Definition. Let a, b Z . We say that a divides b , denoted a | b , - - PDF document

Definition. Let a, b ∈ Z. We say that a divides b, denoted a | b, if there is k ∈ Z such that b = k · a. Then we say that a is a factor of b and that b is a multiple of a.

Fact. For every n ∈ Z the following are true: 1 | n, n | n, and n | 0.

Theorem. Let a, b, c ∈ Z. (i) If a | b and b | c, then a | c. (ii) a | b if and only if |a| | |b|. (iii) If a | b and b = 0, then |a| ≤ |b|.

Theorem. The relation a | b is a partial ordering on N and on N0.

Theorem. (division theorem) Let a, d ∈ Z, d = 0. Then there exist q ∈ Z and r ∈ N0 such that a = qd + r and 0 ≤ r < |d|. The numbers q and r are unique. Definition. The number r is called the remainder and we denote it r = a mod d. The number q is called the quotient.

Fact. Let a ∈ Z and d ∈ N, let q be the quotient of a by d. Then q = a d

• .

Fact. Let a, b ∈ Z, a = 0. Then a | b if and only if b mod |a| = 0, that is, the remainder when dividing b by |a| is 0.

Definition. Let a, b ∈ Z. A number d ∈ N is called a common divisor of numbers a, b if d | a and d | b. A number d ∈ N is called a common multiple of numbers a, b if a | d and b | d.

Definition. Let a, b ∈ Z. We define their greatest common divisor, denoted gcd(a, b), as the largest element of the set of common divisors, if at least one of a, b is not zero. Otherwise we define gcd(0, 0) = 0. We say that numbers a, b ∈ Z are coprime if gcd(a, b) = 1. We define their least common multiple, denoted lcm(a, b), as the smallest element of the set of their common multiples, if a, b are both not zero. Otherwise we set lcm(a, 0) = lcm(0, b) = 0.

Fact. Let a, b ∈ Z. Then gcd(a, b) = gcd(|a|, |b|) and lcm(a, b) = lcm(|a|, |b|).

Theorem. Let a, b ∈ Z. Then lcm(a, b) · gcd(a, b) = |a| · |b|.

Fact. Let a ∈ N. Then gcd(a, 0) = a, lcm(a, 0) = 0 and gcd(a, a) = lcm(a, a) = a.

Lemma. Let a > b ∈ N, let q, r ∈ N0 satisfy a = qb + r and 0 ≤ r < b. Then the following are true: (i) d ∈ N is a common divisor of a, b if and only if it is a common divisor of b, r. (ii) gcd(a, b) = gcd(b, r).

Euklid’s algorithm for finding gcd(a, b) for a > b ∈ N. Version 1. Initiation: r0 := a, r1 := b, k := 0. Step: k := k + 1, rk−1 = qk · rk + rk+1 Repeat until rk+1 = 0. Then gcd(a, b) = rk. Version 2. procedure gcd(a, b: integer) repeat r := a mod b; a := b; b := r; until b = 0;

• utput: a;

Theorem. (Bezout theorem/identity) Let a, b ∈ Z. Then there are A, B ∈ Z such that gcd(a, b) = Aa + Bb.

Extended Euclid algorithm for finding gcd(a, b) = Aa + Bb for a > b ∈ N. Version 1. Initiation: r0 := a, r1 := b, k := 0, A0 := 1, A1 := 0, B0 := 0, B1 := 1. Step: k := k + 1, qk := rk−1 rk

• ,

rk+1 := rk−1 − qkrk, Ak+1 := Ak−1 − qkAk, Bk+1 := Bk−1 − qkBk. Repeat until rk+1 = 0. Then gcd(a, b) = rk = Aka + Bkb. Version 2. procedure gcd-Bezout(a, b: integer) A0 := 1; A1 := 0; B0 := 0; B1 := 1; repeat q := a b

• ;

r := a − qb; ra := A0 − qA1; rb := B0 − qB1; a := b; b := r; A0 := A1; A1 := ra; B0 := B1; B1 := rb; until b = 0;

• utput: gcd(a, b) = a = A0a + B0b;

Definition. By a linear diophantine equation of two variables we mean any equation of the form ax+by = c with unknowns x, y, where a, b, c ∈ Z and only integer solutions are allowed.

