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Selected Chapters from Number Theory and Algebra A project under construction Franz Rothe Department of Mathematics University of North Carolina at Charlotte Charlotte, NC 28223 [email protected] December 28, 2012 Allnumalg\numalg.tex Contents I Number Theory 3 1 Euclid’s Number Theory 1.1 The Euclidean algorithm . . . . . 1.2 How many owl-primes are there? 1.3 Prime numbers . . . . . . . . . . 1.4 Rational and irrational . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Fermat Primes 2.1 A short paragraph about Fermat numbers 2.2 More about Fermat numbers . . . . . . . 2.3 Pseudo random Number Generation . . . 2.4 Pseudo prime and Carmichael numbers . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 4 7 9 12 . . . . 13 13 14 22 22 3 Arithmetic of the Complex Numbers 3.1 Arctan identities . . . . . . . . . . . . . . 3.2 Gaussian primes . . . . . . . . . . . . . . . 3.3 Sums of two squares . . . . . . . . . . . . 3.4 Roots . . . . . . . . . . . . . . . . . . . . 3.5 The solution of the reduced cubic equation 3.6 The square root of a complex number . . . 3.7 Pythagorean triples . . . . . . . . . . . . . 3.8 Roots of unity . . . . . . . . . . . . . . . . II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Algebra 26 26 27 31 36 39 43 44 47 55 1 Polynomials 1.1 Rational zeros of polynomials . . . 1.2 Gauss’ Lemma . . . . . . . . . . . 1.3 Eisenstein’s irreducibility criterium 1.4 Descartes’ Rule of Signs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Tschebychev polynomials 55 55 56 60 63 67 3 Polynomials over the Complex Numbers 3.1 A proof of the Fundamental Theorem of Algebra 3.2 Meditation on Descartes’ rule of signs . . . . . . 3.3 Estimation of zeros with Rouché’s Theorem . . 3.4 Perron’s irreducibility criterium . . . . . . . . . 3.5 The limiting case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 71 72 77 78 79 4 The Residue Theorem and its Consequences 82 4.1 The residue theorem and the logarithmic residue theorem . . . . . . . . . 82 4.2 Comparing two functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 2 Part I Number Theory 3 1 Euclid’s Number Theory 1.1 The Euclidean algorithm 10 Problem 1.1. Calculate the greatest common divisor of 765 and 567. Find integers s and t such that gcd (765, 567) = s · 765 − t · 567. Answer. The extended Euclidean algorithms is used to calculate the greatest common divisor. In an additional parallel calculation, one gets the greatest common divisor as integer combination of the two given numbers. row 0: row 1: row 2: row 3: row 4: row 5: row 5+1: 765 567 = 1 rem 567 : 198 = 2 rem 198 : 171 = 1 rem 171 : 27 = 6 rem 27 : 9 = 3 rem 198 171 27 9 0 1 0 0 1 1 1 2 3 3 4 20 27 63 85 Indeed, gcd (765, 567) = 9 = (−20) · 765 + 27 · 567. Hence s = −20 and t = −27. Remark. The optional extra row 5 + 1 does not contain a division, only sM +1 and tM +1 are calculated. This is a convenient check, since 63 · 765 − 85 · 567 = 0. 10 Problem 1.2. Use the last problem to calculate the least common multiple lcm (765, 567). Answer. lcm (765, 567) = 765 · 567 = 765 · 63 = 48 195 gcd (765, 567) 10 Problem 1.3. Given that x ≡ 19 mod 765 and x ≡ 1 mod 567 Find the smallest positive solution for x. Answer. One can determine the unknown integer x modulo the least common multiple of lcm (765, 567). To get a solution, one needs p and q such that x = 19+p·765 = 1+q·567. Since 19−1 = 2 is an integer, the congruence is solvable. We need to multiply the result 9 of the problem above by this integer 2 and get 9 = (−20) · 765 + 27 · 567 18 = (−40) · 765 + 54 · 567 18 + 40 · 765 = 54 · 567 19 + 40 · 765 = 1 + 54 · 567 30 619 = 30 619 4 We get a solution x = 30 619, which turns out to be the smallest one, in this example. 1 10 Problem 1.4. Calculate the greatest common divisor of 367 and 47. Find integers s and t such that gcd (367, 47) = s · 367 + t · 47. Answer. The extended Euclidean algorithms is used to calculate the greatest common divisor. In an additional parallel calculation, one gets the greatest common divisor as integer combination of the two given numbers. Here is the example: row 0: row 1: row 2: row 3: row 4: row 5: row 5+1: 367 : 47 = 7 rem 47 : 38 = 1 rem 38 : 9 = 4 rem 9: 2 = 4 rem 2: 1 = 2 rem 1 0 0 1 1 7 1 8 5 39 21 164 47 367 32 9 2 1 0 Indeed, gcd (367, 74) = 1 = (−21) · 367 + 164 · 47. The optional extra row 5 + 1 does not contain a division, only sM +1 and tM +1 are calculated. This is a convenient check, since 47 · 367 − 367 · 47 = 0. 10 Problem 1.5. Find integers s and t such that 22s + 8t = 2. Find the least common multiple of 8 and 22. Answer. One can get immediately gcd (22, 8) = 2. Hence the least common multiple of 22 and 8 is 88. The extended Euclidean algorithm gives the greatest common divisor as linear combination: row row row row 0: 1: 2: 3: 22 : 8 = 2 rem 8 : 6 = 1 rem 6 : 2 = 3 rem 6 2 0 1 0 1 1 0 1 2 3 Indeed, gcd (22, 8) = 2 = (−1) · 22 + 3 · 8. 10 Problem 1.6. Given that x≡3 mod 8 and x ≡ 5 mod 22 determine the unknown integer x modulo the least common multiple of 8 and 22. Answer. One needs s and t such that 3 + 8s = 5 + 22t. This works for 3 + 8 · 3 = 5 + 22 · 1 = 27. Hence both x ≡ 27 mod 8 and x ≡ 27 mod 22 what implies x ≡ 27 mod 88 1 One does not always get immediately the smallest solution. The values p and q are not needed any more. 5 10 Problem 1.7. Calculate and prime factor the number 215 −1−(23 −1)(25 −1). Answer. (2p − 1) is always a divisor of (2p·q − 1 for any p and q. The prime factoring is 32 767 15 3 5 − 1 = 7 · 31 · 150 = 2 · 3 · 52 · 7 · 31 = 32 550 2 − 1 − (2 − 1)(2 − 1) = 7 · 31 · 7 · 31 Proposition 1.1 (Euclid’s Lemma). If a prime number divides the product of two integers, the prime number divides at least one of the two integers. Proof. Let p be the prime number, and the integers 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, we get gcd (a, p) = 1. By the extended Euclidean algorithm, there exist integers s, t such that 1 = sa + tp Hence b ab = s + tb p p Because p divides ab, the right hand side is an integer. Hence p divides b, as to be shown. 10 Problem 1.8. Prove that a nonconstant integer polynomial assumes infinitely many composite values P (n), for appropriately chosen natural numbers n. Proposition 1.2. A nonconstant integer polynomial assumes infinitely many composite values P (n), for appropriately chosen natural numbers n. Proof. Let d ≥ 1 be the degree of the given polynomial. By the fundamental theorem of algebra, the polynomial P can assume any value at most d times. 2 Hence there exist at most 3d natural numbers n for which P (n) is either 0, −1 or 1. Let a be natural number such that P (a) 6= 0, ±1. Let p be any prime factor of P (a) and put P (a) = pq. It is easy to see that for all integer k, the difference P (a + kp) − P (a) is divisible by p. Indeed, this follows by the binomial theorem. For any given polynomial with integer coefficients cl P (x) = P (a + kp) − P (a) = d X l=0 d X cl x l cl (a + kp)l − al l=0 = d X l=0 " d # l l X X X l l−j l cl a (kp)j = p · cl al−j k j pj−1 j j j=1 j=1 l=0 P (a + kp) − P (a) = p · m 2 Indeed, we only need to know for this proof that any value can be assumed at most as many times as the degree of the polynomial. 6 where m is an integer abbreviating the last double sum. Since P (a) = pq, we conclude P (a + kp) = p(m + q) is divisible by p for all integer k. There exist at most 3d integers k for which P (a + kp) is either 0, −p or p. Hence there exist infinitely many integers k such that a + kp ≥ 0 and P (a + kp) 6= 0, ±p is divisible by p and hence composite. 