+by 1=7-2·(10-7) = -2·10+3·7 1=7-2·3 Solve for r. Plug it in. Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 1 - 2 244=2·117+10 117=11·10+7 x y 244 117 117 10 10 7 1=-2·10+3·(117-11·10) = 3·117-35·10 1=7-2·(10-7) = -2·10+3·7 3 10=7+3 7 3 4 5 7=2·3+1 3 1 1 0 3=3·1+0 Códigos y Criptografía gcd = ax+by 1=7-2·3 Solve for r. Plug it in. Francisco Rodríguez Henríquez Extended Euclidean Algorithm gcd(244,117): Step x = qy + r 0 - 1 2 3 244=2·117+10 117=11·10+7 x y gcd = ax+by 244 117 117 10 10 1= 3·117-35·(244- 2·117) = -35·244+73·117 7 1=-2·10+3·(117-11·10) = 3·117-35·10 1=7-2·(10-7) = -2·10+3·7 10=7+3 7 3 7=2·3+1 4 3=3·1+0 5 y Criptografía Códigos 3 1 1 0 1=7-2·3 Solve forRodríguez r. PlugHenríquez it in. Francisco Extended Euclidean Algorithm inverse of 244 modulo 117 gcd(244,117): Step x = qy + r 0 - 1 2 3 244=2·117+10 117=11·10+7 x y gcd = ax+by 244 117 117 10 10 1= 3·117-35·(244- 2·117) = -35·244+73·117 7 1=-2·10+3·(117-11·10) = 3·117-35·10 1=7-2·(10-7) = -2·10+3·7 10=7+3 7 3 7=2·3+1 4 3=3·1+0 5 y Criptografía Códigos 3 1 1 0 1=7-2·3 Solve forRodríguez r. PlugHenríquez it in. Francisco Extended Euclidean Algorithm Summary: Extended Euclidean algorithm works by keeping track of how remainder r results from dividing x by y. Last such equation gives gcd in terms of last x and y. By repeatedly inserting r into the last equation, one can get the gcd in terms of bigger and bigger values of x,y until at the very top is reached, which gives the gcd in terms of the inputs a,b. Códigos y Criptografía Francisco Rodríguez Henríquez Extended Euclidean Algorithm Input two positive integers a and b with a ≥ b. Output d = gcd(a, b) and integers x, y satisfying ax+by =d. 1. if (b = 0) { d = a; x = 1; y = 0; Fact: This algorithm has a Running time of O((lg n)2) return(d, x, y); bit operations. } 2. x2 = 1; x1 = 0; y2 = 0; y1 = 1. 3. while (b >0) { q a / b; r a % b; x x2 qx1 ; y y2 qy1 ; a b; b r; x2 x1 ; x1 x; y2 y1 ; y1 y; } 4. d = a; x = x2; y = y2; return(d, x, y); Códigos y Criptografía Francisco Rodríguez Henríquez Extended Euclidean Algorithm Example: Let a = 4864 and b = 3458. Hence gcd(a, b) = 38 and (4864)(32) + (3458) (-45) = 38. q 1 r 1406 x 1 y -1 a b 4864 3458 3458 1406 x2 1 0 x1 0 1 y2 0 1 y1 1 -1 2 2 5 1 646 114 76 38 -2 5 -27 32 3 -7 38 -45 1406 646 114 76 646 114 76 38 1 -2 5 -27 -2 5 -27 32 -1 3 -7 38 3 -7 38 -45 2 0 -91 128 38 0 32 -91 -45 128 Códigos y Criptografía Francisco Rodríguez Henríquez Quizz !! 1. Prove that there are an infinite number of prime numbers. 2. Prove that e has an inverse modulo N if and only if e and N are relatively prime. Códigos y Criptografía Francisco Rodríguez Henríquez Finite fields: definitions and operations FP finite field operations : Addition, Squaring, multiplication and inversion Códigos y Criptografía Francisco Rodríguez Henríquez What is a Group? An Abelian group <G, +> is an abstract mathematical object consisting of a set G together with an operation * defined on pairs of elements of G, here denoted by +: : G G G : a, b a b In order to qualify as an Abelian group, the operation has to fulfill the following conditions: i. Closed: ii. Associative: iii. Commutative: iv. Neutral element: v. Inverse elements: Códigos y Criptografía a, b G : a b G a, b, c G : a b c a (b c) a, b G : a b b a 0 G, a G : a 0 a a G, b G : a b 0 Francisco Rodríguez Henríquez What is a Group? • Example: The best-known example of an Abelian Group is <Z, +> • Example: The additive group Z15 uses the integers from 0 to 14. Some examples of additions in Z15 are: (10 + 12) mod 15 = 22 mod 15 = 7 • In Z15, 10 + 12 = 7 and 4 + 11 = 0. Notice that both calculations have answers between 0 and 14. • Additive Inverses – Each number x in an additive group has an additive inverse element in the group; that is an integer -x such that x + (-x) = 0 in the group. In Z15, -4 =11 since (4 + 11) mod 15 = 15 mod 15 = 0. Códigos y Criptografía Francisco Rodríguez Henríquez Rings (1/2) 1. 2. 