Elementary Number Theory


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1 Elementary Number Theory Ahto Buldas December 3, 2016 Ahto Buldas Elementary Number Theory December 3, / 1
2 Division For any m > 0, we define Z m = {0, 1,... m 1} For any n, m Z (m > 0), there are unique q Z and r Z m such that: n = qm + r, where r is called the reminder (of n modulo m) and is denoted by r = n mod m. If r = 0, we say that m divides n (or n is divisible by m) and write m n. If 0 n < m, then r = n; if m n < 2m, then r = n m Z m, etc. If m n < 0, then r = n + m; if 2m n < m, then r = n + 2m, etc. Ahto Buldas Elementary Number Theory December 3, / 1
3 Equivalence of Numbers modulo m If a mod m = b mod m (i.e. if a b = km for a k Z, or m (a b)), then we write a b (mod m), and say that a and b are equivalent modulo m. For example 1 2 (mod 3), 7 1 (mod 3), 2 12 (mod 5), etc. Ahto Buldas Elementary Number Theory December 3, / 1
4 Z m as a Number Domain We can define addition and multiplication in Z m denoted by ja in the next way: For example, in Z 3 : and in Z 5 : a b = (a + b) mod m, a b = (a b) mod m. 2 2 = 2 2 = 1, 1 2 = 0, 2 3 = 0, 3 3 = 1 = 3 2 and 3 4 = 2. Ahto Buldas Elementary Number Theory December 3, / 1
5 Properties of the Function mod m: Z Z m mod m is a projector: (a mod m) mod m = a mod m. mod m preserves the operations (i.e. is a homomorphism): If a = a mod m, b = b mod m ja c = c mod m, then Conclusion 1: When computing a + b = c = a b = c a b = c = a b = c. a + b (c + d (e + f))... mod m we can reduce mod m whenever we want. Conclusion 2: and are somewhat similar to ordinary + and Ahto Buldas Elementary Number Theory December 3, / 1
6 Properties of the Z m Number Domain Though and differ from + and, we mostly use + and if this will not cause confusion. The following properties hold in Z m : Commutativity: a + b = b + a, a b = b a Associativity: (a + b) + c = a + (b + c), (a b) c = a (b c) Zero: a + 0 = 0 + a = a, a 0 = 0 a = 0 Unit: a 1 = 1 a = a Distributivity: (a + b) c = a c + b c, Ahto Buldas Elementary Number Theory December 3, / 1
7 Somewhat Unusual Properties of Z m The inverse a of an element a Z m is m a Z m, because: a + (m a) = m 0 (mod m). Zero divisors: the product of two nonzero elements can be zero. For example, in Z 6 : (mod 6). Not every element a has an inverse a 1 in Z m : a a 1 1. For example, zero divisors never have inverses. Ahto Buldas Elementary Number Theory December 3, / 1
8 Motivation from Cryptography In cryptography, the operations should be invertible, because any encrypted message should later be decrypted. Both mod addition and multiplication are extensively used in cryptography. Modular addition is invertible, i.e. a x = b is always solvable. Modular multiplication is not always invertible, i.e. a x = b can be unsolvable. For example, 2 x 5 (mod 6) is not solvable. The equation 2 x 5 (mod 7) is solvable: x = 6, because 2 6 = 12 5 (mod 7). Ahto Buldas Elementary Number Theory December 3, / 1
9 Greatest Common Divisor By the greatest common divisor gcd(a, b) of two nonnegative numbers a and b (not both zero!) we mean the largest d that divides both numbers, i.e.: gcd(a, b) = max{d: d a and d b}. Theorem An element a Z m is invertible if and only if gcd(a, m) = 1. Ahto Buldas Elementary Number Theory December 3, / 1