1.2 How many owl-primes are there? The values of the polynomial n2 − n + 41 for n = 1, 2, 3, . . . 40 all turn out to be primes. This curious observation goes back to Euler. 10 Problem 1.9. As a warm-up, find the three smallest primes larger than 41 which are not in the list of 40 primes just mentioned. Answer. These are the first three primes not in the range: 59, 67, 73 6= n2 − n + 41 for any natural number n. We now address the question which other primes besides 41 have the curious property noted by Euler. Definition 1.1 (Owl-prime). I call a prime p an owl-prime 3 if the values of the polynomial n2 − n + p for n = 1, 2, 3, . . . p − 1 all turn out to be primes. 10 Problem 1.10. As a warm-up, find the three smallest owl-primes. Now we have the obvious question: are there further owl-primes between 5 and 41? Proposition 1.3. For any prime p > 5 equivalent are (a) For any natural number n, the values n2 − n + p are divisible neither by 2, 3 nor 5. (b) p is congruent to either 11 or 17 modulo 30. Remark. Note this proposition gives only a necessary condition for a prime being an owl-prime. Reason. Clearly n2 − n + p is odd since n2 − n is always even, and every prime p > 2 is odd. If we go through the natural numbers n = 1, 2, 3, . . . , the value of n2 − n is either divisible by three, or it is congruent to 2 modulo 3. The first case occurs for n ≡ 0 or n ≡ 1 modulo 3. The second case occurs for n ≡ 2 mod 3. To assure that n2 − n + p is never divisible by 3 for any n, we need to have p ≡ 2 mod 3. From the following small table, we get the possible values of n − n2 mod 5 and n − n2 mod 7: 3 I have chosen this name in honest of Euler 7 n 1 2 3 4 5 6 7 n − n2 0 −2 −6 −12 −20 −30 −42 n − n2 mod 5 0 3 4 3 0 0 3 n − n2 mod 7 0 5 1 2 1 5 0 We see 0, 3, 4 are the possible values of n − n2 mod 5. To avoid that n2 − n + p is divisible by 5, we need that n − n2 and p are different modulo 5. Hence p has to be congruent to either 1 or 2 modulo 5. 10 Problem 1.11. If a number p is odd, p ≡ 2 mod 3, and p ≡ 1 mod 5, determine the number p modulo 30. Similarly, assume p is odd, p ≡ 2 mod 3, and p ≡ 2 mod 5, and determine p modulo 30. Answer. In the first case, we get p ≡ 11 mod 30. In the second case, we get p ≡ 17 mod 30. 10 Problem 1.12. Find the next two owl-primes larger than 5. 10 Problem 1.13. What are the values possible for p mod 7 of an owl-prime p. Answer. We have already obtained the possible values of n − n2 mod 7 are 0, 1, 2, 5. Let p > 7 be any prime. To avoid that n2 − n + p is divisible by 7, we need that n − n2 and p are different modulo 7. Hence p has to be congruent to either 3, 4 or 6 modulo 7. From the last remark, we already see that 47 is not an owl-prime. Indeed, 7 divides n − n + 47 for n = 2— since both 47 ≡ 5 mod 7 and 2 − 22 ≡ 5 mod 7. 2 Open Problem. Show that 2, 3, 5, 11, 17 and 41 are the only owl-primes, or find a larger owl-prime. 8 1.3 Prime numbers Proposition 1.4 (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 ab = s + tb c c The second line results by multiplication of both sides with cb . Because c divides ab, the 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 G := gcd (ab, ac) Hence the integer (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 1.2 (prime number). A prime number is an integer p ≥ 2, 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 p1 = 2 , p 2 = 3 , p 3 = 5 , p 4 = 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 αi ≥ 0 and βi ≥ 0, with index i = 1, 2, . . . and only finitely many terms nonzero such that Y Y β (1.2) a= piαi , b = pi 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. 9 Proposition 1.5 (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 ab b = s + tb p p Because p divides ab, the right hand side is an integer. Hence p divides b, as to be shown. Proposition 1.6 (Monotonicity). Let a and b have the prime decompositions Y β Y pi i (1.3) a= piαi , b = i≥1 i≥1 The number b is a divisor of a if and only if βi ≤ αi for all i ≥ 1. Reason. If βi ≤ αi for all i ≥ 1, then a = qb with Y α −β q= pi i i i≥1 and hence b is a divisor of a. Conversely, assume that b is a divisor of a. We need to show that βi ≤ αi for all i ≥ 1. Proceed by induction on b. If b = 1, then βi = 0 for all i ≥ 1, and the assertion is true. Here is the induction step ”b < n 7→ b = n”: Let pi be any prime factor of b ≥ 2, which means that βi ≥ 1. Because pi divides b and b divides a, the prime pi divides a. By Euclid’s Lemma Proposition 1.5, pi is a divisor of one of the primes pj occurring in the prime decomposition of a. Hence αj ≥ 1. Because different primes cannot divide each other, this implies i = j and pi = pj . Hence pbj < n is a divisor of paj . By the induction assumption, this implies βi ≤ αi for all i 6= j, as well as βj − 1 ≤ αj − 1 and hence βi ≤ αi for all i ≥ 1. Proposition 1.7 (Uniqueness of prime decomposition). The prime decomposition of any positive integer is unique. Reason. Assume a= Y piαi and a = i≥1 Y piβi i≥1 Because a divides a, the fact given above both tells that βi ≤ αi and αi ≤ βi for all i ≥ 1. Hence βi = αi for all i ≥ 1. 10 Proposition 1.8. Let a and b have the prime decompositions (1.3). The prime decompositions of the greatest common divisor and least common multiple are Y min[α ,β ] i i (1.4) gcd (a, b) = pi i≥1 lcm (a, b) = (1.5) Y max[αi ,βi ] pi i≥1 10 Problem 1.14. Check these formulas for a = 1001, b = 4221. Proof of Proposition 1.8. Let g be the righthand side of equation (1.5). We need to check properties (i) and (ii) defining the greatest common divisor. (i) g divides both a and b. Check. This is clear, because both min[αi , βi ] ≤ αi and min[αi , βi ] ≤ βi 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= Y piγi i≥1 be the prime decomposition of h. Because h divides both a and b, monotonicity implies that both γi ≤ αi and γi ≤ βi for all i ≥ 1. Hence γi ≤ min[αi , βi ] 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[αi , βi ] ≥ αi and max[αi , βi ] ≥ βi 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. 11 Check. Let (1.7) k= Y piγi i≥1 be the prime decomposition of k. Because k is a multiple of both a and b, monotonicity implies that both γi ≥ αi and γi ≥ βi for all i ≥ 1. Hence γi ≥ max[αi , βi ] for all i ≥ 1, which easily implies that k is a multiple of l. 1.4 Rational and irrational Proposition 1.9. For any natural numbers r ≥ 1 and a ≥ 1, the r-th root a rational number if it is even an integer. √ r a is only Proof. The assertion is clear for r = 1 or a = 1, hence we may assume r ≥ 2 and a ≥ 2. Now assume that √ m r a= n r n a = mr 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 pr divides nr , which in turn divides r an = mr . Hence pr divides mr . Hence, by Euclid’s Lemma p divides m, as claimed. As already explained, the assertion follows. Proposition 1.10. For any natural numbers r ≥ 2 and any a ≥ 2, which is not the √ r r-th power of an integer, the root 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 = mr 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. 