Favorite Topics from Complex Arithmetic, Analysis and Related Algebra
|
|
- Raymond Little
- 5 years ago
- Views:
Transcription
1 Favorite Topics from Complex Arithmetic, Analysis and Related Algebra construction at 09FALL/complex.tex Franz Rothe Department of Mathematics University of North Carolina at Charlotte Charlotte, NC 3 frothe@uncc.edu September 7, 011 Contents 1 Some Number Theory 1.1 Prime numbers Rational and irrational Algebra of the Complex Numbers 5.1 Arctan identities Gaussian primes Sums of two squares Roots The solution of the reduced cubic equation The square root of a complex number Pythagorean triples Roots of unity Polynomials Rational zeros of polynomials Gauss Lemma Eisenstein s irreducibility criterium Descartes Rule of Signs
2 Polynomials over the Complex Numbers 5.1 A proof of the Fundamental Theorem of Algebra Meditation on Descartes rule of signs Estimation of zeros with Rouché s Theorem Perron s irreducibility criterium The limiting case The Residue Theorem and its Consequences The residue theorem and the logarithmic residue theorem Comparing two functions
3 1 Some Number Theory 1.1 Prime numbers Proposition 1.1 (Euclidean Property). If a number c divides the product ab and gcd (c, a) = 1, then c divides b. Standard proof. By the extended Euclidean algorithm, there exist integers s, t such that 1 = sa + tc b c = sab c + tb The second line results by multiplication of both sides with b. Because c divides ab, the c right hand side is an integer. Hence c divides b, as to be shown. Second proof. Both numbers a and c are divisors of both products ab and ac. Hence both a and c are divisors of the greatest common divisor Hence the integer G := gcd (ab, ac) (1.1) q := ac G is a divisor of both a and c. Hence q is a divisor of gcd (a, c) = 1, which was assumed to be one. Hence q = 1, and ac = G is a divisor of ab. This implies that c is a divisor of b, as to be shown. Definition 1.1 (prime number). A prime number is an integer p, which is divisible only by 1 and itself. Euclid and many other mathematicians have shown that there exist infinitely many prime numbers. We put them into the increasing sequence p 1 =, p = 3, p 3 = 5, p = 7,... It is rather easy to see that for every positive integer, there exists a decomposition into prime factors. Let a and b be any positive integers. There exist sequence α i 0 and β i 0, with index i = 1,,... and only finitely many terms nonzero such that (1.) a = p α i i, b = p β i i i 1 i 1 The uniqueness of the prime decomposition turns out harder to prove. Astonishingly, the proof depends on Euclid s lemma, the proof of which in turn relies on the extended Euclidean algorithm.
4 Proposition 1. (Euclid s Lemma). If a prime number divides the product of two integers, the prime number divides at least one of the two integers. Reason. Let p be the prime number, and the integers be a and b. We assume that p divides the product ab, but p does not divide a. We need to show that p divides b. Because p does not divide a, the definition of a prime number implies gcd (a, p) = 1. By the extended Euclidean algorithm, there exist integers s, t such that 1 = sa + tp Hence b p = sab p + tb Because p divides ab, the right hand side is an integer. Hence p divides b, as to be shown. Proposition 1.3 (Monotonicity). Let a and b have the prime decompositions (1.3) a = p α i i, b = p β i i i 1 i 1 The number b is a divisor of a if and only if β i α i for all i 1. Reason. If β i α i for all i 1, then a = qb with q = i 1 p α i β i i and hence b is a divisor of a. Conversely, assume that b is a divisor of a. We need to show that β i α i for all i 1. Proceed by induction on b. If b = 1, then β i = 0 for all i 1, and the assertion is true. Here is the induction step b < n b = n : Let p i be any prime factor of b, which means that β i 1. Because p i divides b and b divides a, the prime p i divides a. By Euclid s Lemma Proposition 1., p i is a divisor of one of the primes p j occurring in the prime decomposition of a. Hence α j 1. Because different primes cannot divide each other, this implies i = j and p i = p j. Hence b p j < n is a divisor of a p j. By the induction assumption, this implies β i α i for all i j, as well as β j 1 α j 1 and hence β i α i for all i 1. Proposition 1. (Uniqueness of prime decomposition). The prime decomposition of any positive integer is unique. Reason. Assume a = p α i i and a = p β i i i 1 i 1 Because a divides a, the fact given above both tells that β i α i and α i β i for all i 1. Hence β i = α i for all i 1. 3
5 Proposition 1.5. Let a and b have the prime decompositions (1.3). The prime decompositions of the greatest common divisor and least common multiple are (1.) (1.5) gcd (a, b) = i 1 lcm (a, b) = i 1 p min[α i,β i ] i p max[α i,β i ] i 10 Problem 1.1. Check these formulas for a = 1001, b = 1. Proof of Proposition 1.5. Let g be the righthand side of equation (1.5). check properties (i) and (ii) defining the greatest common divisor. We need to (i) g divides both a and b. Check. This is clear, because both min[α i, β i ] α i and min[α i, β i ] β i for all i 1. (ii) If any positive integer h divides both a and b, then h divides the greatest common divisor g. Check. Let (1.6) h = i 1 p γ i i be the prime decomposition of h. Because h divides both a and b, monotonicity implies that both γ i α i and γ i β i for all i 1. Hence γ i min[α i, β i ] for all i 1, which easily implies that h is a divisor of g. Let l be the righthand side of equation (1.5). We need to check properties (i) and (ii) defining the least common multiple. (i) The number l is a multiple of both a and b. Check. This is clear, because both max[α i, β i ] α i and max[α i, β i ] β i for all i 1. (ii) If any positive integer k is a multiple of both a and b, the integer k is a multiple of the least common multiple l.
6 Check. Let (1.7) k = i 1 p γ i i be the prime decomposition of k. Because k is a multiple of both a and b, monotonicity implies that both γ i α i and γ i β i for all i 1. Hence γ i max[α i, β i ] for all i 1, which easily implies that k is a multiple of l. 1. Rational and irrational Proposition 1.6. For any natural numbers r 1 and a 1, the r-th root r a is only a rational number if it is even an integer. Proof. The assertion is clear for r = 1 or a = 1, hence we may assume r and a. Now assume that r a = m n n r a = m r with natural m, n. We show that any prime number p dividing n has to divide m, too. Hence after cancelling common factors, we get n = 1, and hence the root is an integer. Now assume that the prime p divides n. Hence p r divides n r, which in turn divides an r = m r. Hence p r divides m r. Hence, by Euclid s Lemma p divides m, as claimed. As already explained, the assertion follows. Proposition 1.7. For any natural numbers r and any a, which is not the r-th power of an integer, the root r a is irrational. Especially, the roots of the primes are all irrational. Proof. If the root r a is rational, it is even an integer m, and hence a = m r is the r-th power of m. Take the contrapositive: If a is not the r-th power of an integer, the root r a is irrational. Algebra of the Complex Numbers Definition.1 (Gaussian integer ). A complex number with integer real- and imaginary part is called a Gaussian integer. 5
7 .1 Arctan identities What is remarkable about the following products? (a) ( + i) (3 + i) = 5 + 5i (b) ( + i) (7 i) = 5 + 5i (c) (3 + i) (7 + i) = i (d) (5 + i) (39 i) = i Each time, the real- and imaginary parts turn out to be equal. Do there exists more examples with this special property? For any positive x > 0, the principal argument of the complex number x + i is Arg (x + i) = arctan 1. Too, Arg (1 + i) = arctan 1 = π. Taking the arguments of the x formulas (a) through (d), one gets: (a) (b) arctan 1 + arctan 1 3 = π arctan 1 arctan 1 7 = π (c) (d) arctan arctan 1 7 = π arctan arctan 1 39 = π Of course, left- and right-hand side of these formulas could still differ by an integer multiple of π. But it is clear that this cannot happen in the given four examples. Do there exist more similar formulas? Open Problem (May be difficult). How many solutions has the equation (.1) c + 1 = N with natural numbers N and c? Open Problem (May be difficult). Are there more than four solutions to the equation (.) M (c k + 1) = N k=1 with natural numbers N and M, and c k for k = 1,..., M? 6
8 . Gaussian primes It is straightforward to see that the Gaussian integers are a ring. The units in any ring are defined as its invertible elements. Clearly there are just four units 1, i, 1, i in the ring of Gaussian integers. Furthermore, division with remainder is possible. Hence the Euclidean algorithm works in the Gaussian integers. Consequently, any Gaussian integer can be decomposed uniquely into a product of irreducible elements unique up multiplication by the four units. Hence the Gaussian integers are a unique factorization domain, abbreviated UFD. The irreducible elements are called Theorem.1 (Gaussian primes). The Gaussian primes are: (a) the number 1 + i. (b) the usual primes p = 3, 7, 11, 19,... which are p 3 mod. (c) all pairs p + iq, p iq where p + q = r is a prime r 1 mod. Indeed, for every usual prime r = 5, 13, 17, 9,... which is r 1 mod there exists a unique decomposition into a sum of two integer square. The decomposition r = p + q becomes unique by the additional requirement that the (negative or positive) integer p 1 mod and q is even and positive. Here is a table with the smallest examples. Too, I list the squares a + ib = (p + iq). r p q a b At that point, my heating system had been repaired, and I stopped. 10 Problem.1. Prove that there are infinitely many primes p 3 mod. 7
