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Math 537 — Elementary Number Theory

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Math 537 — Elementary Number Theory MATH 537 Class Notes Ed Belk Fall, 2014 1 Week One 1.1 Lecture One Instructor: Greg Martin, Office Math 212 Text: Niven, Zuckerman & Montgomery Conventions: N will denote the set of positive integers, and N0 the set of nonnegative integers. Unless otherwise stated, all variables are assumed to be elements of N. x1.2 { Divisibility Definition: Let a; b 2 Z with a 6= 0. Then a is said to divide b, denoted ajb, if there exists some c 2 Z such that ac = b. If in addition a 2 N, then a is called a divisor of b. Properties of Divisibility: For all a; b; c 2 Z with a 6= 0, one has: • If ajb then ±aj ± b • 1jb; bjb; aj0 • If ajb and bja then a = ±b • If ajb and ajc, then aj(bx + cy) for any x; y 2 Z If we assume that a and b are positive, we also have • If ajb then a ≤ b The Division Algorithm: Let a; b 2 N. Then there exist unique natural numbers q and r such that: 1. b = aq + r, and 2. 0 ≤ r < a Proof: We prove existence first; consider the set R = fb − an : n 2 N0g \ N0: By the well-ordering axiom, R has a least element r, and we define q to be the nonnegative integer q such that b − aq = r. Then b = aq + r and r ≥ 0; moreover, if r ≥ a then one has 0 ≤ r − a = (b − aq) − a = b − a(q + 1) < b − aq + r; contradicting the minimality of r 2 R, and we are done. 1 Now, suppose q0 and r0 are such that we have b = aq + r = aq0 + r0: Without loss of generality we may assume than r ≥ r0. Then r − r0 = (b − aq) − (b − aq0) = a(q0 − q) ) aj(r − r0); but 0 ≤ r − r0 ≤ r < a, and so the above equation is a contradiction unless r − r0 = 0, and the result is immediate. Greatest Common Divisor: Given any two integers a and b not both equal to zero, we define their greatest common divisor (commonly abbreviated gcd) to be the largest d 2 N such that dja and djb; we write d = (a; b). Note that because a and b each have only finitely many divisors, the gcd is always well-defined. Theorem 1.1.1 Let a; b 2 Z, not both equal to zero. Then: 1. (a; b) = min S, where S = (fax + by : x; y 2 Zg \ N), and 2. For any c 2 Z such that cja and cjb, we have cj(a; b). The existence of integers x; y so that ax + by = (a; b) as in part (1) is known as B´ezout'sidentity. Proof: 1. Let m = min S, with u and v such that m = au + bv, and let g = (a; b); note that m ≤ a. Since gja and gjb, we know from the properties of divisibility that gjm and so g ≤ m. Now, if m - a then by the division algorithm we may write a = mq + r with 0 < r < m, and thus r = a − mq = a − q(au + bv) = a(1 − qu) + b(−qv) 2 S; and we deduce that r ≥ m = min S, a contradiction; thus mja. In the same fashion we show mjb, and so by definition m ≤ (a; b) = g, and we are done. 2. If cja and cjb, then we know cj(ax + by) for every x; y 2 Z, and in particular for those u; v such that (a; b) = au + bv, whose existence is guaranteed by part 1. 2 1.2 Lecture Two Recall: B´ezout'sidentity states that (a; b) is the smallest positive integer that may be written ax + by, where x; y 2 Z. Proposition 1.2.1 For a; b 2 N, one has (ma; mb) = m(a; b). a b 1 a b Corollary 1: If dja; djb, then d ; d = d (a; b); in particular, (a;b) ; (a;b) = 1. Proof: Set g = (a; b), so that we may write ax + by = g; for some x; y 2 Z. Then mg = (ma)x + (mb)y; thus mg ≥ (ma; mb): Furthermore, gja and so mgjma; similarly mgjmb, thus mg ≤ (ma; mb), and we are done. Definition: Two integers a and b are called relatively prime (or coprime) if (a; b) = 1. nb. We observe that (a; b) = 1 if and only if there exist x; y such that ax+by = 1. The corresponding statement with (a; b) = k > 1 is not, in general, true, however it is the case that ax + by = k ) (a; b)jk: Proposition 1.2.2 If (a; n) = (b; n) = 1, then (ab; n) = 1. Proof: Suppose we have u; v; x; y so that au + nv = bx + ny = 1; then we have 1 = 1 · 1 = (au + nv)(bx + ny) = ab(ux) + n(auy + bvx + nvy); and the result is immediate. [Aside: Compare with the analagous result in commutative algebra. If R is a commutative, unital ring and I; J; K ⊂ R are ideals such that I + K = J + K = R, then IJ + K = R.] Proposition 1.2.3 If ajc; bjc, and (a; b) = 1, then abjc. (Note that this is not, in general, true for (a; b) > 1, e.g. a = b = c = 2.) Proof: Choose m; n; x; y so that c = am = bn and ax + by = 1. Then c = cax + cby = (bn)ax + (am)by = ab(nx + my); and we deduce that abjc. Theorem 1.2.4 (Theorem 1.10, Niven) If djab and (b; d) = 1, then dja. Proof: Exercise. nb. If dja; djb, then djb + ax for any x 2 Z. In fact, the condition is also necessary, as b = (b + ax) − x(a). The Euclidean Algorithm: How can we find the gcd of two integers, for example 537 and 105? By the division algorithm, we have 537 = 5 · 105 + 12, and so by the above note we know (537; 105) = (105; 12). Repeating this process, we see 105 = 8 · 12 + 9 ) (105; 12) = (12; 9); 12 = 1 · 9 + 3 ) (12; 9) = (9; 3); 3 9 = 3 · 3 + 0 ) (9; 3) = (3; 0) = 3: Thus (537; 105) = 3. Notation: The least common multiple of a and b is denoted lcm(a; b) or, more commonly, [a; b]. Exercise: Show that (a; b)[a; b] = ab. x1.3 { Primes Definition: A natural number n is called prime if it has exactly two divisors. n is called composite if there exists some d with 1 < d < n such that djn. The integer n = 1 is neither prime nor composite. Notation: Unless otherwise stated, p will denote a prime number. Lemma 1.2.5 (Euclid's lemma) If pjab, then pja or pjb. Proof: Suppose p - b. Then (p; b) = 1, and so by theorem 1.2.4 we know that pja. Theorem 1.2.6 (The Fundamental Theorem of Arithmetic) Every n 2 N; n > 2 may be written as the product of primes; moreover this expression is unique up to reordering of the factors. Proof: (existence) We use strong induction. The case n = 2 is trivial from the definition of a prime, therefore suppose n > 2. If n is prime we have the trivial factorization n = n, otherwise we may write n = ab, with 1 < a < n and 1 < b < n. By the inductive hypothesis we may write a = p1p2 ··· pk; b = q1q2 ··· ql, with each pi; qj prime, and the result is immediate. (uniqueness) Let n 2 N and suppose we have n = p1p2 ··· pk = q1q2 ··· ql; each pi; qj prime. Since p1jq1q2 ··· ql we have by lemma 1.2.5 that p1jq1 or p1jq2 ··· ql. Repeating this process as many times as necessary, we find qt such that p1jqt, and by relabelling the qj if necessary we will assume t = 1. Since p1 6= 1 this implies that p1 = q1, as q1 has no other factors. We then cancel p1 = q1 on both sides of the equation and we have p2p3 ··· pk = q2q3 ··· ql: We apply the same argument to this expression to obtain p2 = q2; p3 = q3, and so on; it follows that k = l, and we are done. 4 2 Week Two 2.1 Lecture Three Doing a linear algebra problem backwards. Consider the augmented matrix 1 0 537 ; 0 1 105 x 537 this system clearly has solution = : Moreover, from basic linear algebra we know that the application y 105 of elementary row operations to this augmented system will not change the solution; therefore, with R1;R2 x 537 respectively denoting the first and second row of the matrix, we observe that = is also a solution y 105 to the augmented matrices 1 −5 12 (R ! R − 5R ); 0 1 105 1 1 2 1 −5 12 (R ! R − 8R ); −8 41 9 2 2 1 9 −46 3 (R ! R − R ); −8 41 9 1 1 2 9 −46 3 (R ! R − 3R ): −35 179 0 2 2 1 Thus we have the matrix equation 9 −46 537 3 = : −35 179 105 0 The first entry of this equation indicates that 9(537) + (−46)(105) = 3 = (537; 105), while the entries in the 105 537 second row of the matrix are −35 = − (537;105) and 179 = (537;105) . This operation is known as the extended Euclidean algorithm. Lemma 2.1.1 Let a; b 2 N and use the division algorithm to write b = aq + r with 0 ≤ r < a. Then ajb if and only if r = 0. Proof: If r = 0 then b = aq and we are done.
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