12 2 Fermat Primes 2.1 A short paragraph about Fermat numbers Definition 2.1 (Fermat number, Fermat prime). The numbers n Fn = 22 + 1 for any integer n ≥ 0 are called Fermat numbers. A Fermat number which is prime is called a Fermat prime. Lemma 2.1 (Fermat). If p is any odd prime, and p − 1 is a power of two, then p = Fn is a Fermat prime. Proof. Assume p = 1 + 2a and a ≥ 3 odd. Because of the fatorization a = 2n · (2b + 1) one can factor n ·(2b+1) p = 1 + 22 n n n ·2 = (1 + 22 )(1 − 22 + 22 n ·2b − · · · + 22 ) If p is an odd prime, the only possibility is b = 0 and p = Fn . The only known Fermat primes are Fn for n = 0, 1, 2, 3, 4. They are 3, 5, 17, 257, and 65537, see4 . Fermat numbers and Fermat primes were first studied by Pierre de Fermat, who conjectured 1654 in a letter to Blaise Pascal that all Fermat numbers are prime, but told he had not been able to find a proof. Indeed, the first five Fermat numbers are easily shown to be prime. However, Fermat’s conjecture was refuted by Leonhard Euler in 1732 when he showed, as one of his first number theoretic discoveries: 5 F5 = 22 + 1 = 232 + 1 = 4294967297 = 641 · 6700417. Later, Euler proved that every prime factor p of Fn must have the form p = i · 2n+1 + 1. Almost hundred years later, Lucas showed that even p = j · 2n+2 + 1. These results give a vague hope that one could factor Fermat numbers and perhaps settle Fermat’s conjecture to the negative. There are no other known Fermat primes Fn with n > 4. However, little is known about Fermat numbers with large n. In fact, each of the following is an open problem: 1. Is Fn composite for all n > 4? By this anti-Fermat hypothesis 3, 5, 17, 257, and 65537 would be the only Fermat primes. 2. Are there infinitely many Fermat primes? (Eisenstein 1844) 3. Are there infinitely many composite Fermat numbers? 4 sequence A019434 in OEIS 13 4. Are all Fermat numbers square free? As of 2012, the next twenty-eight Fermat numbers, F5 through F32 , are known to be composite. As of February 2012, only F0 to F11 have been completely factored. For complete information, see Fermat factoring status by Wilfrid Keller, on the internet at http://ww.prothsearch.net/fermat.html Main Theorem (Gauss-Wantzel Theorem). A regular polygon with n sides is constructible if and only if n = 2h p1 · p2 · · · ps where p2 · · · ps is a product of different Fermat primes. If the anti-Fermat hypothesis is true, there are exactly five Fermat primes, and hence exactly 31 regular constructible polygons with an odd number of sides. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 2.2 n f actored 3 F0 5 F1 15 3·5 17 F2 51 3 · 17 85 5 · 17 255 3 · 5 · 17 257 F3 771 3 · 257 1 285 5 · 257 3 855 3 · 5 · 257 4 369 17 · 257 13 107 3 · 17 · 257 21 845 5 · 17 · 257 65 535 3 · 5 · 17 · 257 65 537 F4 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 n f actored 196 611 3 · 65 537 327 685 5 · 65 537 983 055 3 · 5 · 65 537 1 114 129 17 · 65 537 3 342 387 3 · 17 · 65 537 5 570 645 5 · 17 · 65 537 16 711 935 3 · 5 · 17 · 65 537 16 843 009 257 · 65 537 50 529 027 3 · 257 · 65 537 84 215 045 5 · 257 · 65 537 252 645 135 3 · 5 · 257 · 65 537 286 331 153 17 · 257 · 65 537 858 993 459 3 · 17 · 257 · 65 537 1 431 655 765 5 · 17 · 257 · 65 537 4 294 967 295 3 · 5 · 17 · 257 · 65 537 More about Fermat numbers 10 Problem 2.1. Prove by induction that F0 · F1 · · · Fn−1 = Fn − 2 for all natural numbers n ≥ 1. Answer. Basic step: For n = 1, both sides of the formula are equal to 3, since 3 = F0 = F1 − 2 = 5 − 2. 14 Induction step ”n → n + 1”: The formula is assumed to hold for n as written. It is shown for n replaced by n + 1, as follows: F0 · F1 · · · Fn = [F0 · F1 · · · Fn−1 ] · Fn by recursive definition of the product = (Fn − 2)Fn by the induction assumption n n n n+1 2 = 22 − 1 22 + 1 = 22 − 12 = 22 − 1 = Fn+1 We have checked the asserted formula step for n + 1. From the basic step and the induction step together, we conclude by the principle of induction that the formula holds for all n ≥ 1. 10 Problem 2.2. Use the last problem to conclude Goldbach’s Theorem, which states that any two different Fermat numbers are relatively prime. Conclude there exist infinitely many primes. Answer. Assume that p is a common prime factor of the Fermat numbers Fk and Fn with 0 ≤ k < n. We could conclude that p is a divisor of F0 · F1 · · · Fn−1 = Fn − 2, and hence of both Fn − 2 and of Fn . This is only possible for p = 2. But all Fermat numbers are odd and cannot have divisor 2. This argument excludes that any two Fermat numbers have a common prime factor. For any natural number n, let pn be the smallest (or any) prime factor of Fn . The primes pn are all different since the Fermat numbers are relatively prime. Hence there exist infinitely many primes. Corollary. In any arithmetic sequence 1 + i 2k with i = 1, 2, 3, . . . there exist infinitely many primes. Reason. Any prime factor of Fn has the form pn = 1+i 2n+2 . Hence all pn with n ≥ k −2 are contained in the arithmetic sequence 1 + i 2k with i = 1, 2, 3, . . . . Proposition 2.1 (Euler 1770). Any prime factor of the Fermat number Fn has the form pn = 1 + i 2n+1 . Reason. Assume that p is a prime factor of the Fermat number Fn . Let h > 1 be the smallest integer such that 2h ≡ 1 mod p. This power h is also called the order of 2 modulo p. The definition of Fn implies n 22 ≡ −1 2n 2 ≡ −1 n+1 ≡1 mod Fn 2n+1 ≡1 mod p mod Fn and 22 mod p and 2 Since h is a divisor of 2n+1 but not 2n , we get h = 2n+1 . Fermat’s Little Theorem implies 2p−1 ≡ 1 mod p, and hence h is a divisor of p − 1. Together we conclude that 2n+1 is a divisor of p − 1, as to be shown. It took more than hundred years, until Édouard Lucas improved Euler’s result. 15 Proposition 2.2 (Lucas 1878). Any prime factor of the Fermat number Fn with n ≥ 2 has the form pn = 1 + j 2n+2 . Reason. We continue the reasoning from Euler’s Proposition 2.1. Assume that p is a prime factor of the Fermat number Fn and n ≥ 2. As a consequence of Euler’s Proposition, we see that 8 is a divisor of p − 1. Hence one can calculate the Legendre symbol (p−1)(p+1) 2 8 = (−1) =1 p and one gets from Euler’s criterium 2 p−1 2 2 ≡ =1 p mod p Hence the order h = 2n+1 of 2 modulo p is actually a divisor of (p − 1)/2. In the end we conclude that 2n+2 is a divisor of p − 1, as to be shown. Remark. It turns out occur many times that the Fermat number Fn has a prime factor 1 + j 2n+2 with j odd. This is known to happen for n = 5, 6, 7, 9, 10, 11, 12, 15, 17, 18, 19, 21, 23, 25, 27, 29, 30, 31, 32 and many more cases with n ≥ 33. In this sense, Lucas’ result turns out to be optimal. On the other hand, F8 has no prime factor with j odd. Theorem 2.1 (Pépin’s test (1877)). For n ≥ 1, the Fermat number Fn is prime if and only if 3(Fn −1)/2 ≡ −1 mod Fn In 1905 and 1909, J.C. Morehead and A.E.Western used Pépins test to prove that F7 and F8 are composite. As of 2001, no factor is known for the Fermat numbers Fn with n = 14, 20, 22, 24. These numbers were proved composite only with Pépin’s test. Here is a proof and some results related to this test. Proposition 2.3 (Sufficiency of a Pépin-like test). Assume there exists a natural number a such that a(Fn −1)/2 ≡ −1 mod Fn Then the Fermat number Fn is prime. Proof. The assumption implies (2.1) a(Fn −1)/2 ≡ −1 mod Fn and 16 aFn −1 ≡ 1 mod Fn Let p be any prime factor of the Fermat number Fn . Let h > 1 be the smallest integer such that ah ≡ 1 mod p. Since the congruences (2.1) hold modulo p, too, we see that h divides Fn − 1 but not (Fn − 1)/2, and hence h = Fn − 1. The Little Fermat Theorem implies ap−1 ≡ 1 mod p and hence h is a divisor of p − 1. We conclude 1 + h = Fn ≤ p ≤ Fn Hence Fn = p is a prime, as to be shown. Lemma 2.2. No matter whether Fn is prime or not, the Jacobi symbols are a Fn = = −1 a Fn for a = 3 and all n ≥ 1, as well as a = 5 and 7 for all n ≥ 2. The case a = 3, n ≥ 1. We calculate the Legendre symbol n n Fn 1 + 22 mod 3 1 + (−1)2 2 = = = ≡ 2(3−1)/2 ≡ −1 3 3 3 3 mod 3 From the quadratic reciprocity for the Jacobi symbols we get,—no matter whether Fn is prime or not: (Fn −1)(3−1) Fn Fn 3 4 = = (−1) = −1 Fn 3 3 since Fn − 1 is divisible by 4. The case a = 5, n ≥ 2. We calculate the Legendre symbol ! ! 