9 Answer. Let p 1,... p N be any primes congruent to 3 modulo. The number has the following properties: (a) P is odd. (b) P 3 mod. P := 1 + (c) P is not divisible by p ν for any ν = 1... N. By (a) and (b), P cannot be the product of primes which are all congruent to 1 modulo. Hence there exists a prime p congruent to 3 modulo dividing P. By item (c), this prime p is different from p 1,... p N. Hence there exist at least N + 1 different primes equivalent to 3 modulo. Since we know that such primes exist, there do exist infinitely many. 10 Problem.. Prove that there are infinitely many primes r 1 mod. Answer. Let r 1,... r N be any primes congruent to 1 modulo. Define the number P := 1 + Indeed P has in the Gaussian integers the factoring [ ] [ ] N N P = 1 + i r ν 1 i r ν ν=1 Neither of the two factor is divisible by 1 + i nor any prime p 1 mod. Hence we get the following properties: (a) P is odd. (b) P 1 mod. (c) P is not divisible by r ν for any ν = 1... N. N ν=1 N ν=1 (d) P is not divisible by any prime p 3 mod. By (a) and (d), P is the product of primes congruent to 1 modulo only. By item (c), all these factors are different from r 1,... r N. Hence there exist at least N + 1 different primes equivalent to 1 modulo. Since we know that such primes exist, there do exist infinitely many. p ν r ν ν=1
10 10 Problem.3 (The little forgotten theorem). Assume that (.3) z = a + ib = a + b e iπk l with real integers a, b, k, l such that a and b are relatively prime, and k and l are relatively prime. How many different nonzero values can z 0 assume? Prove that there are non besides the obvious eight ones. Answer. The complex number z can only assume the eight values 1, i, 1, i and ±1±i, but no other values. Because of the assumption gcd (a, b) = 1, the extended Euclidean algorithm yields integers s, t such that sa tb = 1, and hence (a + ib)(s + it) = 1 + i(at + bs). Hence no real prime can divide a + ib. We now repeatedly use the fact that the Gaussian integers are a unique factorization domain. Assume the Gaussian prime p + iq divides a + ib. Ruling out all other cases, we show the only possibility is p + iq = 1 + i. Because of the remark above, the case of a real prime p 3 mod, and q = 0, is impossible. Hence we are left with the task to rule out the case of a Gaussian prime p + iq with 5 p + q 1 mod. In the l-th power of the given equation (.3), (.) (a + ib) l = (a + b ) l the two sides sides are both Gaussian integers as well as real, hence real positive integers. Equation (.) implies that p + iq divides the natural number (a + b ) l. Hence p iq divides this same number, too. Since the Gaussian prime p iq divides (a+ib) l, unique factorization implies that p iq divides a + ib, too. Since 5 p + q 1 mod. the two numbers p + iq and p iq are relatively prime. Hence we conclude there product (p + iq)(p iq) = p + q divides a + ib, too. But we have already shown that no real prime divides a + ib. Hence the only possibilities left are a + ib = i u (1 + i) s with u = 0, 1,, 3 and s = 0, 1. Theorem. (The forgotten little theorem). A Gaussian integer does only have an argument iπk with integer k, l an argument rational in degree measurement if it l lies on the x-axis or on the y-axis, or on one of the two lines of slope ±1 through the origin. Proof. Assume that (.5) z = a + ib = a + b e iπk l with real integers a, b, k, l such that k and l are relatively prime. Let d := gcd (a, b) and define a := a d and b := b d. By the last problem, a + ib = i u (1 + i) s with u = 0, 1,, 3 and s = 0, 1. Multiplying both sides by d yields a + ib = i u (1 + i) s d with u = 0, 1,, 3 and s = 0, 1, and d any natural number. These are just the Gaussian integers on the four lines Z, i Z, (1 + i) Z, and (1 i) Z, as to be shown. 9
11 Corollary. Let a + ib be a Gaussian integer with a 0, b 0. Both real- and imaginary parts of (a + ib) l are nonzero if either l odd or a b. If l > 0 is even, the following happens: l Re (a + ib) l Im (a + ib) l l mod Re (a + ib) l = 0 a = b Im (a + ib) l 0 l 0 mod Re (a + ib) l 0 Im (a + ib) l = 0 a = b 10 Problem.. Give the reason for Corollary.. Proof of Corollary.. For any nonzero Gaussian integer a + ib, and integer l > 0, the equation (.6) (a + ib) l = i u (a + b ) l is equivalent to Re (a + ib) l = 0 for u odd, but equivalent to Im (a + ib) l = 0 for u even. In all cases, the equation (.7) (a + ib) l = (a + b ) l follows. Hence, as explained in the solution of Problem.3, we conclude that either a = 0, or b = 0, or a = b. Since we have excluded the first two possibilities, we conclude that a = ±b and hence equation.6 implies a l (1 ± i) l = i u l/ a l. Hence l = m is even and (±i) m = i u. Hence either l mod, and m, u both odd, and Re (a + ib) l = 0 or l 0 mod, and m, u both even, and Im (a + ib) l = Problem.5. (a + ib)l+ R l (a, b) := Re a b (a + ib)l+ S l (a, b) := Im ab(a b ) Calculate R 0, S 0, R, S. Prove that for all l 0 mod, R l and S l are integer homogenous and symmetric polynomials in a and b. Answer. (a + ib) R 0 (a, b) = Re = a b a b a b = 1 (a + ib)6 R (a, b) = Re = a6 15a b + 15a b b 6 = a 1a b + b a b a b (a + ib) S 0 (a, b) = Im ab(a b ) = a3 b ab 3 ab(a b ) = (a + ib) S (a, b) = Im ab(a b ) = a7 b 56a 5 b a 3 b 5 ab 7 = a a b + b ab(a b ) 10
12 After some pain with the binomial formulas, one gets the general case with l = m: m +m ( ) m + R m (a, b) = ( 1) s a m+t b m t s s=0 t= m m +m ( ) ( 1) s m + S m (a, b) = a m+t b m t s + 1 s=0 t= m Since the binomial coefficients ( ) even odd are even, these are integer polynomials the other assertions are easy to check, too. 10 Problem.6 (Calculation to the bones). Calculate R m (1, 1) and S m (1, 1), confirming they are nonzero. Answer. R l + (a + b )abs l = Re (a + ib)l+ + (a + b )Im (a + ib) l a b With a = b = 1 and l = m, we get = Re (a + ib)l+ Re i(a ib)(a + ib) l+1 a b [a + ib i(a ib)](a + ib)l+1 = Re a b (1 i)(a + ib)l+1 = Re a + b R m (1, 1) + S m (1, 1) = Re (1 i)(1 + i) m+1 / = ( ) m With a = 1 + ε = z and b = 1, we get in the limit z 1 using l Hôpital s rule (1 + ε i)m+1 R m (1, 1) S m (1, 1) = lim Re (1 i) ε 0 ε d (z i)m+1 = Re (1 i) d z = Re (m + 1)(1 i) m+1 = (m + 1)( ) m Both formulas together imply S m = m( ) m 1. Actually, we check that for all m 0 R m = (m + 1)( ) m, S m = (m + )( ) m.3 Sums of two squares From the decomposition of a Gaussian integer into Gaussian primes, one can quite easily see how many solutions the integer equation m = a +b has for any given natural number m. 11
13 10 Problem.7. We begin with some examples. For this illustration, I count only the solutions of m = a + b with integers 0 a b, but give these solutions completely. Give a list for m = 1 through 30. Answer. m list of a + b no solution no solution 7 no solution no solution 1 no solution m list of a + b no solution no solution no solution 3 no solution no solution = no solution no solution no solution It is still hard to find a pattern. Since the square of the absolute value of a complex number is the sum of the squares of its real- and imaginary parts, we can use complex arithmetic. Here are some further examples for illustration. (a) 1 + i = 1 + = 5 (b) 3 + i = 3 + = 13 (c) The square of the absolute value of (1 + i) ( 3 + i) = 7 i, and (1 + i) ( 3 i) = 1 i implies 5 13 = 7 + = 1 + = 65. (d) Squaring part (c) yields (1 + i) ( 3 + i) = ( 3 + i)(5 1i) = i (1 + i) ( 3 i) = ( 3 + i)(5 + 1i) = 63 16i The square of the absolute value yields 5 13 = = = 65 Are there more ways to solve a + b = 65? Indeed, there are still two further solutions obtained as absolute squares of 13( 3+i) = 39+5i and 5(5 1i) = 5 60i. Last not least, 65 is a perfect square = = = 65 1
14 (e) The absolute square of (1 + i)(1 + i) = 1 + 3i yields 10 = Problem.. Prove that a natural number is a perfect square if and only if it has an odd number of divisors. Answer. If d [1, m) is a divisor of m N, then m ( m, m] is different one. In d that way, all divisors of m appear in pairs. The only exception is m in the case that m is a perfect square. Proposition.1. The integer equation m = a + b is solvable for a given natural number m with integers a, b if and only if, in the prime decomposition of m, all primes p 3 mod appear with even multiplicity. In other words, m can be factored as m = s g c where g has only prime factors congruent 3 mod, and (.) c = has only prime factors congruent 1 mod. N ν=1 Theorem.3. The number of different ways one can decompose m = s g c into squares of integers a, b such that m = a + b is τ(c). Here τ(c) = r sν ν N (1 + s ν ) ν=1 is the number of divisors of c. The numbers a + ib from the solutions are exactly the Gaussian integers (.9) a + ib := i u (1 + i) s g N (p ν + iq ν ) σν (p ν iq ν ) τν ν=1 All possible choices are obtained with u = 0, 1,, 3 and σ ν, τ ν 0 with σ ν + τ ν = s ν for ν = 1... N. For example m = 65 is decomposed as 65 = 5 13 and has 9 divisors. The possible 13
15 values of a + ib as solution of (.9) with unit factor i u = 1 are (1 + i) ( 3 + i) = ( 3 + i)(5 1i) = i, (1 + i) ( 3 + i)( 3 i) = ( 3 + i) 13 = i, (1 + i) ( 3 i) = ( 3 + i)(5 + 1i) = 63 16i, (1 + i)(1 i)( 3 + i) = 5 (5 1i) = 5 60i, (1 + i)(1 i)( 3 + i)( 3 i) = 5 13 = 65, (1 + i)(1 i)( 3 i) = 5 (5 + 1i) = i, (1 i) ( 3 + i) = ( 3 i)(5 1i) = i, (1 i) ( 3 + i)( 3 i) = ( 3 i) 13 = 39 5i, (1 i) ( 3 i) = ( 3 i)(5 + 1i) = 33 56i, As one sees there are still pairs of conjugate complex solutions, and one real solution since 65 is a perfect square. Remark. If m is not a perfect square, there are τ(c) essentially different solutions of m = a + b with 1 a b. If m is a perfect square, there are τ(c) 1 essentially different solutions of m = a + b with 1 a b. 10 Problem.9. Prove that a Gaussian integer a + ib with prime decomposition (.9) satisfies gcd (a, b) = 1 if and only if s = 0 or s = 1; g = 1; either σ ν = 0 or τ ν = 0 for all ν = 1,... N. Answer. A Gaussian integer has relatively prime real and imaginary parts if and only if it is not divisible by any real prime. But the three conditions just imply: a + ib is not divisible by ; a + ib is not divisible by any prime factor congruent to 3 mod ; a + ib is not divisible by any real prime congruent to 5 mod. Proof of Proposition.1 and Theorem.3. For any given natural number m, let d be the divisor of m assembling all prime factors congruent 3 mod. Let c be the divisor of m assembling all prime factors congruent 1 mod. The factor c is completely decomposed into its prime factors as in formula (.). In the ring of Gaussian integers, the usual prime factoring of m will split further into irreducible factors (.10) m = s dc = i 3s (1 + i) s d N (p ν + iq ν ) sν (p ν iq ν ) sν ν=1 1