3. A ring <R, +, *> consists of a set R with 2 operations defined on its elements, here denoted by + and *. In order to qualify as a ring, the operations have to fulfill the following conditions: The structure <R, +> is an Abelian group. The operations * is closed, and associative over R. There is a neutral element for * in R. The two operations + and * are related by the law of distributivity: a, b, c R : a b c a c b c 4. A ring <R, +. *> is called a commutative ring if the operation * is commutative. Códigos y Criptografía Francisco Rodríguez Henríquez Rings (2/2) • The integer numbers, the rational numbers, the real numbers and the complex numbers are all rings. • An element x of a ring is said to be invertible if x has a multiplicative inverse in R, that is, if there is a unique e R such that: xu ux 1 • 1 is called the unit element of the ring. Códigos y Criptografía Francisco Rodríguez Henríquez What is a Field? • A structure <F, +, *> is called a field if F is a ring in which the multiplication is commutative and every element except 0 has a multiplicative inverse. We can define the field F with respect to the addition and the multiplication if: F is a commutative group with respect to the addition. • F 0 is a commutative group with respect to the multiplication. The distributive laws mentioned for rings, hold. Códigos y Criptografía Francisco Rodríguez Henríquez What is a Field? • A field is a set of elements with two custom-defined arithmetic operations: most commonly, addition and multiplication. The elements of the field are an additive abelian group, and the non-zero elements of the field are a multiplicative abelian group. This means that all elements of the field have an additive inverse, and all non-zero elements have a multiplicative inverse. • A field is called finite if it has a finite number of elements. The most commonly used finite fields in cryptography are the field Fp (where p is a prime number) and the field F2m. Códigos y Criptografía Francisco Rodríguez Henríquez Finite Fields • A finite field or Galois field denoted by GF(q=pn), is a field with characteristic p, and a number q of elements. As we have seen, such a finite field exists for every prime p and positive integer n, and contains a subfield having p elements. This subfield is called ground field of the original field. • For the rest of this class, we will consider only the two most used cases in cryptography: q=p, with p a prime and q=2m. The former case, GF(p), is denoted as the prime field, whereas the latter, GF(2m), is known as the finite field of characteristic two or simply binary field. Códigos y Criptografía Francisco Rodríguez Henríquez Finite Fields • A finite field is a field with a finite number of elements. The number of elements in a finite field is called the order of the field. Fields of the same order are isomorphic: they display exactly the same algebraic structure differing only in the representation of the elements. Códigos y Criptografía Francisco Rodríguez Henríquez The field Fp • The finite field Fp (p a prime number) consists of the numbers from 0 to p1. Its operations are addition and multiplication. All calculations must be reduced modulo p. • It is mandatory to select p as a prime number in order to guarantee that all the non-zero elements of the field have a multiplicative inverse. • Other operations in Fp (such as division, subtraction and exponentiation) can be derived from the definitions of addition and multiplication. Códigos y Criptografía Francisco Rodríguez Henríquez The field Fp Example: Some calculations in the field F23 include 10*4 - 11 mod 23 = 29 mod 23 = 6 7-1 mod 23 = 10 (since 7 * 10 mod 23= 70 mod 23 = 1) (29) / 7 mod 23 = 512 / 7 mod 23 = 6 * 7-1 mod 23 = 6 * 10 mod 23 = 14 Códigos y Criptografía Francisco Rodríguez Henríquez Congruences Definition: Let a, b, n be integers with n ≠ 0. We say that a b mod n , (read: a is congruent to b mod n). If (a-b) is a multiple (positive or negative) of n, i.e., a = b + nk, for some integer k. Examples: 32=7 mod 5, -12 = 37 mod 7. Proposition: Let a, b, c, d, n be integers with n ≠ 0. i. ii. iii. iv. a = 0 mod n iff n|a. a = a mod n; a = b mod n iff a = b mod n. If a = b mod n and b = c mod n, then a = c mod n. a = b mod n and c = d mod n. Then a ± c = b ± d mod n, ac = bd mod n Códigos y Criptografía Francisco Rodríguez Henríquez