2n−1 2n−1 Fn 1+4 mod 5 1 + (−1) 2 = = = ≡ 2(5−1)/2 ≡ −1 5 5 5 5 The case a = 7, n ≥ 2. The Legendre symbols obey the recurrence ! ! n−2 n−2 Fn 1 + 16 2 mod 7 1 + 22 mod 7 Fn−2 = = = 7 7 7 7 17 mod 5 for all n ≥ 2, which allows an induction starting with 17 3 = ≡ 3(7−1)/2 ≡ −1 mod 7 7 7 5 ≡ 5(7−1)/2 ≡ −1 mod 7 7 From the quadratic reciprocity for the Jacobi symbols we get,—no matter whether Fn is prime or not: (Fn −1)(a−1) Fn a Fn 4 = = (−1) = −1 Fn a a since Fn − 1 is divisible by 4 and a = 5, 7 are odd. Proposition 2.4 (Necessity of a Pépin-like test). Assume the Fermat number Fn is prime. Then a(Fn −1)/2 ≡ −1 mod Fn holds for a = 3 in the case n ≥ 1, and a = 5 and 7 in the case n ≥ 2. Proof. Because of the assumption that Fn is prime, we can use Euler’s criterium and get from Euler’s criterium Fn −1 Fn 2 ≡ a = −1 mod Fn a as claimed. Theorem 2.2 (Lucas-Lehmer). Any number m ≥ 2 is prime if and only if there exists a primarity witness. A witness is a natural number a with the following two properties: (i) am−1 ≡ 1 mod m; (ii) a(m−1)/p 6≡ 1 mod m for all prime divisors p of m − 1. If m = Fn is a Fermat number, the only prime divisor of m − 1 is the number 2. Thus we obtain sufficiency for the following Pépin-like test: Proposition 2.5 (A more general Pépin-like test). Assume there exists a natural number a such that aFn −1 ≡ 1 mod Fn but Then the Fermat number Fn is prime. 18 a(Fn −1)/2 6≡ 1 mod Fn Proof of the Lucas-Lehmer Theorem. The Theorem is true for m = 2 because assumption (i) holds for a = 1, and assumption (ii) is an empty truth. Assume that a is a witness for m ≥ 3 and let h > 1 be the smallest integer such that ah ≡ 1 mod m. Assumption (i) yields that h is a divisor of m − 1. By assumption (ii), we conclude h = m − 1. Moreover, the witness a is relatively prime to m. The Euler-Fermat Theorem tells that aφ(m) ≡ 1 mod m where φ(m) is the Euler totient function. Hence the order h is a divisor of φ(m). The inequalities m − 1 = h ≤ φ(m) ≤ m − 1 imply φ(m) = m − 1. Hence m is a prime number. Proth’s Theorem is a slight generalization of Pépin’s test. Theorem 2.3 (Proth’s Theorem 1878). To test whether the number N is prime, one chooses a base a relatively prime to N for which the Jacobi symbol is a = −1 N Sufficient condition for primarity: Assume additionally that N = k · 2m + 1 with k ≤ 1 + 2m . If (2.2) a(N −1)/2 ≡ −1 mod N then N is prime. Necessary condition for primarity: No restriction on k needs to be assumed. If the above congruence (2.2) does not hold, then the number N is composite. Reason for the sufficient condition: Assume that p is a prime factor of the given number N . Let h > 1 be the smallest integer such that ah ≡ 1 mod p. The assumed congruence (2.2) implies that h is a divisor of N − 1 = k · 2m but not (N − 1)/2. The Little Fermat Theorem implies that h is a divisor of p − 1. Hence we get the inequalities 2m ≤ h ≤ p − 1 The assumption k ≤ 1 + 2m implies N ≤ 22m + 2m + 1 < (1 + 2m )2 . Hence √ N < 1 + 2m ≤ p ≤ N holds for each prime divisor of N . Since any composite number has a prime divisor less or equal its square root, this is only possible if N is a prime number. 19 The necessary condition . —is a direct consequence of Euler’s Theorem about the Legendre symbol. Here are some further less important remarks about Fermat numbers. Lemma 2.3. For any Fermat number Fn 6= 3, 5 2(Fn −1)/2 ≡ 1 mod Fn Proof. By definition n 22 ≡ −1 mod Fn n 22·2 ≡ 1 mod Fn and For all n ≥ 2 we know that n + 1 ≤ −1 + 2n and hence n 2n+1 = 2 · 2n ≤ 2−1+2 = (Fn − 1)/2 n 22·2 ≡ 2(Fn −1)/2 ≡ 1 mod Fn Lemma 2.4. For any Fermat number Fn 6= 3, 5, 17 (F −1)/2 Fn−1n ≡1 mod Fn Proof. Just calculate: n−1 2 Fn−1 = (1 + 22 n−1 )2 = 1 + 2 · 22 (F −1)/2 Fn−1n n n−1 + 22 ≡ 21+2 (1+2n−1 )(Fn −1)/4 ≡ 2 mod Fn mod Fn Since n ≥ 3, we know that n ≤ n − 4 + 2n and hence n 2 · 2n = 2n+1 ≤ (1 + 2n−1 )2−2+2 = (1 + 2n−1 )(Fn − 1)/4 hence n 22·2 ≡ 1 (1+2n−1 )(F 2 n −1)/4 (F −1)/2 ≡ Fn−1n ≡1 mod Fn implies mod Fn Lemma 2.5. Any Fermat number Fn is a pseudo prime for the base 2 as well as the base Fn−1 . In 1964, Rotkiewicz showed that the product of any number of prime or composite Fermat numbers will be a Fermat pseudo prime to the base 2. 20 Lemma 2.6. Let n > m ≥ 2. No matter whether the Fermat numbers are prime or not, the Jacobi symbols are Fm Fn = =1 Fm Fn Proof. n m Fn = 1 + 22 = 1 + 22 2n−m n−m ≡ 1 + (−1)2 ≡2 mod Fm for n > m. We calculate the Jacobi symbol (Fm −1)(Fm +1) Fn 2 8 =1 = = (−1) Fm Fm since Fm − 1 is divisible by 8 for m ≥ 2. From the quadratic reciprocity for the Jacobi symbols we get,—no matter whether Fn is prime or not: (Fn −1)(Fm −1) Fn Fm Fn 4 = (−1) =1 = Fn Fm Fm again since Fm − 1 is divisible by 8 for m ≥ 2. Lemma 2.7. The number 2 is not a primitive root of any Fermat prime except 3. If Fn 6= 3, 5, 17 is a Fermat prime, none of the Fermat numbers 17 ≤ Fm < Fn with 2 ≤ m < n is a primitive roots modulo Fn . Lemma 2.8. Any Fermat prime Fn has (Fn − 1)/2 primitive roots. Equivalent are: 1. r is a primitive root of Fn . 2. r is a quadratic non-residue of Fn . 3. r(Fn −1)/2 ≡ −1 mod Fn The primitive roots of any Fermat prime Fn 6= 3 are are r ≡ 3 · 9j with j = 1, 2, . . . , (Fn − 1)/2. 21 mod Fn 2.3 Pseudo random Number Generation Very large Fermat primes are of particular interest in data encryption. (see Linear congruential generator, RANDU) Fermat primes are particularly useful in generating pseudo-random sequences of numbers in the range 1 . . . N , where N is a power of 2. The most common method used is to take any seed value between 1 and P −1, where P is a Fermat prime. Now multiply this by a number A, which is greater than the square root of P and is a primitive root modulo P (i.e., it is not a quadratic residue). Then take the result modulo P . The result is the new value for the random number generator. Vj+1 = (A × Vj ) mod P This procedure is useful in computer science since most data structures have members with 2X possible values. For example, a byte has 256 = 28 possible values 0 . . . 255. Therefore to fill a byte (or any number of bytes, indeed) with random values, a random number generator which produces values 1 to 256 can be used, subtracting 1 from each output. This method produces only pseudo random values as, after P − 1 repetitions, the sequence repeats. A poorly chosen multiplier A which is not a primitive root results in the sequence repeating sooner than P − 1. 2.4 Pseudo prime and Carmichael numbers Definition 2.2 (Pseudo prime and Carmichael number). Any number m ≥ 2 for which (2.3) am−1 ≡ 1 mod m is called a pseudo prime of base a ≥ 2. Any number m ≥ 2 such that equation (2.3) holds for all a relatively prime to m is called a Carmichael number. The Carmichael numbers are those which cannot be proved to be composite with the help of the Little Fermat Theorem—if one agrees to use only basis a relatively prime to m. Lemma 2.9. Let a be any positive or negative integer. If k divides l, then ak − 1 divides al − 1. Moreover gcd [ak − 1, al − 1] = agcd (k,l) − 1 for any k and l. Proof. To check the first part, assume that l = ks. The geometric series with quotient q := ak yields al − 1 = q s − 1 = (q − 1)(1 + q + q 2 + · · · + q s−1 ) 22 Hence al − 1 is divisible by ak − 1 = q − 1. We see from this first step that agcd (k,l) − 1 is a divisor of gcd [ak − 1, al − 1]. The converse divisibility can be checked via the Euclidean algorithm. By the extended Euclidean algorithm, there exist natural numbers s and t such that sk − tl = gcd (k, l) (possibly after switching k with l). Assume that d divides both ak − 1 and al − 1. By the first part of the Lemma, the number d divides both ask − 1 and atl − 1, and