16 where the real factoring of d is not needed here. Assume that m = a +b. Because of (a+ib)(a ib) = a +b = m, we conclude that a + ib is a divisor of m. The uniqueness of prime factorization implies equation (.9) holds for some values u {0, 1,, 3} and 0 σ ν s ν, 0 τ ν s ν for ν = 1... N (Why?). We now multiply equation (.9) with the conjugate complex equation and get a + b = s g N ν=1 (p ν + iq ν ) σν+τν (p ν iq ν ) σν+τν We compare all factors with those in equation (.10) and use the uniqueness of prime factorization. Hence g = d and σ ν + τ ν = s ν for ν = 1... N, as claimed. Too, we see that no equation m = a + b can hold unless the factor d is a perfect square. Too, the uniqueness of the prime factoring implies that all solutions given by equation (.9) are different. It is straightforward to check that equation (.9) yields solutions of the equation m = a + b. Hence we get exactly solutions. τ(c) = N (1 + s ν ) ν=1 10 Problem.10. Assemble different examples to illustrate Proposition.1 and Theorem.3. Examples. For different values of m, here is a table listing the factors s, d, c, the prime decomposition of c and the number of divisors τ(c). Furthermore, we list the possible values of a + ib form solution (.9). For abbreviation, I put the unit factor i u = 1, and ignore the obvious conjugate complex ones. m s d c factor c τ(c) list of a + ib i i (1 + i)(1 + i) = 1 + 3i (1 + i)( 3 + i) = 5 i i i no solution i (1 + i)( 3 ± i) = 7 i, 1 i (1 + i)(1 + i)( 3 ± i) = 3 11i, 9 7i i(1 + i)( 3 ± i) = 1i, 16 + i 15
17 . Roots Definition. (argument). The argument of a complex number z 0 is the set of angles which in polar coordinates yield z: arg z = {θ R : z = z e iθ } The principle value of the argument Arg z of a complex number z 0 is the unique value for the argument lying in the interval ( π, +π], including +π but excluding π. Remark. All values of the argument differ by integer multiples of π. Hence there always exists a unique principle argument. Nevertheless, the choice of a principle argument is just an arbitrary convention. Especially, the set of arguments arg z depends continuously on z 0, but the principle argument does not depend continuously on z. Lemma.1 (Principle value). The principle argument of a complex number z = x + iy 0 is given by the formula arctan y if x > 0; x π if x = 0 and y > 0; (.11) Arg z = π if x = 0 and y < 0; π + arctan y if x < 0 and y > 0; x π + arctan y if x < 0 and y < 0 x Remark. Several computer languages provide Arg (x + iy) as a two variable function arctan(x, y). 10 Problem.11. Give simple examples to illustrate all cases in Lemma.1. Proof of Lemma.1. The principle argument Arg z of any complex number z 0 is defined by restricting the argument of z = z e iarg z to the interval Arg z ( π, π]. Separating real- and imaginary parts yields x = z cos Arg z, y = z sin Arg z Hence y = tan Arg z unless x = 0. But since both cos(θ ± π) = cos θ and sin(θ ± π) = x sin θ, the tangent function has period π. Hence the value of tan Arg z determines the argument only modulo π not yet modulo π. The given cases result from comparing the signs of sin and cos with those of x and y. 16
18 10 Problem.1. The angle α of the tangent to the graph of a differentiable function y = f(x) with the x-axis is always given by α = arctan f (x) = arctan dy dx Why does one not distinguish the cases occurring in Lemma.1? Answer. The angle between the tangent and the x-axis is a directed angle between two lines, and hence always lies in the interval ( π, π ). But the principle argument of a complex number z is a directed angle between two rays the positive x-axis and the ray from the origin 0 to the point z. Hence it can take any value in the interval ( π, π], including +π and excluding π. Definition.3 (right semi-plane). Define the right semi-plane as the set RS = {z = x + iy C : x > 0 or (x = 0 and y 0) } Remark. The right semi-plane is neither open nor closed. I use the word half plane only for an open set. 10 Problem.13. What is the exact interval to which the principle value of the argument for any z 0 in the right semi-plane is restricted. Answer. For any z 0 in the right semi-plane, the principle value of the argument lies in the half open interval Arg z ( π, π]. Conversely, Arg z ( π, π ] implies z RS. Definition. (Principle value). The principle argument Arg z of any complex number z 0 is defined by restricting the argument of z = z e iarg z to the interval Arg z ( π, π]. The principle n-th root is define as P.V. n z := n z e iarg z n The principle logarithm of any complex number z 0 is defined as Ln z := ln z + iarg z 10 Problem.1. Prove that for all a, b RS and all n Arg ab = Arg a + Arg b for a, b RS and a, b 0 P.V. n ab = P.V. n a P.V. n b 17
19 Answer. For the product of any nonzero a, b RS, the principal values Arg ab ( π, π], and Arg a + Arg b ( π, π], too. Hence Arg ab = Arg a + Arg b + πk holds with k = 0, since both sides do not differ by an integer multiple of π. Now the second assertion easily follows P.V. n ab = n Arg ab i ab e n = P.V. n a P.V. n b = n Arg a+arg b i ab e n = n a e i Arg a n n b e i Arg b n Here are some examples of principle third roots. (a) (b) (c) P.V. 3 i = P.V. 3 i = 1 + i P.V i 3 i = 1 + i (d) P.V. 3 1 = 1 + i 3 Note that the principal third root of any negative real, is not the negative real value. 10 Problem.15. In which examples does equality hold for the principle values. Show enough of your calculations. (a) P.V. 3 i P.V. 3 i? = P.V. 3 1 (b) P.V. 3 i P.V. 3 i? = P.V. 3 1 (c) P.V. 3 1+i P.V. 3 1+i [ ] =? P.V. 3 1+i [ ] (d) P.V. 3 1+i P.V. 3 1+i? = P.V. 3 1+i 1
20 Answer. (a) (b) (c) [ ] P.V. 3 i P.V i i = = 1 + i 3 [ ] P.V. 3 i P.V. 3 3 i i = = 1 i 3 [ ] 1 + i 1 + i 1 + i P.V. 3 P.V. 3 = = i = P.V. 3 1 = 1 + i 3 P.V. 3 1 = 1 + i 3 (d) 1 + i 1 + i P.V. 3 P.V. 3 = [ P.V. 3 e πi P.V. 3 [ 1 + i ] = P.V. 3 i = ] = [ e πi ] = e πi 6 = 3 i [1 ] + i P.V. 3 = P.V. 3 i = e πi 6.5 The solution of the reduced cubic equation Given is the equation (.1) x 3 + b x + c = 0 Since the quadratic term is lacking, it is called a reduced or depressed cubic. We want to find all three solutions, including the complex ones. Putting x := u b leads to the 3u equation ( ) 3 b (.13) u 3 + c = 0 3u Hence z = u 3 satisfies the quadratic resolvent equation (.1) z + cz b3 7 = 0 The two solutions of equation (.1) are (.15) z 1, = c ± c + b3 7 We get a resolvent equation with real or complex solutions, depending on whether the discriminant (.16) D = c + b3 7 19
21 is positive or negative. In the first case, the original cubic has only one or (exceptionally) two real solutions. The second case with complex solutions of the resolvent equation (.1) is called the casus irreducibilis. In that case, the original cubic has three real solutions. Hence one is forced to go through some complex arithmetic just in the case where all final results are real. The solution procedure turns out to be different for these two cases. Case 1: D 0 The cubic has one real solution or in the special case D = 0 two real solutions. From Viëta s formula (.17) z 1 z = b3 7 b 3 3 = z 3 z 1 Hence the real solution of the cubic (.1) is (.1) x 1 = u b 3u = 3 z z = 3 c + c + b c c + b3 7 which is just Cardano s formula. The two complex solutions are (.19) x = 3 z 1 ω + 3 z ω x 3 = 3 z 1 ω + 3 z ω where (.0) ω = 1 + i 3 = cos π 3 + i sin π 3 is the primitive third root of unity. Simplifying (.1) and (.19) yields (.1) x,3 = 3 z z ± 3 z 1 3 z 3 i Case : D < 0 The cubic has three real solutions. In this case, the resolvent equation (.1) has complex solutions. As already Rafael Bombelli discovered (157), this leads to a third root of a complex number in Cardano s formula (.1). Actually, this case can only occur for b < 0. To cover the general case, extraction of the third root has to be done via polar coordinates. Viëta s formula (.17) and 0
22 equation (.15) imply (.) z 1 = z 1 z = b3 7 z 1, = c b ± i 3 7 c b 3 = (cos θ ± i sin θ) 7 The argument θ can be calculated more simple from the real parts. One gets cos θ, and finally θ: (.3) c b = 3 7 cos θ c θ = arc cos b 3 7 Hence equations (.) and (.0) imply (.) 3 z1 = b 3 3 z1 ω = b 3 ( cos θ 3 + i sin θ ) 3 ( cos θ + π + i sin θ + π ) 3 3 and equations (.17) and (.1) imply that b x 1 = 3 cos θ 3 (.5) b θ + π x = cos 3 3 b θ π x 3 = cos 3 3 are the three real solutions of the original cubic. 10 Problem.16. Calculate the roots in the limiting case D 0+. Answer. x 1 = 3 c, x = x 3 = x 1 10 Problem.17. Calculate the roots in the limiting case D 0 in terms of c. Distinguish the cases c < 0 and c > 0. In both cases b < 0. 1
23 Answer. In the approach D 0, one gets c = b3 7, 3 c = c b = 3 7 Distinguish the cases c < 0 and c > 0. c b = 3 7 cos θ b 3 Case c < 0. Case c > 0. One gets cos θ = 1, and hence θ = 0. The result (.5) implies x 1 = 3 c, x = x 3 = x 1 One gets cos θ = 1, and hence θ = π. The result (.5) implies x = 3 c, x 1 = x 3 = x 1 10 Problem.1. Use Cardano s formula and Bombelli s method to get one solutions of x 3 30x 36 = 0. Answer. Cardano s formula for a solution of x 3 + bx + c = 0 yields in the present case (.6) x 1 = 3 c + c + b = i i c c + b3 7 To calculate the third root, let i = u+iv and 1+6i = (u+iv) 3. We compare real- and imaginary parts and get the system 1 = u 3 3uv = u(u 3v ) 6 = 3u v v 3 = (3u v )v With the help of the factoring, one finds the integer solution u + iv = 3 + i. Hence Cardano s formula gives x 1 = Problem.19. Find the two other solutions of the equation x 3 30x 36 = 0. Answer. The simplest way is division of the polynomial. One gets (x 3 30x 36)/(x 6) = x + 6x + 6. Hence the two other solutions are x,3 = 3 ± 3.