Fermat’s Petit Theorem Theorem: Let p be a prime. i. If gcd( a, p) 1, then a p 1 1mod p ii. If r s mod p 1, then a r a s mod p, a In other words, when working modulo a prime p, exponents can be reduced modulo p-1. iii. In particular a p a mod p, a Códigos y Criptografía Francisco Rodríguez Henríquez Euler Theorem Theorem: Let n ≥ 2 be an integer. Then, If gcd( a, n) 1, then a n 1mod n If n is a product of distinct primes, and if r s mod n, then a r a s mod n In other words, when working modulo such an n, exponents can be reduced modulo (n). A special case of Euler’s theorem is Fermat’s petit theorem. Códigos y Criptografía Francisco Rodríguez Henríquez Euler and Fermat’s theorems examples Examples: 1. What are the last three digits of 7803 Equivalent to work mod 1000 (why?). Since (1000)=1000(1-1/2)(1-1/5)=400, we have 7803 = (7400)273=(1) 273=73=343 mod 1000. (why?) 2. Compute 23456 mod 5. From Fermat’s petit theorem we know that 24=1 mod 5. Therefore, 23456 = (24)864 = (1) 864 = 1 mod 5 Códigos y Criptografía Francisco Rodríguez Henríquez The order of an element in the field Fp The order of an element in F, is defined as the smallest positive integer k such that k=1 mod p. Any finite field always contains at least one element, called a primitive element, which has order p-1. From Euler’s theorem we know that for any element in F, p p 1 1mod p, Using the above result, one can easily prove that the order of any element in F must divide (p)=p-1, i.e., ord ( )| (p)= ord ( )| p-1. Códigos y Criptografía Francisco Rodríguez Henríquez Primitive Elements: how many? Fact: Suppose that is a primitive element in F. Then b = i mod n is also a primitive element in F iff gcd(i, (n))=1. It follows that the number of primitive elements in F is ¿Cuál es el otro? ((n)). Example: Consider the powers of 3 mod 7: 31=3;32=2; 33=6;34=4;35=5;36=1. There are ((7)) = 2 primitive elements in F7 Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem Fairy Tale: Chinese Emperor used to count his army by giving a series of tasks. 1. All troops should form groups of 3. Report back the number of soldiers that were not able to do this. 2. Now form groups of 5. Report back. 3. Now form groups of 7. Report back. 4. Etc. At the end, if product of all group numbers is sufficiently large, can ingeniously figure out how many troops. Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem mod 3: N mod 3 = 1 Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem mod 5: N mod 5 = 2 Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem mod 7: N mod 7 = 2 Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem Secret inversion formula (for N < 105 = 3·5·7): N a (mod 3) N b (mod 5) N c (mod 7) Implies that N = (-35a + 21b + 15c) mod 105. So in our case a = 1, b = 2, c = 2 gives: N = (-35·1 + 21·2 + 15·2) mod 105 = (-35 + 42 + 30) mod 105 = 37 mod 105 = 37 Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem How can we find the secret formula? For any x, a, b, and c satisfying x a (mod 3) x b (mod 5) x c (mod 7) Chinese Remainder Theorem says that this is enough information to uniquely determine x modulo 3·5·7. Proof, gives an algorithm for finding x –i.e. the secret formula. Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem Theorem: Suppose that gcd(m, n) = 1. Given a and b, there exists exactly one solution x (mod mn) to the simultaneous congruences x a mod m, x b mod n. Proof [sketch]: There exist integers s, t such that ms+nt=1 (why?). Then ms=1 mod n and nt =1 mod m (why?). Let x = bms +ant. Then, x ant a mod m, x bms b mod n Suppose x1 is another solution, then c = (x-x1) is a multiple of both, m and n (why?). But then provided that m and n are relatively primes then c is also a multiple of mn. Hence, any two solutions x to the system of congruences are congruent mon mn as claimed. Códigos y Criptografía Francisco Rodríguez Henríquez Chinese Remainder Theorem THM (CRT): Let m1, m2, … , mn be pairwise relatively prime positive integers. Then there is a unique solution x in [0,m1·m2···mn-1] to the system of congruences: x a1 (mod m1 ) x a2 (mod m2 ) x an (mod mn ) Códigos y Criptografía Francisco Rodríguez Henríquez