hence their difference (ask − 1) − (atl − 1) = (ask−tl − 1)atl = (agcd (k,l) − 1)atl The base a and atl are relatively prime to ak − 1 and hence to d. Hence d divides agcd (k,l) − 1. Since this reasoning applies for any common divisor of ak − 1 and al − 1, we conclude that gcd [ak − 1, al − 1] is a divisor of agcd (k,l) − 1. Proposition 2.6. A number m is a pseudo prime with base a if and only if it has the following property: (*) if the prime power ps divides m, then ps divides agcd (m−1,p−1) − 1. Lemma 2.10. Especially, if a number m is a pseudo prime with base a and the prime power ps divides m, then ps divides ap−1 − 1. Sufficiency of (*). Assume that the base a satisfies the assumption (*). Let ps be any prime power dividing m and put g := gcd (p − 1, m − 1). Since g divides m − 1, we conclude by the Lemma 2.9 that ag − 1 divides am−1 − 1. The assumption (*) tells that ps divides ag − 1 and hence am−1 − 1 by the Lemma 2.9. The simultaneous congruences am−1 ≡ 1 mod ps for all prime power divisors ps of m together imply am−1 ≡ 1 mod m. Necessity of (*). We assume that m is a pseudo prime with base a and thus am−1 ≡ 1 mod m. To check property (*), let ps be any prime power dividing m. This assumption s implies am−1 − 1 ≡ 0 mod ps , too. The Euler-Fermat Theorem yields aφ(p ) − 1 = s−1 ap (p−1) − 1 ≡ 0 mod ps . Hence agcd (m−1,p s−1 (p−1)) ≡1 mod ps Since p is a divisor of m, it is not a divisor of m − 1. Thus m − 1 and ps−1 are relatively prime and hence gcd [m − 1, ps−1 (p − 1)] = gcd (m − 1, p − 1) Hence agcd (m−1,p−1) ≡ 1 mod ps for all prime powers ps dividing m, as to be shown. Lemma 2.11. A Carmichael number cannot be divisible neither by any odd prime square nor by 4. 23 Proof. Assume that m is a Carmichael number and divisible by the prime power ps . By the second part of Proposition 2.11, we know that ps divides ap−1 − 1. We now show that it is impossible that any prime square divides m. Excluding the case p ≥ 3 is odd and s ≥ 2: In the case m that is divisible by an odd prime square, we choose a to be a primitive root modulo p2 , which we know to exist since p is odd. In that case p2 is not a divisor of ap−1 − 1, contradicting that ps divides ap−1 − 1. Excluding the special case p = 2 and s ≥ 2: We need to exclude that m is divisible by 4. We choose a = 3. Now 4 is not a divisor of 2 = ap−1 − 1, contradicting that ps divides ap−1 − 1. Proposition 2.7. Assume that the number m has the property (i) if the prime p divides m, then p − 1 divides m − 1; Assume furthermore that (a) the base a is relatively prime to m; (b) if the prime power ps divides m, then ps divides ap−1 − 1. Then m is a pseudo-prime of base a. Proof. Let ps be any prime power divisor of m. By assumption (b), ps divides ap−1 − 1. By assumption (i), p − 1 divides m − 1 and hence the Lemma 2.9 implies that ap−1 − 1 divides am−1 − 1. Together we see that ps divides am−1 − 1. The simultaneous congruences am−1 ≡ 1 mod ps for all prime power divisors ps of m together imply am−1 ≡ 1 mod m. Proposition 2.8. Assume assumption (i) from Proposition 2.7 holds for the composite number Y m= psi i with different primes pi . The number of bases a for which the m is a pseudo prime is equal to Y −1 + (pi − 1) A power of two is never a pseudo prime. Proof. For all odd prime factors p, we choose a primitive root r modulo ps . Choose any integers 0 ≤ t < p − 1, not all zero. By the Chinese Remainder Theorem, the system of simultaneous congruences s−1 a ≡ rt·p mod ps 24 for all prime factors p has a solution, unique modulo m. It is now easy to check by means of the Euler-Fermat Theorem that a satisfies the assumptions (a) and (b) of Proposition 2.7. Hence a is a base for the pseudo prime m. The procedure exhausts all possible choices of the bases a. Proposition 2.9. A number m is a Carmichael number if and only if it has the following two properties: (i) if the prime p divides m, then p − 1 divides m − 1; (ii) the number m is square free. Necessity. We assume that m is a Carmichael number and thus am−1 ≡ 1 mod m for all a relatively prime to m. To check property (i), let p ≥ 3 be any odd prime factor of m. We choose a to be a primitive root modulo p. By the Lemma 2.9 gcd [am−1 − 1, ap−1 − 1] = agcd (m−1,p−1) − 1 By assumption am−1 − 1 ≡ 0 mod p and the Little Fermat Theorem yields ap−1 − 1 ≡ 0 mod p. Hence agcd (m−1,p−1) ≡ 1 mod p Since a is a primitive root, this implies gcd (m − 1, p − 1) = p − 1. Hence p − 1 is a divisor of m − 1, confirming item (i). The property (ii) has been check above already. Sufficiency. Let a be any base relatively prime to m and p be a prime divisor of m. By the Little Fermat Theorem a prime p divides ap−1 − 1. Since we have assume that p − 1 divides m − 1, the Lemma 2.9 yields that ap−1 − 1 divides am−1 − 1. Since m is assumed to be square free, the simultaneous congruences am−1 ≡ 1 mod p for all prime divisors p of m together imply am−1 ≡ 1 mod m, as claimed. Much more has been found out about Carmichael numbers. There exist only finitely Carmichael numbers with three prime factors which can be constructed. The remaining ones have at least four prime factors. Alford et. al. (1994) have proved that the number C(n) of Carmichael numbers less than n has the asymptotics C(n) ∼ n2/7 for large n. Proposition 2.10. The Carmichael numbers with three prime factors are (6k + 1)(12k + 1)(18k + 1) were k has to be chosen such that all three factors are primes. 25 3 Arithmetic of the Complex Numbers Definition 3.1 (Gaussian integer ). A complex number with integer real- and imaginary part is called a Gaussian integer. 3.1 Arctan identities What is remarkable about the following products? (a) (2 + i) · (3 + i) = 5 + 5i (b) (2 + i)2 · (7 − i) = 25 + 25i (c) (3 + i)2 · (7 + i) = 50 + 50i (d) (5 + i)4 · (239 − i) = 114244 + 114244i 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 x1 . Too, Arg (1 + i) = arctan 1 = π4 . Taking the arguments of the formulas (a) through (d), one gets: (a) (b) (c) (d) 1 1 + arctan 2 3 1 1 2 arctan − arctan 2 7 1 1 2 arctan + arctan 3 7 1 1 4 arctan + arctan 5 239 arctan π 4 π = 4 π = 4 π = 4 = Of course, left- and right-hand side of these formulas could still differ by an integer multiple of 2π. 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 (3.1) c2 + 1 = 2N 2 with natural numbers N and c? Open Problem (May be difficult). Are there more than four solutions to the equation (3.2) M Y (c2k + 1) = 2N 2 k=1 with natural numbers N and M ≥ 2, and ck ≥ 2 for k = 1, . . . , M ? 26 3.2 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 3.1 (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 4. (c) all pairs p + iq, p − iq where p2 + q 2 = r is a prime r ≡ 1 mod 4. Indeed, for every usual prime r = 5, 13, 17, 29, . . . which is r ≡ 1 mod 4 there exists a unique decomposition into a sum of two integer square. The decomposition r = p2 + q 2 becomes unique by the additional requirement that the (negative or positive) integer p ≡ 1 mod 4 and q is even and positive. Here is a table with the smallest examples. Too, I list the squares a + ib = (p + iq)2 . r p q a b 5 1 2 −3 4 13 −3 2 5 −12 17 1 4 −15 8 29 5 2 21 20 37 1 6 −35 12 41 5 4 9 40 53 −7 2 45 −28 61 5 6 −11 60 73 −3 8 −53 −48 89 5 8 −39 80 97 9 4 65 72 101 1 10 −99 20 109 −3 10 −91 −60 113 −7 8 −15 −58 137 −11 4 105 −88 At that point, my heating system had been repaired, and I stopped. 10 Problem 3.1. Prove that there are infinitely many primes p ≡ 3 mod 4. 