24 Remark. Too, we can find the two other roots by introducing the third root of unity ω = 1+i 3 into Cardano s formula. This has to be done in a consistent way and produces the solutions x = ω i + ω 3 1 6i = 1 + i 3 x 3 = ω i + ω 3 1 6i = 1 i 3 which is the same answer as above. (3 + i) + 1 i 3 (3 + i) i 3.6 The square root of a complex number (3 i) = 3 3 (3 i) = Proposition.. A branch with Re z 0 for the square root of any complex number z = x + iy is (.7) x + iy = x + y + x + i sign (y) x + y x The square root w = z of any complex number z = x + iy has two branches. The second branch is w = w. Reason. Name the square root in question x + iy =: u + iv. Squaring yields x + iy = (u + iv). Separate the real- and imaginary part to get The absolute value squared is Add and substrate x = u v, y = uv x + y = x + iy = u + iv = (u + v ) u + v = x + y u v = x u x + y = + x v x + y = x Of the last two expression, one takes real square roots. u 0 has been assumed. One needs still to determine the sign of v. But y = uv and u > 0 imply sign y = (sign u)(sign v) = sign v. In the special case u = 0, both signs of v give a correct result for the square root. This special case corresponds to y = 0 and x 0 the negative numbers z 0. 3
25 (a) 10 Problem.0. Calculate the square roots: 3 + i, i, + 6i 6i, 3 i, 1 96i (b) (c) 3i, 16 1i, 6 (d) 6 + i, + 3i, i (e) i, i, 3i A Gaussian integer that is the square of another Gaussian integer is called a perfect square. Which of the numbers under the square roots are perfect squares, which are not?.7 Pythagorean triples Any three integers such that a + b = c are called a Pythagorean triple. 10 Problem.1. Show that for any Gaussian integer p + iq, the real and imaginary parts and the absolute value of its square are a Pythagorean triple. Answer. Let the square be a+ib := (p+iq). Its absolute value is c := a+ib = p+iq. Clearly a + b = c holds. 10 Problem.. Show that for any Gaussian integer a+ib 0, it is impossible that a + ib and b + ia are both perfect squares. Theorem. (Pythagorean triples). A nonzero Gaussian integer a + ib = (p + iq) is a perfect square if and only if the following three conditions hold (i) the sum a + b is a perfect square c ; (ii) gcd (a, b) =: g is either (a) a perfect square: g = d, or (b) twice a perfect square: g = d ; (iii) assuming t, but not t+1 divides d, we get two cases: (iii a) if g = d, then t a is odd and t b is even (iii b) if g = d, then t 1 a is even and t 1 b is odd. Corollary. Exactly one of the Gaussian integers a+ib = (p+iq) and b+ia = i(p iq) is a perfect square if and only if (i) the real and imaginary parts and the absolute value of a+ib are a Pythagorean triple; (ii) the greatest common divisor gcd (a, b) is either a perfect square or twice a perfect square.
26 Remark. It is always true that the imaginary part b of a perfect square is even, but this condition is sufficient only in the special case gcd (a, b) = 1. Remark. Condition (iii) in theorem. is only needed to determine which one of the two numbers either a + ib or b + ia is a perfect square. In Problem., we have already checked that these two numbers cannot be both perfect squares, unless they are both zero. Example.1. These are perfect squares, corresponding to t = 0, 1, in (iii a) or (iii b): (a) 3 + i = 1 + i, i = + i, + 6i = + i,... (b) 6i = 1 + 3i, 3 i = + 6i, 1 96i = + 1i,... but these are not perfect squares: ( 1 + 3i) 3i =, 16 1i = ( 1 + 3i), 6 = ( + 6i), i = (1 + i), + 3i = ( + i), i = ( + i),... Here is the most simple example: (a) =, 16 =, 6 =,... (b) i = 1 + i, i = + i, 3i = + i,... Proof of necessity in theorem.. Necessity is easier to prove. We assume that a+ib = (p + iq) is a perfect square. Squaring the absolute values gives c = a + b = a + ib = p + iq = (p + q ) Separating real- and imaginary parts yields the Pythagorean triple a = p q, b = pq, c = p + q Hence condition (i) holds. To check (ii), let d := gcd (p, q). Clearly d divides both a and b, and hence d divides g := gcd (a, b). Let By these divisions we get p := p d, q := q d, a := a d, b := b d, c := c d, a = p q, b = p q, c = p + q, gcd (p, q ) = 1 We show that gcd (a, b ) =: h can only be 1 or. Indeed, any odd prime dividing b needs to divides either p or q, but not both. Hence it cannot divide a = p q. 5
27 Neither can divide both a and b. Indeed, assuming divides b, implies one number among p and q is even, the other one is odd. Hence a is odd. Hence we see that h = 1 or h = are the only possibilities left. Therefore gcd (a, b) = d gcd (a, b ) = hd is either a perfect square or twice a perfect square, as claimed by condition (ii). Finally, we need to confirm that either case (iii a) or (iii b) occurs. Again, t 0 be the integer for which t, but not t+1 divides d. These two cases can occur: (iii a): g = d and h = 1 = gcd (a, b ) = gcd ((p q )(p + q ), p q ). Hence p ± q are both odd, and hence one number among p and q is odd, but the other one is even. Hence a is odd, but divides b. Similarly, t a is odd and t b is even, since these are odd multiples of a and b. (iii b): g = d and h = gcd (a, b ) = gcd ((p q )(p + q ), p q ) =. Hence p ± q are both even, and different from case (a) p and q are both odd. Hence a is divisible by, but b mod. Hence, t 1 a is even and t 1 b is odd, since these are odd multiples of a and b. Proof of sufficiency theorem.. We assume that the integers a and b satisfy the four conditions (i) through (iii). Let g = gcd (a, b) = hd with h = 1 or h = and as above In cases (a) and (b) one gets a := a d, b := b d, c := c d (iii a) if g = d, then gcd (a, b ) = 1, and a is odd, b is even. Hence c is odd, and c ±a are both integers. (iii b) if g = d, then gcd (a, b ) =, a is even and b is odd. Hence c is odd. As explained in Proposition., one obtains a square root of a complex number by the formula c + a (.7) a + ib = c a + i sign (b) c In both cases (iii a) and (iii b), ±a and b are integers. Furthermore a + b = c implies [ ] b = c + a c a The two factors are relatively prime: [ c + a gcd ], c a = 1 6
28 Indeed, the greatest common divisor of c ±a is a divisor of h = gcd (a, b ), which is either 1 or. But is the case (b) where h =, both numbers c ±a turn out to be odd. It is a (too easy) exercise to show that uniqueness of the prime factorization implies Lemma.. If the product of two relatively prime factors is a perfect square, then both factors are perfect squares. Hence we conclude that both roots (E:doit) p = c + a and q = sign (b) c a are integers. In other word, we have shown that a + ib = p + iq is a Gaussian integer. Multiplying by d, we get that a + ib = p + iq is a Gaussian integer, too, as to be shown. 10 Problem.3. Generalize the Lemma. to the product of three or more factors. Are the following conjectures true or false: If the product of three pairwise relatively prime factors is a perfect square, then all three factors are perfect squares. If the product of three relatively prime factors is a perfect square, then the product of any two of them is a perfect square. Remark. Under the additional assumption that gcd (a, b), indeed only the following possibilities occur for perfect squares a + ib = (p + iq) : p q a mod b mod c mod h odd even even odd odd odd 0. Roots of unity 10 Problem.. Find all different powers exactly, expressed using square roots: (a) ω = 1 + i i (b) z =. (c) w = 1 + i. 7
29 You need not show all calculations. Simplify your answers by using the conjugate complex. What is remarkable about these expressions? Answer. It is remarkable that the different powers can be obtained simply by some sign changes of some square roots: (a) (b) (c) ω = 1 + i 3, ω = 1 i 3 = ω, ω 3 = i z =, z = i 10 5, z 3 = z, z = z w = 1 + i, w = i, w 3 = 1 + i, w = 1, w 5 = w 3, w 6 = i, w 7 = w An exact trigonometric table of cos 3 k and sin 3 k for all k = 1,..., 15 can be obtained using the three roots of unity ω, z and w. This trick goes in principle back to Ptolemy! (a) From the primitive third and eighth root of unity ω and w from Problem., we get a product of argument 15. ω 1 w 3 = e πi[ ] = e πi = cos 15 + i sin 15 (b) We can calculate this product in terms of the exact square roots, obtained in Problem.. ω 1 w 3 = 1 i i = i( 3 1) (c) Finally, we compare the real- and imaginary parts obtained form (a) and (b), and use the common denominator. + 6 cos 15 = 6 sin 15 = (d) In this case, we get the table immediately. k 15 cos k 15 sin k