Theorem. A linear diophantine equation ax + by = c has at least one solution if and only if c is a multiple of gcd(a, b).

Theorem. Consider a linear diophantine equation ax + by = c. Let (xp, yp) ∈ Z2 be some particular solution of this equation. A vector (x, y) ∈ Z2 is a solution of this equation if and only if there exists some (xh, yh) ∈ Z2 such that (x, y) = (xp, yp) + (xh, yh) and (xh, yh) solves the associated homogeneous equation.

Theorem. Consider an equation ax + by = 0 for a, b ∈ Z. Then the set of all its integer solutions is

• −k

b gcd(a, b), k a gcd(a, b)

• ; k ∈ Z
• .

Algorithm for finding all integer solutions of the equation ax + by = c.

• 0. Find (for instance using the extended Euclid’s algorithm)

A, B ∈ Z such that gcd(a, b) = Aa + Bb.

• 1. If c is not a multiple of gcd(a, b), then there is no solution.
• 2. The case gcd(a, b) divides c:

a) Multiply the equality aA + bB = gcd(a, b) by c′ = c gcd(a, b), keeping coefficients a, b intact, you will obtain a(Ac′) + b(Bc′) = c and thus one particular solution xp = Ac′, yp = Bc′. b) Cancel gcd(a, b) in the associated homogeneous equation ax + by = 0, obtaining a′x + b′y = 0, that is, a′x = −b′y. This gives xh = −b′k, yh = a′k for k ∈ Z. c) By adding the particular and homogeneous solutions you obtain the general integer solution of the given equation x = Ac′ − b′k = A c gcd(a, b) − b gcd(a, b)k, y = Bc′ + a′k = B c gcd(a, b) + a gcd(a, b)k; k ∈ Z.

Algorithm 2 for finding all integer solutions of the equation ax + by = c.

• 1. Guess gcd(a, b). Try to cancel this number in the given equation.

If it is not possible, that is, if c is not a multiple of gcd(a, b), then there is no solution.

• 2. Case gcd(a, b) divides c:

Divide the given equation by gcd(a, b). You will obtain a new dio- phantine equation a′x + b′y = c′, where a′, b′ are coprime. a) Find (for instance using the extended Euclid’s algorithm) A, B ∈ Z such that gcd(a′, b′) = 1 = Aa′ + Bb′, so a′A + b′B = 1. Multiply this equality by c′ keeping coefficients intact, you will obtain a′(Ac′) + b′(Bc′) = c′ and thus one particular solution xp = Ac′, yp = Bc′, that is, a vector (Ac′, Bc′). b) Solve the associated homogeneous equation a′x + b′y = 0, that is, a′x = −b′y, which gives xh = −b′k, yh = a′k, in other words the pair (−b′k, a′k) for k ∈ Z. c) By adding the particular and homogeneous solutions you obtain the set of all integer solutions {(Ac′ − kb′, Bc′ + ka′); k ∈ Z}.

Definition. Let a ∈ N, a = 1. We say that it is a prime if the only natural numbers that divide it are 1 and a. We say that a is a composite number if it is not a prime.

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.

Lemma. Let a1, . . . , am ∈ N and p be a prime. If p | (a1a2 · · · am), then there is i such that p | ai.

Theorem. (Fundamental theorem of arithmetics, prime decompo- sition) Let n ∈ N. Then there exist primes p1, p2, . . . , pm and exponents k1, k2, . . . , km ∈ N0 so that n = pk1

1 · pk2 2 · · · pkm m = m

• i=1

pki

i .

If we also demand that p1 < p2 < . . . < pm and ki > 0, then this decomposition is unique.