27 Answer. Let p1 , . . . pN be any primes congruent to 3 modulo 4. The number P := −1 + 4 N Y pν ν=1 has the following properties: (a) P is odd. (b) P ≡ 3 mod 4. (c) P is not divisible by pν for any ν = 1 . . . N . By (a) and (b), P cannot be the product of primes which are all congruent to 1 modulo 4. Hence there exists a prime p∗ congruent to 3 modulo 4 dividing P . By item (c), this prime p∗ is different from p1 , . . . pN . Hence there exist at least N + 1 different primes equivalent to 3 modulo 4. Since we know that such primes exist, there do exist infinitely many. 10 Problem 3.2. Prove that there are infinitely many primes r ≡ 1 mod 4. Answer. Let r1 , . . . rN be any primes congruent to 1 modulo 4. Define the number P := 1 + 4 N Y rν2 ν=1 Indeed P has in the Gaussian integers the factoring # # " " N N Y Y rν P = 1 + 2i rν · 1 − 2i ν=1 ν=1 Neither of the two factor is divisible by 1 + i nor any prime p ≡ −1 mod 4. Hence we get the following properties: (a) P is odd. (b) P ≡ 1 mod 4. (c) P is not divisible by rν for any ν = 1 . . . N . (d) P is not divisible by any prime p ≡ 3 mod 4. By (a) and (d), P is the product of primes congruent to 1 modulo 4 only. By item (c), all these factors are different from r1 , . . . rN . Hence there exist at least N + 1 different primes equivalent to 1 modulo 4. Since we know that such primes exist, there do exist infinitely many. 28 10 Problem 3.3 (The little forgotten theorem). Assume that √ iπk (3.3) z = a + ib = a2 + b2 e 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 6= 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 4, 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 ≤ p2 + q 2 ≡ 1 mod 4. In the 2l-th power of the given equation (3.3), (a + ib)2l = (a2 + b2 )l (3.4) the two sides sides are both Gaussian integers as well as real, hence real positive integers. Equation (3.4) implies that p + iq divides the natural number (a2 + b2 )l . Hence p − iq divides this same number, too. Since the Gaussian prime p − iq divides (a + ib)2l , unique factorization implies that p − iq divides a + ib, too. Since 5 ≤ p2 + q 2 ≡ 1 mod 4. the two numbers p + iq and p − iq are relatively prime. Hence we conclude there product (p + iq)(p − iq) = p2 + q 2 divides a + ib, too. But we have already shown that no real prime divides a + ib. Hence the only possibilities left are a + ib = iu (1 + i)s with u = 0, 1, 2, 3 and s = 0, 1. Theorem 3.2 (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 (3.5) z = a + ib = √ a2 + b 2 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 a0 := ad and b0 := db . By the last problem, a0 + ib0 = iu (1 + i)s with u = 0, 1, 2, 3 and s = 0, 1. Multiplying both sides by d yields a + ib = iu (1 + i)s d with u = 0, 1, 2, 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. 29 Corollary. Let a + ib be a Gaussian integer with a 6= 0, b 6= 0. Both real- and imaginary parts of (a + ib)l are nonzero if either l odd or |a| = 6 |b|. If l > 0 is even, the following happens: l Re (a + ib)l Im (a + ib)l l ≡ 2 mod 4 Re (a + ib)l = 0 ⇔ |a| = |b| Im (a + ib)l 6= 0 l l ≡ 0 mod 4 Re (a + ib) 6= 0 Im (a + ib)l = 0 ⇔ |a| = |b| 10 Problem 3.4. Give the reason for Corollary 3.2. Proof of Corollary 3.2. For any nonzero Gaussian integer a + ib, and integer l > 0, the equation p (3.6) (a + ib)l = iu (a2 + b2 )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 (a + ib)4l = (a2 + b2 )2l (3.7) follows. Hence, as explained in the solution of Problem 3.3, 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 3.6 implies al (1 ± i)l = iu 2l/2 |a|l . Hence l = 2m is even and (±i)m = iu . Hence either l ≡ 2 mod 4, and m, u both odd, and Re (a + ib)l = 0— or l ≡ 0 mod 4, and m, u both even, and Im (a + ib)l = 0. 10 Problem 3.5. (a + ib)l+2 a2 − b 2 (a + ib)l+4 Sl (a, b) := Im 2ab(a2 − b2 ) Rl (a, b) := Re Calculate R0 , S0 , R4 , S4 . Prove that for all l ≡ 0 mod 4, Rl and Sl are integer homogenous and symmetric polynomials in a2 and b2 . Answer. (a + ib)2 a2 − b 2 = =1 a2 − b 2 a2 − b 2 (a + ib)6 a6 − 15a4 b2 + 15a2 b4 − b6 R4 (a, b) = Re 2 = = a4 − 14a2 b2 + b4 2 2 2 a −b a −b 4a3 b − 4ab3 (a + ib)4 S0 (a, b) = Im = =2 2ab(a2 − b2 ) 2ab(a2 − b2 ) (a + ib)8 8a7 b − 56a5 b3 + 56a3 b5 − 8ab7 S4 (a, b) = Im = = 4a4 − 24a2 b2 + 4b4 2 2 2 2 2ab(a − b ) 2ab(a − b ) R0 (a, b) = Re 30 After some pain with the binomial formulas, one gets the general case with l = 4m: m X +m X s 4m + 2 a2m+2t b2m−2t R4m (a, b) = (−1) 2s s=0 t=−m m X +m X (−1)s 4m + 4 2m+2t 2m−2t S4m (a, b) = a b 2 2s + 1 s=0 t=−m Since the binomial coefficients even are even, these are integer polynomials—the other odd assertions are easy to check, too. 10 Problem 3.6 (Calculation to the bones). Calculate R4m (1, 1) and S4m (1, 1), confirming they are nonzero. Answer. 2 2 Rl + (a + b )2abSl−4 Re (a + ib)l+2 + (a2 + b2 )Im (a + ib)l = a2 − b 2 l+2 Re (a + ib) − Re i(a − ib)(a + ib)l+1 = a2 − b 2 [a + ib − i(a − ib)](a + ib)l+1 = Re a2 − b 2 (1 − i)(a + ib)l+1 = Re a+b With a = b = 1 and l = 4m ≥ 4, we get R4m (1, 1) + 4S4m−4 (1, 1) = Re (1 − i)(1 + i)4m+1 /2 = (−4)m With a = 1 + ε = z and b = −1, we get in the limit z → 1— using l’Hôpital’s rule (1 + ε − i)4m+1 ε→0 ε d (z − i)4m+1 = Re (1 − i) dz = Re (4m + 1)(1 − i)4m+1 = (4m + 1)(−4)m R4m (1, −1) − 4S4m−4 (1, −1) = lim Re (1 − i) Both formulas together imply S4m−4 = 2m(−4)m−1 . Actually, we check that for all m≥0 R4m = (2m + 1)(−4)m , S4m = (2m + 2)(−4)m 3.3 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 = a2 +b2 has for any given natural number m. 31 10 Problem 3.7. We begin with some examples. For this illustration, I count only the solutions of m = a2 + b2 with integers 0 ≤ a ≤ b, but give these solutions completely. Give a list for m = 1 through 30. m list of a2 + b2 16 02 + 42 17 12 + 42 18 32 + 32 19 no solution 20 22 + 42 21 no solution 22 no solution 23 no solution 24 no solution 2 25 0 + 52 = 32 + 42 26 12 + 52 27 no solution 28 no solution 29 22 + 52 30 no solution m list of a2 + b2 1 02 + 12 2 12 + 12 3 no solution 4 02 + 22 5 12 + 22 6 no solution 7 no solution Answer. 8 22 + 22 9 02 + 32 10 12 + 32 11 no solution 12 no solution 13 22 + 32 14 02 + 22 15 12 + 22 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 + 2i|2 = 12 + 22 = 5 (b) |3 + 2i|2 = 32 + 22 = 13 (c) The square of the absolute value of (1 + 2i) · (−3 + 2i) = −7 − 4i, and (1 + 2i) · (−3 − 2i) = 1 − 8i implies 5 · 13 = 72 + 42 = 12 + 82 = 65. (d) Squaring part (c) yields (1 + 2i)2 · (−3 + 2i)2 = (−3 + 4i)(5 − 12i) = 33 + 56i (1 + 2i)2 · (−3 − 2i)2 = (−3 + 4i)(5 + 12i) = −63 − 16i The square of the absolute value yields 52 · 132 = 332 + 562 = 632 + 162 = 652 Are there more ways to solve a2 + b2 = 652 ? Indeed, there are still two further solutions obtained as absolute squares of 13(−3 + 4i) = −39 + 52i and 5(5 − 12i) = 25 − 60i. Last not least, 652 is a perfect square. 392 + 522 = 252 + 602 = 02 + 652 = 652 32 (e) The absolute square of (1 + 2i)(1 + i) = −1 + 3i yields 10 = 12 + 32 . 10 Problem 3.8. 