30 10 Problem.5. Calculate cos 15 + i sin 15 directly by formulas cos θ = 1 + cos θ and confirm that you get the same values. Answer. Similarly, one gets sin 15 = cos cos 30 = 3, sin θ = 1 cos θ = + 3 = + 3. By squaring both sides, one checks that indeed + 3 = 6 +, 3 = 6 10 Problem.6. The following steps can be elaborated to obtain an exact trigonometric table of cos 9 k and sin 9 k for all k = 1,..., 0. (a) Find natural numbers p, q such that p 5 + q = 1 0. (b) With z, w from Problem., calculate the argument θ in z p w q = e πiθ (c) Hint: calculate i i (d) Calculate z p w q with the exact square roots, obtained in Problem.. (e) Use Euler s (.) e πi π π 0 = cos + i sin 0 0 and get exact expressions for cos 9 and sin 9. Use the common denominator in your results. Answer. (a) The natural numbers p, q have to satisfy p + 5q = 1. solution is p = 3, q = 5. The smallest (b) With z, w from Problem., z p w q = e πi[ ] = e πi 0 9
31 (d) With the same z, w from Problem., using z 5 = 1 and w = i, we can simplify and get i 10 5 z 3 w 5 = z w = 1 + i = + i (e) Comparing real- and imaginary part from parts (b) and (c) yields (.9) cos 9 = (.30) sin 9 = 10 Problem.7. Give the exact expression for the perimeter of a regular 0- gon, inscribed into the unit circle. Compare with π. How large is the deviation? Answer. The circumference of a regular 0-gon is [ 10 0 sin 9 = ] Bur π so only one decimal point. But the relative error of less than a percent is a beginning. In a similar way, we can use the twelfth and fifth root of unity, and get sin and cos for the multiples of 6. (a) The natural numbers p, q have to satisfy 1p 5q = ±1. The smallest solution is p = 3, q = 7. Dividing by 60 yields = 1 60 (b) With the twelfth and fifth roots of unity τ and z we get z 3 τ 7 = e πi[ ] = e πi 60 = cos 6 + i sin 6 (d) We can calculate this product with the exact values of the square roots, simplify and get i i z 3 τ 7 = z 3 τ 1 = = + i
32 (e) Comparing real- and imaginary part from parts (b) and (c) yields (.31) (.3) cos 6 = sin 6 = Problem.. Give the exact expression for the perimeter of a regular 30- gon, inscribed into the unit circle. Compare with π. How large is the deviation? Answer. The circumference of a regular 30-gon is 60 sin 6 = 15 [ ] But π so just one decimal point coincides. 10 Problem.9. Prove that (.33) cos = (.3) sin = Here is a table assembling the information we have gathered so far. θ cos θ sin θ Problem.30. An exact trigonometric table of cos 3 k and sin 3 k for all k = 1,..., 15 can be obtained with the same trick which goes in principle back to Ptolemy! (a) Calculate With ω, z and w from Problem., get the argument θ in ωz 1 w 1 = e πiθ = cos πθ + i sin πθ 31
33 (b) Calculate ωz 1 w 1 with the exact square roots, obtained in Problem.. Get a readable result, do not distribute every term. (c) Get exact expressions for cos 3 and sin 3. Solution. (a) Since = 1, we get the argument 10 ωz 1 w 1 = e πi[ ] π π = cos + i sin (b) With the exact square roots from Problem., [ ωz 1 w i ] [1 ] 3 i 5 1 i = ωwz = = i( 3 + 1) = ( 3 1)( 5 1) + ( 3 + 1) i i ( 3 + 1)( 5 1) ( 3 1) (e) We get exact expressions ( 3 1)( 5 1) + ( 3 + 1) cos 3 = 16 ( 3 + 1)( 5 1) ( 3 1) sin 3 = Problem.31. Calculate the circumference of a regular 60-gon. Which approximation of π do we get? How many decimal points are correct? Answer. The circumference of an 60-gon is 10 sin 3 = 15 One gets two correct decimal points. [ ( 3 + 1)( 5 1) ( 3 1) 5 + ] 5 π Problem.3. Gather the entire table with cos k3 and sin k3 for all k = Go on as far as you get. 3
34 [ ] k [ ω k z k w k = e πik i( 3 + 1) 5 1 i = [ ω z w = e πi 3 + i 10 = ] [ 5 1 i 10 ] 5 ] k = i 5 i + i [ ω z w = e πi 1 i ] [ ] i = ωz = = i 15 + i 3 + i = [ ω 7 z 7 w 7 = e 1πi 10 = ωwz i( ] [ 3 1) = 5 1 i 10 ] 5 ( 3 + 1)( 5 + 1) + ( 3 1) 5 5 ( 3 + 1)( 5 + 1) + ( 3 + 1) i Further values were obtained by trial and error for the signs of the roots. 33
35 θ cos θ sin θ 3 ( 3 1)( 5 1)+( 3+1) 5+ 5 ( 3+1)( 5 1) ( 3 1) ( 3+1)( 5+1)+( 3 1) ( 3 1)( 5 1)+( 3+1) ( 3+1)( 5+1)+( 3+1) ( 3+1)( 5+1) ( 3 1) ( 3+1)( 5 1)+( 3 1) ( 3+1)( 5+1)+( 3+1) ( 3+1)( 5+1)+( 3+1) ( 3+1)( 5 1)+( 3 1) ( 3+1)( 5+1) ( 3 1) ( 3+1)( 5+1)+( 3+1) ( 3 1)( 5 1)+( 3+1) ( 3+1)( 5+1)+( 3 1) ( 3 1)( 5 1)+( 3+1) ( 3+1)( 5 1) ( 3 1)
36 3 Polynomials 3.1 Rational zeros of polynomials Proposition 3.1. Assume x is a real or complex zero of a monic 1 integer polynomial x r + a r 1 x r a 1 x + a 0 = 0 where a r 1,... a 1, a 0 are integers. If x is rational, then it is even an integer, and a divisor of the constant term a 0, too. Proof. Assume x = m n yields where m and n are relatively prime integers. Multiplying by nr m r n r + a m r 1 r 1 n + + a m r 1 1 n + a 0 = 0 m r + a r 1 m r 1 n + + a 1 mn r 1 + a 0 n r = 0 m r = ( a r 1 m r a 1 mn r + a 0 n r 1) n Now we argue the same way as in the proof of Proposition 1.6: Assume that any prime p divides n. By the last line, the same prime divides m r. Hence, by Euclid s Lemma p divides m. This cannot occur after cancellation of the fraction m. Hence n = 1, and n the rational root x is an integer. To get the last claim, we reuse the formulas from above, now with n = 1, and get m r + a r 1 m r a 1 m + a 0 = 0 m ( m r 1 + a r 1 m r + + a 1 ) = a0 Hence any zero m has to be a divisor of the constant coefficient a 0. Proposition 3.. Assume x is a real or complex zero of any polynomial a r x r + a r 1 x r a 1 x + a 0 = 0 where a r 0, a r 1,... a 1, a 0 are integers. If the root x = m is rational with m and n n relatively prime, then m is a divisor of a 0 and n is a divisor of a r. Proof. a r m r + a r 1 m r 1 n + + a 1 mn r 1 + a 0 n r = 0 ( a r m r 1 + a r 1 m r n + + a 1 n r 1) m = a 0 n r a r m r = ( a r 1 m r a 1 mn r + a 0 n r 1) n 1 A polynomial is called monic if the coefficient of the leading term x r is one. Hence n is a divisor of m r and hence n = 1, since n and m are relatively prime. 35
37 The second line implies that n is a divisor of a r m r. Since gcd (n, m) = 1, the Euclidean Property stated in Proposition 1.1 implies that n is a divisor of a r. The last line implies that m is a divisor of a 0 n r. Since gcd (n, m) = 1, the Euclidean Property stated in Proposition 1.1 implies that m is a divisor of a 0. Rule: The rational solutions of an integer polynomial a r x r + + a 0 = 0, are among the positive or negatives of the rational numbers top and bottom of which are divisors of top and bottom of the solution of a r X + a 0 = Gauss Lemma Proposition 3.3 (Linear Gauss Lemma). Suppose that an integer polynomial has a rational zero m, with n, m are relatively prime. Then it can be factored into integer n polynomials (3.1) a r x r + + a 0 = (nx m)(b r 1 x r b 0 ) If additionally gcd (a r,..., a 0 ) = 1, then gcd (b r 1,..., b 0 ) = 1, too. Proposition 3. (Monic Linear Gauss Lemma). Suppose that a monic integer polynomial has a rational zero. Then the root is an integer, and the polynomial can be factored into monic integer polynomials x r + + a 0 = (x m)(x r b 0 ) Proof. We may assume from the start that gcd (a r,..., a 0 ) = 1. The factoring of equation (3.1) holds with rational coefficients b r 1 = B r 1,..., b B 0 = B 0. We choose the B smallest possible denominator B. In that case, the assumption that gcd (a r,..., a 0 ) = 1 implies that gcd (B r 1,..., B 0 ) = 1, too. 3 After multiplying with the least common denominator B of the b 0,... b r 1, comparison of coefficients yields Ba 0 = mb 0 Ba 1 = nb 0 mb 1 Ba = nb 1 mb Ba r 1 = nb r mb r 1 Ba r = nb r 1 Assume towards a contradiction that a prime p divides all Ba 0,... Ba r. Either p does not divide n or p does not divide m. 3 Indeed, a common prime divisor of all B r 1,..., B 0 could not divide B and hence would be a common divisor of all a r,..., a 0, what we just have ruled out. 36