10 Computing gcd(a, b): Euclid s Algorithm For a > b 0: gcd(a, b) = { a if b = 0 gcd(b, a mod b) if b 0 (1) The work of Euclid s algorithm can be represented as a sequence: gcd(r 0, r 1 ) = gcd(r 1, r 2 ) =... = gcd(r m 1, r m ) = gcd(r m, 0), where r 0 = a, r 1 = b, and r k+1 = r k 1 mod r k < r k for any k > 1. This algorithm stops (an m with r m+1 = 0 exist), because otherwise r 0 > r 1 > r 2 >... > r k >... is an infinite decreasing sequence of natural numbers, which does not exist. Ahto Buldas Elementary Number Theory December 3, / 1
11 Correctness of Euclid s Algorithm Clearly gcd(a, 0) = a. We prove gcd(a, b) = gcd(b, a mod b), if a > b > 0. If D a,b = {d: d a and d b} is the set of all common divisors of a and b: gcd(a, b) = max D a,b and gcd(b, a mod b) = max D b,a mod b. It is sufficient to prove that D a,b = D b,a mod b. This is indeed the case, as: If d a ja d b, then d (a mod b) = a kb, and hence D a,b D b,a mod b If d (a mod b) and d b, then also d a, because a = (a mod b) + kb, and hence D a,b D b,a mod b. Ahto Buldas Elementary Number Theory December 3, / 1
12 Efficiency of Euclid s Algorithm Theorem Euclid s algorithm finds gcd(a, b) using 1.44 log 2 b + 1 divisions. Let r 0 > r 1 >... r n 1 > r n be the sequence produced by Euclid s algorithm so that r n = gcd(a, b). Let φ = , i.e. 1 + φ 1 = φ. We show by induction that r k φ n k for 1 k m, i.e. b = r 1 φ n 1. As r k+1 = r k 1 mod r k = r k 1 q k r k, we have r k 1 = q k r k + r k+1, where q k 1 because of r k 1 > r k. Basis: r n = gcd(a, b) 1 = φ 0. As r n+1 = 0 and q n r n = r n 1 > r n, we have q n 2 and hence r n 1 2 > φ 1. Step: If r k+1 φ n k 1 and r k φ n k, then r k 1 = q k r k +r k+1 r k +r k+1 = φ n k 1 +φ n k = φ n k (1+φ 1 ) = φ n k+1. Ahto Buldas Elementary Number Theory December 3, / 1
13 Conclusions Conclusion 1: If a > b 0, then there exist α, β Z such that gcd(a, b) = αa + βb. Conclusion 2: gcd(a, b) = 1 if and only if α, β Z, such that αa + βb = 1. Proof: If gcd(a, b) = 1, then use Conclusion 1. If α, β Z such that αa + βb = 1, (2) d a and d b, then d 1 by (??), i.e. gcd(a, b) = 1. Conclusion 3: If gcd(a, m) = 1, then b Z m, such that b a mod m = 1. Proof: Given α, β Z, so that αa + βm = 1, define b = α mod m. Ahto Buldas Elementary Number Theory December 3, / 1
14 Finding Inverses with Euclid s Algorithm Find 1 3 mod 26. Let a = 3 and b = a b 3 2 a b 8a 1 2 a (b 8a) = 9a b b 8a 1 0 9a b b 8a 2(9a b) = 26a + 3b Hence, = 1, which means (mod 26) Ahto Buldas Elementary Number Theory December 3, / 1
15 Solvability of ax mod n = c Theorem The equation ax mod n = c (where c Z n ) is solvable iff gcd(a, n) c. Proof. If the equation is solvable and d = gcd(a, n), then a, n, k Z so that a = a d, n = n d, and hence d c, because: c = ax mod n = ax kn = a dx kn d = (a x kn )d. If d = gcd(a, n) c, then gcd( a d, n d ) = 1, which means that a d has inverse modulo n d and the equation a d x mod n d = c d is solvable, i.e. k Z: a d x k n d = c d, and hence ax kn = c Z n, which means that ax mod n = c. Ahto Buldas Elementary Number Theory December 3, / 1