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 md ∈ ( m, m]√is different one. In that way, all divisors of m appear in pairs. The only exception is m in the case that m is a perfect square. Proposition 3.1. The integer equation m = a2 + b2 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 4 appear with even multiplicity. In other words, m can be factored as m = 2s g 2 c where g has only prime factors congruent 3 mod 4, and (3.8) c= N Y rνsν ν=1 has only prime factors congruent 1 mod 4. Theorem 3.3. The number of different ways one can decompose m = 2s g 2 c into squares of integers a, b such that m = a2 + b2 is 4τ (c). Here τ (c) = N Y (1 + sν ) ν=1 is the number of divisors of c. The numbers a + ib from the solutions are exactly the Gaussian integers (3.9) u s a + ib := i (1 + i) g N Y (pν + iqν )σν (pν − iqν )τν ν=1 All possible choices are obtained with u = 0, 1, 2, 3 and σν , τν ≥ 0 with σν + τν = sν for ν = 1 . . . N. For example m = 652 is decomposed as 652 = 52 · 132 and has 9 divisors. The possible 33 values of a + ib as solution of (3.9)—with unit factor iu = 1—are (1 + 2i)2 (−3 + 2i)2 = (−3 + 4i)(5 − 12i) = 33 + 56i , 2 (1 + 2i) (−3 + 2i)(−3 − 2i) = (−3 + 4i) · 13 = −39 + 52i , 2 2 (1 + 2i) (−3 − 2i) = (−3 + 4i)(5 + 12i) = −63 − 16i , 2 (1 + 2i)(1 − 2i)(−3 + 2i) = 5 · (5 − 12i) = 25 − 60i , (1 + 2i)(1 − 2i)(−3 + 2i)(−3 − 2i) = 5 · 13 = 65 , 2 (1 + 2i)(1 − 2i)(−3 − 2i) = 5 · (5 + 12i) = 25 + 60i , 2 2 (1 − 2i) (−3 + 2i) = (−3 − 4i)(5 − 12i) = −63 + 16i , 2 (1 − 2i) (−3 + 2i)(−3 − 2i) = (−3 − 4i) · 13 = −39 − 52i , 2 2 (1 − 2i) (−3 − 2i) = (−3 − 4i)(5 + 12i) = 33 − 56i , As one sees there are still pairs of conjugate complex solutions, and one real solution since 652 is a perfect square. Remark. If m is not a perfect square, there are τ (c) essentially different solutions of 2 m = a2 + b2 with 1 ≤ a ≤ b. If m is a perfect square, there are τ (c)−1 essentially different solutions of m = a2 + b2 2 with 1 ≤ a ≤ b. 10 Problem 3.9. Prove that a Gaussian integer a + ib with prime decomposition (3.9) satisfies gcd (a, b) = 1 if and only if s = 0 or s = 1; g = 1; either σν = 0 or τν = 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 2; a + ib is not divisible by any prime factor congruent to 3 mod 4; a + ib is not divisible by any real prime congruent to 5 mod 4. Proof of Proposition 3.1 and Theorem 3.3. For any given natural number m, let d be the divisor of m assembling all prime factors congruent 3 mod 4. Let c be the divisor of m assembling all prime factors congruent 1 mod 4. The factor c is completely decomposed into its prime factors as in formula (3.8). In the ring of Gaussian integers, the usual prime factoring of m will split further into irreducible factors (3.10) s 3s 2s m = 2 dc = i (1 + i) d N Y (pν + iqν )sν (pν − iqν )sν ν=1 34 where the real factoring of d is not needed here. Assume that m = a2 +b2 . Because of (a+ib)(a−ib) = a2 +b2 = m, we conclude that a + ib is a divisor of m. The uniqueness of prime factorization implies equation (3.9) holds for some values u ∈ {0, 1, 2, 3} and 0 ≤ σν ≤ sν , 0 ≤ τν ≤ sν for ν = 1 . . . N (Why?). We now multiply equation (3.9) with the conjugate complex equation and get 2 2 s 2 a +b =2 g N Y (pν + iqν )σν +τν · (pν − iqν )σν +τν ν=1 We compare all factors with those in equation (3.10) and use the uniqueness of prime factorization. Hence g 2 = d and σν + τν = sν for ν = 1 . . . N , as claimed. Too, we see that no equation m = a2 + b2 can hold unless the factor d is a perfect square. Too, the uniqueness of the prime factoring implies that all solutions given by equation (3.9) are different. It is straightforward to check that equation (3.9) yields solutions of the equation m = a2 + b2 . Hence we get exactly 4τ (c) = 4 N Y (1 + sν ) ν=1 solutions. 10 Problem 3.10. Assemble different examples to illustrate Proposition 3.1 and Theorem 3.3. Examples. For different values of m, here is a table listing the factors 2s , 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 (3.9). For abbreviation, I put the unit factor iu = 1, and ignore the obvious conjugate complex ones. m 2s d c factor c τ (c) list of a + ib 5 1 1 5 5 2 −1 + 2i 8 8 1 1 1 1 −2 + 2i 9 1 9 1 1 1 −3 10 2 1 5 5 2 (1 + i)(1 + 2i) = −1 + 3i 26 2 1 13 13 2 (1 + i)(−3 + 2i) = −5 − i 20 4 1 5 5 2 −4 + 2i 36 4 9 1 1 1 6i 39 1 3 13 13 2 no solution 52 4 1 13 13 2 −4 − 6i 65 1 1 65 5 · 13 4 (1 + 2i)(−3 ± 2i) = −7 − 4i , 1 − 8i 130 2 1 65 5 · 13 4 (1 + i)(1 + 2i)(−3 ± 2i) = −3 − 11i , 9 − 7i 260 4 1 65 5 · 13 4 2i(1 + 2i)(−3 ± 2i) = 8 − 14i , 16 + 2i 35 3.4 Roots Definition 3.2 (argument). The argument of a complex number z 6= 0 is the set of angles which in polar coordinates yield z: arg z = {θ ∈ R : z = |z|eiθ } The principle value of the argument Arg z of a complex number z 6= 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 2π. 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 6= 0, but the principle argument does not depend continuously on z. Lemma 3.1 (Principle value). The principle argument of x + iy 6= 0 is given by the formula arctan xy if x > 0; π if x = 0 and y 2 (3.11) Arg z = − π2 if x = 0 and y y π + arctan x if x < 0 and y −π + arctan y if x < 0 and y x a complex number z = > 0; < 0; > 0; <0 Remark. Several computer languages provide Arg (x + iy) as a two variable function arctan(x, y). 10 Problem 3.11. Give simple examples to illustrate all cases in Lemma 3.1. Proof of Lemma 3.1 . The principle argument Arg z of any complex number z 6= 0 is defined by restricting the argument of z = |z| eiArg 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 xy = tan Arg z unless x = 0. But since both cos(θ ± π) = − cos θ and sin(θ ± π) = − sin θ, the tangent function has period π. Hence the value of tan Arg z determines the argument only modulo π—not yet modulo 2π. The given cases result from comparing the signs of sin and cos with those of x and y. 36 10 Problem 3.12. 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 0 (x) = arctan dy dx Why does one not distinguish the cases occurring in Lemma 3.1? Answer. The angle between the tangent and the x-axis is a directed angle between two lines, and hence always lies in the interval (− π2 , π2 ). 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 3.3 (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 3.13. What is the exact interval to which the principle value of the argument for any z 6= 0 in the right semi-plane is restricted. Answer. For any z 6= 0 in the right semi-plane, the principle value of the argument lies in the half open interval Arg z ∈ (− π2 , π2 ]. Conversely, Arg z ∈ (− π2 , π2 ] implies z ∈ RS. Definition 3.4 (Principle value). The principle argument Arg z of any complex number z 6= 0 is defined by restricting the argument of z = |z| eiArg z to the interval Arg z ∈ (−π, π]. The principle n-th root is define as p √ iArg z P.V. n z := n |z| e n The principle logarithm of any complex number z 6= 0 is defined as Ln z := ln |z| + iArg z 10 Problem 3.14. Prove that for all a, b ∈ RS and all n ≥ 2 Arg ab = Arg a + Arg b for a, b ∈ RS and a, b 6= 0 √ √ √ n n P.V. ab = P.V. n a · P.V. b 37 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 + 2πk holds with k = 0, since both sides do not differ by an integer multiple of 2π. Now the second assertion easily follows p p p p √ Arg ab Arg a+Arg b Arg a Arg b n n P.V. ab = n |ab| ei n = n |ab| ei = n |a| ei n · n |b| ei n √ √ n = P.V. n a · P.V. b Here are some examples of principle third roots. √ √ 3+i 3 (a) P.V. i = √2 √ 3−i (b) P.V. 3 −i = 2 s −1 + i 1+i (c) P.V. 3 √ = √ 2 2 √ √ 1+i 3 3 (d) P.V. −1 = 2 Note that the principal third root of any negative real, is not the negative real value. 10 Problem 3.15. 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 rh i q q (c) P.V. (d) P.V. 3 −1+i √ 2 q 3 1+i √ 2 · P.V. · P.V. 3 q 3 ? −1+i √ = 2 ? 