38 Take the second case. Successively, we conclude that p divides Ba 0 = mb 0, B 0, Ba 1 nb 0 = mb 1, B 1, Ba nb 1 = mb, B,..., B r 1, hence all B 0,... B r 1 contrary to the assumption. Hence we conclude gcd (Ba 0,... Ba r ) = 1. In the other case that p does not divide n, we get the same conclusion, now going through the system in reversed order: Successively, we see that the prime p divides Ba r = nb r 1, B r 1, Ba r 1 +mb r 1 = nb r, B r,..., Ba 1 + mb 1 = nb 0, B 0. In both cases, the conclusion gcd (Ba 0,... Ba r ) = 1 holds. Hence B = 1, and in the first place, the rational numbers b 0,... b r 1 need to be integers. Finally, we see that gcd (b r 1,..., b 0 ) = 1. Indeed, since B = 1, a prime p dividing all coefficients b r 1,..., b 0 would divides all coefficients a r,..., a 0, too, contrary to the assumption. Corollary (Gauss Lemma). If an integer polynomial can be factored over the rationals, it can be factored over the integers by adjusting a common factor. Especially, a monic integer polynomial that can be factored over the rationals, can even be factored over the integers into monic integer polynomials. Proof. Suppose that an integer polynomial of degree r can be factored into rational polynomials of degree s 1 and t 1. Multiplying with the least common denominators, we get the integer formula (3.) B (a r x r + + a 0 ) = A (b s x s + + b 0 ) ( ) c t x t + + c 0 where a r x r + + a 0 is obtained from the given integer polynomial by dividing through with the greatest common divisor of the coefficients. Because of using least common denominators everywhere, and cancellation, we get (3.3) gcd (A, B) = 1 gcd (a r,..., a 0 ) = 1 gcd (b s,..., b 0 ) = 1 gcd (c t,..., c 0 ) = 1 Question. Show that A = 1. Answer. Assume any prime number p divides A, and derive a contradiction. By formula (3.) this would imply that p divides all numbers Ba r,... Ba 0. Since gcd (a r,..., a 0 ) = 1, we conclude that p divides B. But this contradicts the assumption gcd (A, B) = 1. Lemma 3.1. If a prime p divides all coefficients of the product of two polynomials with integer coefficients (3.) (b s x s + + b 0 ) ( c t x t + + c 0 ), then p divides either gcd (b s,..., b 0 ) or gcd (c t,..., c 0 ). 37
Selected Chapters from Number Theory and Algebra
Selected Chapters from Number Theory and Algebra A project under construction Franz Rothe Department of Mathematics University of North Carolina at Charlotte Charlotte, NC 83 frothe@uncc.edu December 8,
More informationGaussian integers. 1 = a 2 + b 2 = c 2 + d 2.
Gaussian integers 1 Units in Z[i] An element x = a + bi Z[i], a, b Z is a unit if there exists y = c + di Z[i] such that xy = 1. This implies 1 = x 2 y 2 = (a 2 + b 2 )(c 2 + d 2 ) But a 2, b 2, c 2, d
More information1. multiplication is commutative and associative;
Chapter 4 The Arithmetic of Z In this chapter, we start by introducing the concept of congruences; these are used in our proof (going back to Gauss 1 ) that every integer has a unique prime factorization.
More informationSolutions to Practice Final
s to Practice Final 1. (a) What is φ(0 100 ) where φ is Euler s φ-function? (b) Find an integer x such that 140x 1 (mod 01). Hint: gcd(140, 01) = 7. (a) φ(0 100 ) = φ(4 100 5 100 ) = φ( 00 5 100 ) = (
More informationIntroduction to Abstract Mathematics
Introduction to Abstract Mathematics Notation: Z + or Z >0 denotes the set {1, 2, 3,...} of positive integers, Z 0 is the set {0, 1, 2,...} of nonnegative integers, Z is the set {..., 1, 0, 1, 2,...} of
More information7.2 Applications of Euler s and Fermat s Theorem.
7.2 Applications of Euler s and Fermat s Theorem. i) Finding and using inverses. From Fermat s Little Theorem we see that if p is prime and p a then a p 1 1 mod p, or equivalently a p 2 a 1 mod p. This
More informationPYTHAGOREAN TRIPLES KEITH CONRAD
PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient
More informationNotes: Pythagorean Triples
Math 5330 Spring 2018 Notes: Pythagorean Triples Many people know that 3 2 + 4 2 = 5 2. Less commonly known are 5 2 + 12 2 = 13 2 and 7 2 + 24 2 = 25 2. Such a set of integers is called a Pythagorean Triple.
More informationConsidering our result for the sum and product of analytic functions, this means that for (a 0, a 1,..., a N ) C N+1, the polynomial.
Lecture 3 Usual complex functions MATH-GA 245.00 Complex Variables Polynomials. Construction f : z z is analytic on all of C since its real and imaginary parts satisfy the Cauchy-Riemann relations and
More information8 Primes and Modular Arithmetic
8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.
More information(x 1, y 1 ) = (x 2, y 2 ) if and only if x 1 = x 2 and y 1 = y 2.
1. Complex numbers A complex number z is defined as an ordered pair z = (x, y), where x and y are a pair of real numbers. In usual notation, we write z = x + iy, where i is a symbol. The operations of
More informationContribution of Problems
Exam topics 1. Basic structures: sets, lists, functions (a) Sets { }: write all elements, or define by condition (b) Set operations: A B, A B, A\B, A c (c) Lists ( ): Cartesian product A B (d) Functions
More informationTable of Contents. 2013, Pearson Education, Inc.
Table of Contents Chapter 1 What is Number Theory? 1 Chapter Pythagorean Triples 5 Chapter 3 Pythagorean Triples and the Unit Circle 11 Chapter 4 Sums of Higher Powers and Fermat s Last Theorem 16 Chapter
More informationWORKSHEET ON NUMBERS, MATH 215 FALL. We start our study of numbers with the integers: N = {1, 2, 3,...}
WORKSHEET ON NUMBERS, MATH 215 FALL 18(WHYTE) We start our study of numbers with the integers: Z = {..., 2, 1, 0, 1, 2, 3,... } and their subset of natural numbers: N = {1, 2, 3,...} For now we will not
More informationM381 Number Theory 2004 Page 1
M81 Number Theory 2004 Page 1 [[ Comments are written like this. Please send me (dave@wildd.freeserve.co.uk) details of any errors you find or suggestions for improvements. ]] Question 1 20 = 2 * 10 +
More informationMATH 4400 SOLUTIONS TO SOME EXERCISES. 1. Chapter 1
MATH 4400 SOLUTIONS TO SOME EXERCISES 1.1.3. If a b and b c show that a c. 1. Chapter 1 Solution: a b means that b = na and b c that c = mb. Substituting b = na gives c = (mn)a, that is, a c. 1.2.1. Find
More informationSOLUTIONS TO PROBLEM SET 1. Section = 2 3, 1. n n + 1. k(k + 1) k=1 k(k + 1) + 1 (n + 1)(n + 2) n + 2,
SOLUTIONS TO PROBLEM SET 1 Section 1.3 Exercise 4. We see that 1 1 2 = 1 2, 1 1 2 + 1 2 3 = 2 3, 1 1 2 + 1 2 3 + 1 3 4 = 3 4, and is reasonable to conjecture n k=1 We will prove this formula by induction.
More informationg(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.
6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral
More information2. THE EUCLIDEAN ALGORITHM More ring essentials
2. THE EUCLIDEAN ALGORITHM More ring essentials In this chapter: rings R commutative with 1. An element b R divides a R, or b is a divisor of a, or a is divisible by b, or a is a multiple of b, if there
More informationHMMT February 2018 February 10, 2018
HMMT February 018 February 10, 018 Algebra and Number Theory 1. For some real number c, the graphs of the equation y = x 0 + x + 18 and the line y = x + c intersect at exactly one point. What is c? 18
More informationa + bi by sending α = a + bi to a 2 + b 2. To see properties (1) and (2), it helps to think of complex numbers in polar coordinates:
5. Types of domains It turns out that in number theory the fact that certain rings have unique factorisation has very strong arithmetic consequences. We first write down some definitions. Definition 5.1.