16 How Many Invertible Elements mod m are there? Answer to that question is called the Euler s function ϕ(m). Computing ϕ(m) requires the primefactorization of m. A prime number is a number if it has exactly two divisors. For example: 2, 3, 5, 7, 11, 13, etc. Theorem (Fundamental Theorem of Arithmetics) Every integer m > 0 has a unique prime factorization: p e 1 1 pe pe k k, where p 1 < p 2 <... < p k are prime numbers. For example: 60 = Ahto Buldas Elementary Number Theory December 3, / 1
17 Some Lemmas Lemma 1: Every composite m 2 is a product of primes. Proof: Let m be the smallest composite number that is not a product of primes. Hence, there exist composite numbers m 1, m 2 < m, so that m = m 1 m 2. Hence, m 1 and m 2 are products of primes and so must be m. A contradiction. Lemma 2: If gcd(a 1, b) = 1 = gcd(a 2, b), then gcd(a 1 a 2, b) = 1. Proof: As there are α 1, β 1, α 2, β 2, so that α 1 a 1 + β 1 b = 1 = α 2 a 2 + β 2 b: 1 = (α 1 a 1 + β 1 b) (α }{{} 2 a 2 + β 2 b) = α }{{} 1 α 2 a }{{} 1 a 2 + (β 1 + α 1 a 1 β 2 ) b, }{{} 1 1 α β we have gcd(a 1 a 2, b) = 1. Ahto Buldas Elementary Number Theory December 3, / 1
18 Fundamental Theorem of Arithmetics: Proof Theorem Every composite m 2 has a unique primefactorization p 1 p 2... p k, where p 1 p 2... p k. Proof. Let m be the smallest number that has two different primefactorisations: p 1 p 2... p k = m = q 1 q 2... q l. Hence, p i q j, because otherwise m = m/p i < m also has two different factorizations. Thus, gcd(p 1, q 1 ) = gcd(p 2, q 1 ) =... = gcd(p k, q 1 ) = 1, which by the assumption q 1 m and Lemma 2 implies a contradiction: q 1 = gcd(m, q 1 ) = gcd(p 1 p 2... p k, q 1 ) = 1. Ahto Buldas Elementary Number Theory December 3, / 1
19 Computing the Euler s Function Theorem If m = p e 1 1 pe pe k k ( ϕ(m) = = m p e 1 1 pe is the prime decomposition, then ) (1 1p1 ) ( p e 2 2 pe ) (1 1p2 )... (... ) p e k k p e k 1 k ) (1 1pk. The proof uses the inclusionexclusion principle from counting theory. Ahto Buldas Elementary Number Theory December 3, / 1
20 InclusionExclusion Principle Let P 1,..., P k be subsets of a set M. We want to count those elements of M that belong to none of P n, i.e. we want to compute M\ n P n. If k = 1, then M\ n P n = M P 1. If k = 2, then M\ n P n = M P 1 P 2 + P 1 P 2. If k = 3, then: M\ n P n = M P 1 P 2 P 3 + P 1 P 2 + P 1 P 3 + P 2 P 3 P 1 P 2 P 3. General case: M\ n P n = M Σ 1 + Σ 2 Σ ( 1) i Σ i where Σ i = (j 1,...,j i ) c(i) P j 1... P ji and the summation is over the set c(i) of all icombinations of indices 1, 2,..., k. There are ( k i) of them. Ahto Buldas Elementary Number Theory December 3, / 1
21 InclusionExclusion Principle and Euler s function Let M = Z m, where m = p e 1 1 pe pe k k. Let P n be the set of elements in Z m divisible by p n. Then ϕ(m) = M\ n P n This is because a Z m is invertible iff none of p 1,... p k divides a. P i = m p i, P i P j = m p i p j... P i1... P il = m p i1 p i2...p il. and hence: ϕ(m) = m m... m + m m m... p 1 p k p 1 p 2 p k 1 p k p 1 p 2 p ) ) ) 3 = m (1 1p1 (1 1p2... (1 1pk. Ahto Buldas Elementary Number Theory December 3, / 1
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