More informationNUMBER SYSTEMS. Number theory is the study of the integers. We denote the set of integers by Z:
NUMBER SYSTEMS Number theory is the study of the integers. We denote the set of integers by Z: Z = {..., 3, 2, 1, 0, 1, 2, 3,... }. The integers have two operations defined on them, addition and multiplication,
More information32 Divisibility Theory in Integral Domains
3 Divisibility Theory in Integral Domains As we have already mentioned, the ring of integers is the prototype of integral domains. There is a divisibility relation on * : an integer b is said to be divisible
More informationChapter 5. Number Theory. 5.1 Base b representations
Chapter 5 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,
More information1 Overview and revision
MTH6128 Number Theory Notes 1 Spring 2018 1 Overview and revision In this section we will meet some of the concerns of Number Theory, and have a brief revision of some of the relevant material from Introduction
More informationMath 109 HW 9 Solutions
Math 109 HW 9 Solutions Problems IV 18. Solve the linear diophantine equation 6m + 10n + 15p = 1 Solution: Let y = 10n + 15p. Since (10, 15) is 5, we must have that y = 5x for some integer x, and (as we
More informationCourse 2316 Sample Paper 1
Course 2316 Sample Paper 1 Timothy Murphy April 19, 2015 Attempt 5 questions. All carry the same mark. 1. State and prove the Fundamental Theorem of Arithmetic (for N). Prove that there are an infinity
More informationCoach Stones Expanded Standard Pre-Calculus Algorithm Packet Page 1 Section: P.1 Algebraic Expressions, Mathematical Models and Real Numbers
Coach Stones Expanded Standard Pre-Calculus Algorithm Packet Page 1 Section: P.1 Algebraic Expressions, Mathematical Models and Real Numbers CLASSIFICATIONS OF NUMBERS NATURAL NUMBERS = N = {1,2,3,4,...}
More informationINTEGERS. In this section we aim to show the following: Goal. Every natural number can be written uniquely as a product of primes.
INTEGERS PETER MAYR (MATH 2001, CU BOULDER) In this section we aim to show the following: Goal. Every natural number can be written uniquely as a product of primes. 1. Divisibility Definition. Let a, b
More information1. Factorization Divisibility in Z.
8 J. E. CREMONA 1.1. Divisibility in Z. 1. Factorization Definition 1.1.1. Let a, b Z. Then we say that a divides b and write a b if b = ac for some c Z: a b c Z : b = ac. Alternatively, we may say that
More informationIntroduction to Number Theory
INTRODUCTION Definition: Natural Numbers, Integers Natural numbers: N={0,1,, }. Integers: Z={0,±1,±, }. Definition: Divisor If a Z can be writeen as a=bc where b, c Z, then we say a is divisible by b or,
More information18. Cyclotomic polynomials II
18. Cyclotomic polynomials II 18.1 Cyclotomic polynomials over Z 18.2 Worked examples Now that we have Gauss lemma in hand we can look at cyclotomic polynomials again, not as polynomials with coefficients
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More informationNotes on Complex Analysis
Michael Papadimitrakis Notes on Complex Analysis Department of Mathematics University of Crete Contents The complex plane.. The complex plane...................................2 Argument and polar representation.........................
More informationIn Z: x + 3 = 2 3x = 2 x = 1 No solution In Q: 3x = 2 x 2 = 2. x = 2 No solution. In R: x 2 = 2 x = 0 x = ± 2 No solution Z Q.
THE UNIVERSITY OF NEW SOUTH WALES SCHOOL OF MATHEMATICS AND STATISTICS MATH 1141 HIGHER MATHEMATICS 1A ALGEBRA. Section 1: - Complex Numbers. 1. The Number Systems. Let us begin by trying to solve various
More informationDefinition 6.1 (p.277) A positive integer n is prime when n > 1 and the only positive divisors are 1 and n. Alternatively
6 Prime Numbers Part VI of PJE 6.1 Fundamental Results Definition 6.1 (p.277) A positive integer n is prime when n > 1 and the only positive divisors are 1 and n. Alternatively D (p) = { p 1 1 p}. Otherwise
More informationThe Fundamental Theorem of Arithmetic
Chapter 1 The Fundamental Theorem of Arithmetic 1.1 Primes Definition 1.1. We say that p N is prime if it has just two factors in N, 1 and p itself. Number theory might be described as the study of the
More information2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer?
Chapter 3: Theory of Modular Arithmetic 25 SECTION C Solving Linear Congruences By the end of this section you will be able to solve congruence equations determine the number of solutions find the multiplicative
More informationNumber Theory Solutions Packet
Number Theory Solutions Pacet 1 There exist two distinct positive integers, both of which are divisors of 10 10, with sum equal to 157 What are they? Solution Suppose 157 = x + y for x and y divisors of
More informationExplicit Methods in Algebraic Number Theory
Explicit Methods in Algebraic Number Theory Amalia Pizarro Madariaga Instituto de Matemáticas Universidad de Valparaíso, Chile amaliapizarro@uvcl 1 Lecture 1 11 Number fields and ring of integers Algebraic
More informationSolutions to Assignment 1
Solutions to Assignment 1 Question 1. [Exercises 1.1, # 6] Use the division algorithm to prove that every odd integer is either of the form 4k + 1 or of the form 4k + 3 for some integer k. For each positive
More informationMA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES
MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2018 57 5. p-adic Numbers 5.1. Motivating examples. We all know that 2 is irrational, so that 2 is not a square in the rational field Q, but that we can
More informationCOMPLEX NUMBERS AND QUADRATIC EQUATIONS
Chapter 5 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 5. Overview We know that the square of a real number is always non-negative e.g. (4) 6 and ( 4) 6. Therefore, square root of 6 is ± 4. What about the square
More informationNumber Theory. Final Exam from Spring Solutions
Number Theory. Final Exam from Spring 2013. Solutions 1. (a) (5 pts) Let d be a positive integer which is not a perfect square. Prove that Pell s equation x 2 dy 2 = 1 has a solution (x, y) with x > 0,
More informationMath 118: Advanced Number Theory. Samit Dasgupta and Gary Kirby
Math 8: Advanced Number Theory Samit Dasgupta and Gary Kirby April, 05 Contents Basics of Number Theory. The Fundamental Theorem of Arithmetic......................... The Euclidean Algorithm and Unique
More informationSolutions to Homework for M351 Algebra I
Hwk 42: Solutions to Homework for M351 Algebra I In the ring Z[i], find a greatest common divisor of a = 16 + 2i and b = 14 + 31i, using repeated division with remainder in analogy to Problem 25. (Note
More informationFoundations for Functions Knowledge and Skills: Foundations for Functions Knowledge and Skills:
Texas University Interscholastic League Contest Event: Number Sense This 80-question mental math contest covers all high school mathematics curricula. All answers must be derived without using scratch
More information2 Arithmetic. 2.1 Greatest common divisors. This chapter is about properties of the integers Z = {..., 2, 1, 0, 1, 2,...}.
2 Arithmetic This chapter is about properties of the integers Z = {..., 2, 1, 0, 1, 2,...}. (See [Houston, Chapters 27 & 28]) 2.1 Greatest common divisors Definition 2.16. If a, b are integers, we say
More informationMath Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions
Math Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions Warm-up Problems 1. What is a prime number? Give an example of an even prime number and an odd prime number. A prime number
More informationAlgebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001
Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,
More informationReview Sheet for the Final Exam of MATH Fall 2009
Review Sheet for the Final Exam of MATH 1600 - Fall 2009 All of Chapter 1. 1. Sets and Proofs Elements and subsets of a set. The notion of implication and the way you can use it to build a proof. Logical
More informationMATH 2112/CSCI 2112, Discrete Structures I Winter 2007 Toby Kenney Homework Sheet 5 Hints & Model Solutions
MATH 11/CSCI 11, Discrete Structures I Winter 007 Toby Kenney Homework Sheet 5 Hints & Model Solutions Sheet 4 5 Define the repeat of a positive integer as the number obtained by writing it twice in a
More informationNOTES ON SIMPLE NUMBER THEORY
NOTES ON SIMPLE NUMBER THEORY DAMIEN PITMAN 1. Definitions & Theorems Definition: We say d divides m iff d is positive integer and m is an integer and there is an integer q such that m = dq. In this case,
More informationEXAMPLES OF MORDELL S EQUATION
EXAMPLES OF MORDELL S EQUATION KEITH CONRAD 1. Introduction The equation y 2 = x 3 +k, for k Z, is called Mordell s equation 1 on account of Mordell s long interest in it throughout his life. A natural
More informationTest Codes : MIA (Objective Type) and MIB (Short Answer Type) 2007
Test Codes : MIA (Objective Type) and MIB (Short Answer Type) 007 Questions will be set on the following and related topics. Algebra: Sets, operations on sets. Prime numbers, factorisation of integers
More informationCONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS
CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS JEREMY BOOHER Continued fractions usually get short-changed at PROMYS, but they are interesting in their own right and useful in other areas
More informationALGEBRA. 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers
ALGEBRA CHRISTIAN REMLING 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers by Z = {..., 2, 1, 0, 1,...}. Given a, b Z, we write a b if b = ac for some
More informationLecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman
Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Factorization 0.1.1 Factorization of Integers and Polynomials Now we are going
More informationPRIME NUMBERS YANKI LEKILI
PRIME NUMBERS YANKI LEKILI We denote by N the set of natural numbers: 1,2,..., These are constructed using Peano axioms. We will not get into the philosophical questions related to this and simply assume
More informationPRACTICE PROBLEMS: SET 1
PRACTICE PROBLEMS: SET MATH 437/537: PROF. DRAGOS GHIOCA. Problems Problem. Let a, b N. Show that if gcd(a, b) = lcm[a, b], then a = b. Problem. Let n, k N with n. Prove that (n ) (n k ) if and only if
More informationGlossary. Glossary 981. Hawkes Learning Systems. All rights reserved.
A Glossary Absolute value The distance a number is from 0 on a number line Acute angle An angle whose measure is between 0 and 90 Addends The numbers being added in an addition problem Addition principle
More informationFinite Fields: An introduction through exercises Jonathan Buss Spring 2014
Finite Fields: An introduction through exercises Jonathan Buss Spring 2014 A typical course in abstract algebra starts with groups, and then moves on to rings, vector spaces, fields, etc. This sequence
More informationGlossary. Glossary Hawkes Learning Systems. All rights reserved.
A Glossary Absolute value The distance a number is from 0 on a number line Acute angle An angle whose measure is between 0 and 90 Acute triangle A triangle in which all three angles are acute Addends The
More informationHomework #2 solutions Due: June 15, 2012
All of the following exercises are based on the material in the handout on integers found on the class website. 1. Find d = gcd(475, 385) and express it as a linear combination of 475 and 385. That is
More informationTC10 / 3. Finite fields S. Xambó
TC10 / 3. Finite fields S. Xambó The ring Construction of finite fields The Frobenius automorphism Splitting field of a polynomial Structure of the multiplicative group of a finite field Structure of the
More information6]. (10) (i) Determine the units in the rings Z[i] and Z[ 10]. If n is a squarefree
Quadratic extensions Definition: Let R, S be commutative rings, R S. An extension of rings R S is said to be quadratic there is α S \R and monic polynomial f(x) R[x] of degree such that f(α) = 0 and S
More informationThe primitive root theorem
The primitive root theorem Mar Steinberger First recall that if R is a ring, then a R is a unit if there exists b R with ab = ba = 1. The collection of all units in R is denoted R and forms a group under
More informationAlgebraic. techniques1
techniques Algebraic An electrician, a bank worker, a plumber and so on all have tools of their trade. Without these tools, and a good working knowledge of how to use them, it would be impossible for them
More informationPUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime.
PUTNAM TRAINING NUMBER THEORY (Last updated: December 11, 2017) Remark. This is a list of exercises on Number Theory. Miguel A. Lerma Exercises 1. Show that the sum of two consecutive primes is never twice
More informationComputations/Applications
Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x
More informationAn integer p is prime if p > 1 and p has exactly two positive divisors, 1 and p.
Chapter 6 Prime Numbers Part VI of PJE. Definition and Fundamental Results Definition. (PJE definition 23.1.1) An integer p is prime if p > 1 and p has exactly two positive divisors, 1 and p. If n > 1
More informationQuasi-reducible Polynomials
Quasi-reducible Polynomials Jacques Willekens 06-Dec-2008 Abstract In this article, we investigate polynomials that are irreducible over Q, but are reducible modulo any prime number. 1 Introduction Let
More informationCHAPTER 6. Prime Numbers. Definition and Fundamental Results
CHAPTER 6 Prime Numbers Part VI of PJE. Definition and Fundamental Results 6.1. Definition. (PJE definition 23.1.1) An integer p is prime if p > 1 and the only positive divisors of p are 1 and p. If n
More informationMathematics 136 Calculus 2 Everything You Need Or Want To Know About Partial Fractions (and maybe more!) October 19 and 21, 2016
Mathematics 36 Calculus 2 Everything You Need Or Want To Know About Partial Fractions (and maybe more!) October 9 and 2, 206 Every rational function (quotient of polynomials) can be written as a polynomial
More informationD-MATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 6. Unique Factorization Domains
D-MATH Algebra I HS18 Prof. Rahul Pandharipande Solution 6 Unique Factorization Domains 1. Let R be a UFD. Let that a, b R be coprime elements (that is, gcd(a, b) R ) and c R. Suppose that a c and b c.
More informationElementary Algebra Chinese Remainder Theorem Euclidean Algorithm
Elementary Algebra Chinese Remainder Theorem Euclidean Algorithm April 11, 2010 1 Algebra We start by discussing algebraic structures and their properties. This is presented in more depth than what we
More informationThis is a recursive algorithm. The procedure is guaranteed to terminate, since the second argument decreases each time.
8 Modular Arithmetic We introduce an operator mod. Let d be a positive integer. For c a nonnegative integer, the value c mod d is the remainder when c is divided by d. For example, c mod d = 0 if and only
More informationA field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties:
Byte multiplication 1 Field arithmetic A field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties: F is an abelian group under addition, meaning - F is closed under
More informationComplex numbers, the exponential function, and factorization over C
Complex numbers, the exponential function, and factorization over C 1 Complex Numbers Recall that for every non-zero real number x, its square x 2 = x x is always positive. Consequently, R does not contain
More informationMODEL ANSWERS TO HWK #10
MODEL ANSWERS TO HWK #10 1. (i) As x + 4 has degree one, either it divides x 3 6x + 7 or these two polynomials are coprime. But if x + 4 divides x 3 6x + 7 then x = 4 is a root of x 3 6x + 7, which it
More informationCongruent Number Problem and Elliptic curves
Congruent Number Problem and Elliptic curves December 12, 2010 Contents 1 Congruent Number problem 2 1.1 1 is not a congruent number.................................. 2 2 Certain Elliptic Curves 4 3 Using
More informationSolutions to Practice Final 3
s to Practice Final 1. The Fibonacci sequence is the sequence of numbers F (1), F (2),... defined by the following recurrence relations: F (1) = 1, F (2) = 1, F (n) = F (n 1) + F (n 2) for all n > 2. For
More informationLECTURE NOTES IN CRYPTOGRAPHY
1 LECTURE NOTES IN CRYPTOGRAPHY Thomas Johansson 2005/2006 c Thomas Johansson 2006 2 Chapter 1 Abstract algebra and Number theory Before we start the treatment of cryptography we need to review some basic
More informationMATH Fundamental Concepts of Algebra
MATH 4001 Fundamental Concepts of Algebra Instructor: Darci L. Kracht Kent State University April, 015 0 Introduction We will begin our study of mathematics this semester with the familiar notion of even
More informationChapter 14: Divisibility and factorization
Chapter 14: Divisibility and factorization Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Summer I 2014 M. Macauley (Clemson) Chapter
More informationQUADRATIC RINGS PETE L. CLARK
QUADRATIC RINGS PETE L. CLARK 1. Quadratic fields and quadratic rings Let D be a squarefree integer not equal to 0 or 1. Then D is irrational, and Q[ D], the subring of C obtained by adjoining D to Q,
More informationNumber Theory. Introduction
Number Theory Introduction Number theory is the branch of algebra which studies the properties of the integers. While we may from time to time use real or even complex numbers as tools to help us study
More information2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer?
Chapter 3: Theory of Modular Arithmetic 25 SECTION C Solving Linear Congruences By the end of this section you will be able to solve congruence equations determine the number of solutions find the multiplicative
More informationMathematical Olympiad Training Polynomials
Mathematical Olympiad Training Polynomials Definition A polynomial over a ring R(Z, Q, R, C) in x is an expression of the form p(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0, a i R, for 0 i n. If a n 0,
More information7. Prime Numbers Part VI of PJE
7. Prime Numbers Part VI of PJE 7.1 Definition (p.277) A positive integer n is prime when n > 1 and the only divisors are ±1 and +n. That is D (n) = { n 1 1 n}. Otherwise n > 1 is said to be composite.
More informationPELL S EQUATION, II KEITH CONRAD
PELL S EQUATION, II KEITH CONRAD 1. Introduction In Part I we met Pell s equation x dy = 1 for nonsquare positive integers d. We stated Lagrange s theorem that every Pell equation has a nontrivial solution
More informationAn Introduction to Proof-based Mathematics Harvard/MIT ESP: Summer HSSP Isabel Vogt
An Introduction to Proof-based Mathematics Harvard/MIT ESP: Summer HSSP Isabel Vogt Class Objectives Field Axioms Finite Fields Field Extensions Class 5: Fields and Field Extensions 1 1. Axioms for a field
More informationFactorization in Integral Domains II
Factorization in Integral Domains II 1 Statement of the main theorem Throughout these notes, unless otherwise specified, R is a UFD with field of quotients F. The main examples will be R = Z, F = Q, and
More information. As the binomial coefficients are integers we have that. 2 n(n 1).
Math 580 Homework. 1. Divisibility. Definition 1. Let a, b be integers with a 0. Then b divides b iff there is an integer k such that b = ka. In the case we write a b. In this case we also say a is a factor
More information. In particular if a b then N(
Gaussian Integers II Let us summarise what we now about Gaussian integers so far: If a, b Z[ i], then N( ab) N( a) N( b). In particular if a b then N( a ) N( b). Let z Z[i]. If N( z ) is an integer prime,
More informationLecture Notes on DISCRETE MATHEMATICS. Eusebius Doedel
Lecture Notes on DISCRETE MATHEMATICS Eusebius Doedel c Eusebius J. Doedel, 009 Contents Logic. Introduction............................................................................... Basic logical
More information( 3) ( ) ( ) ( ) ( ) ( )
81 Instruction: Determining the Possible Rational Roots using the Rational Root Theorem Consider the theorem stated below. Rational Root Theorem: If the rational number b / c, in lowest terms, is a root
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)
More informationExercises Exercises. 2. Determine whether each of these integers is prime. a) 21. b) 29. c) 71. d) 97. e) 111. f) 143. a) 19. b) 27. c) 93.
Exercises Exercises 1. Determine whether each of these integers is prime. a) 21 b) 29 c) 71 d) 97 e) 111 f) 143 2. Determine whether each of these integers is prime. a) 19 b) 27 c) 93 d) 101 e) 107 f)
More informationMATH 117 LECTURE NOTES
MATH 117 LECTURE NOTES XIN ZHOU Abstract. This is the set of lecture notes for Math 117 during Fall quarter of 2017 at UC Santa Barbara. The lectures follow closely the textbook [1]. Contents 1. The set
More information