1+i √ = 2 P.V. P.V. 3 rh 3 −1+i √ 2 1+i √ 2 i2 38 2 Answer. "√ #2 √ √ √ 3+i 1+i 3 1+i 3 3 P.V. i · P.V. i = = P.V. −1 = = 2 2 2 #2 "√ √ √ √ √ √ 3−i 1−i 3 1+i 3 3 3 3 P.V. −i · P.V. −i = 6= P.V. −1 = = 2 2 2 s s 2 −1 + i −1 + i 1+i 3 3 √ √ P.V. · P.V. = √ = i 6= 2 2 2 s √ 2 √ −1 + i 3−i 3 3 √ = P.V. −i = P.V. 2 2 √ 3 (a) (b) (c) √ 3 (d) s P.V. 3 3.5 1+i √ · P.V. 2 s 3 p i2 h 2πi i2 3 πi 1+i h 2πi √ = P.V. e 8 = = e 24 = e 6 2 s 2 √ πi 1+i 3 3 √ P.V. = P.V. i = e 6 2 The solution of the reduced cubic equation Given is the equation (3.12) x3 + b x + c = 0 Since the quadratic term is lacking, it is called a reduced or depressed cubic. We want b to find all three solutions, including the complex ones. Putting x := u − 3u leads to the equation 3 b 3 (3.13) u − +c=0 3u Hence z = u3 satisfies the quadratic resolvent equation (3.14) z 2 + cz − b3 =0 27 The two solutions of equation (3.14) are r c c2 b3 (3.15) z1,2 = − ± + 2 4 27 We get a resolvent equation with real or complex solutions, depending on whether the discriminant c2 b3 (3.16) D= + 4 27 39 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 (3.14) 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 z1 z2 = − (3.17) b3 27 √ b 3 − √ z2 = 3 3 z1 Hence the real solution of the cubic (3.12) is (3.18) √ √ b x1 = u − = 3 z1 + 3 z2 s s 3u r r 3 2 3 3 b b3 c c c c2 + + − − + = − + 2 4 27 2 4 27 which is just Cardano’s formula. The two complex solutions are √ √ x2 = 3 z1 ω + 3 z2 ω 2 (3.19) √ √ x3 = 3 z1 ω 2 + 3 z2 ω where (3.20) √ −1 + i 3 2π 2π ω= = cos + i sin 2 3 3 is the primitive third root of unity. Simplifying (3.18) and (3.19) yields √ √ √ √ 3 z + 3 z 3 z − 3 z √ 1 2 1 2 (3.21) x2,3 = − ± 3i 2 2 Case 2: D < 0 The cubic has three real solutions. In this case, the resolvent equation (3.14) has complex solutions. As already Rafael Bombelli discovered (1572), this leads to a third root of a complex number in Cardano’s formula (3.18). 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 (3.17) and 40 equation (3.15) imply −b3 27 r c −b3 c2 =− ±i − 2 27 4 r −b3 (cos θ ± i sin θ) = 27 |z1 |2 = z1 z2 = (3.22) z1,2 The argument θ can be calculated more simple from the real parts. One gets cos θ, and finally θ: r −b3 c − = cos θ 2 27 (3.23) −c θ = arc cos q 3 2 −b 27 Hence equations (3.22) and (3.20) imply r √ −b θ θ 3 z1 = cos + i sin 3 3 3 r (3.24) √ −b θ + 2π θ + 2π 3 + i sin z1 ω = cos 3 3 3 and equations (3.17) and (3.18) imply that r −b θ x1 = 2 cos 3 r 3 −b θ + 2π (3.25) x2 = 2 cos 3 r 3 −b θ − 2π x3 = 2 cos 3 3 are the three real solutions of the original cubic. 10 Problem 3.16. Calculate the roots in the limiting case D & 0+. Answer. r c x1 x1 = 2 3 − , x 2 = x3 = − 2 2 10 Problem 3.17. 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. 41 Answer. In the approach D % 0−, one gets r r c2 b3 3 |c| −b =− , = 4 27 r 2 3 3 |c| −b = 2 r 27 −b3 c cos θ − = 2 27 Distinguish the cases c < 0 and c > 0. Case c < 0. One gets cos θ = 1, and hence θ = 0. The result (3.25) implies r c x1 x1 = 2 3 − , x 2 = x3 = − 2 2 Case c > 0. One gets cos θ = −1, and hence θ = π. The result (3.25) implies r x1 c x2 = −2 3 − , x1 = x3 = − 2 2 10 Problem 3.18. Use Cardano’s formula and Bombelli’s method to get one solutions of x3 − 30x − 36 = 0. Answer. Cardano’s formula for a solution of x3 + bx + c = 0 yields in the present case s s r r 2 3 3 3 c b c2 b3 c c (3.26) x1 = − + + + − − + 2 4 27 2 4 27 √ √ 3 3 = 18 + 26i + 18 − 26i √ To calculate the third root, let 3 18 + 26i = u+iv and 18+26i = (u+iv)3 . We compare real- and imaginary parts and get the system 18 = u3 − 3uv 2 = u(u2 − 3v 2 ) 26 = 3u2 v − v 3 = (3u2 − v 2 )v With the help of the factoring, one finds the integer solution u + iv = 3 + i. Hence Cardano’s formula gives x1 = 6. 10 Problem 3.19. Find the two other solutions of the equation x3 −30x−36 = 0. Answer. The simplest way is division of the polynomial. One gets√(x3 − 30x − 36)/(x − 6) = x2 + 6x + 6. Hence the two other solutions are x2,3 = −3 ± 3. 42 Remark. Too, we can find the two other roots by introducing the third root of unity √ −1+i 3 into Cardano’s formula. This has to be done in a consistent way and ω = 2 produces the solutions √ √ √ √ √ −1 + i 3 −1 − i 3 3 2 3 x2 = ω 18 + 26i + ω 18 − 26i = (3 + i) + (3 − i) = −3 − 3 2 √ 2 √ √ √ √ −1 − i 3 −1 + i 3 (3 + i) + (3 − i) = −3 + 3 x3 = ω 2 3 18 + 26i + ω 3 18 − 26i = 2 2 which is the same answer as above. 3.6 The square root of a complex number √ Proposition 3.2. A branch with Re z ≥ 0 for the square root of any complex number z = x + iy is sp sp 2 2 p x +y +x x2 + y 2 − x (3.27) x + iy = + i sign (y) 2 2 √ The square root w = z of any complex number z = x + iy has two branches. The second branch is w2 = −w. √ Reason. Name the square root in question x + iy =: u + iv. Squaring yields x + iy = (u + iv)2 . Separate the real- and imaginary part to get x = u2 − v 2 , y = 2uv The absolute value squared is x2 + y 2 = |x + iy|2 = |u + iv|4 = (u2 + v 2 )2 Add and substrate p u2 + v 2 = x2 + y 2 u2 − v 2 = x p x2 + y 2 + x u2 = 2 p 2 x + y2 − x v2 = 2 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 = 2uv 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. 43 10 Problem 3.20. Calculate the square roots: √ √ √ (a) −3 + 4i , −12 + 16i , −48 + 64i √ √ √ (b) −8 − 6i , −32 − 24i , −128 − 96i √ √ √ (c) −4 − 3i , −16 − 12i , −64 − 48 √ √ √ (d) −6 + 8i , −24 + 32i , −96 + 128i √ √ √ (e) 2i , 8i , 32i 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? 3.7 Pythagorean triples Any three integers such that a2 + b2 = c2 are called a Pythagorean triple. 10 Problem 3.21. 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)2 . Its absolute value is c := |a+ib| = |p+iq|2 . Clearly a2 + b2 = c2 holds. 10 Problem 3.22. Show that for any Gaussian integer a+ib 6= 0, it is impossible that a + ib and b + ia are both perfect squares. Theorem 3.4 (Pythagorean triples). A nonzero Gaussian integer a + ib = (p + iq)2 is a perfect square if and only if the following three conditions hold (i) the sum a2 + b2 is a perfect square c2 ; (ii) gcd (a, b) =: g is either (a) a perfect square: g = d2 , or (b) twice a perfect square: g = 2d2 ; (iii) assuming 2t , but not 2t+1 divides d, we get two cases: (iii a) if g = d2 , then 4−t a is odd and 4−t b is even— (iii b) if g = 2d2 , then 2−2t−1 a is even and 2−2t−1 b is odd. Corollary. Exactly one of the Gaussian integers a+ib = (p+iq)2 and b+ia = i(p−iq)2 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. 44 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 3.4 is only needed to determine which one of the two numbers—either a + ib or b + ia—is a perfect square. In Problem 3.22, we have already checked that these two numbers cannot be both perfect squares, unless they are both zero. Example 3.1. These are perfect squares, corresponding to t = 0, 1, 2 in (iii a) or (iii b): √ √ √ (a) −3 + 4i = 1 + 2i , −12 + 16i = 2 + 4i , −48 + 64i = 4 + 8i , . . . √ √ √ (b) −8 − 6i = −1 + 3i , −32 − 24i = −2 + 6i , −128 − 96i = −4 + 12i , . . . but these are not perfect squares: √ √ √ √ √ (−1 + 3i) 2 √ −4 − 3i = , −16 − 12i = (−1 + 3i) 2 , −64 − 48 = (−2 + 6i) 2 , . . . 2 √ √ √ √ √ √ −6 + 8i = (1 + 2i) 2 , −24 + 32i = (2 + 4i) 2 , −96 + 128i = (4 + 8i) 2 , . . . Here is the most simple example: √ √ √ (a) 4 = 2 , 16 = 4 , 64 = 8 , . . . √ √ √ (b) 2i = 1 + i , 8i = 2 + 2i , 32i = 4 + 4i , . . . Proof of necessity in theorem 3.4 . Necessity is easier to prove. We assume that a+ib = (p + iq)2 is a perfect square. Squaring the absolute values gives c2 = a2 + b2 = |a + ib|2 = |p + iq|4 = (p2 + q 2 )2 Separating real- and imaginary parts yields the Pythagorean triple a = p2 − q 2 , b = 2pq , c = p2 + q 2 Hence condition (i) holds. To check (ii), let d := gcd (p, q). Clearly d2 divides both a and b, and hence d2 divides g := gcd (a, b). Let p0 := q a b c p , q 0 := , a0 := 2 , b0 := 2 , c0 := 2 , d d d d d By these divisions we get a0 = p02 − q 02 , b0 = 2p0 q 0 , c0 = p02 + q 02 , gcd (p0 , q 0 ) = 1 We show that gcd (a0 , b0 ) =: h can only be 1 or 2. Indeed, any odd prime dividing b0 needs to divides either p0 or q 0 , but not both. Hence it cannot divide a0 = p02 − q 02 . 45 Neither can 4 divide both a0 and b0 . Indeed, assuming 4 divides b0 , implies one number among p0 and q 0 is even, the other one is odd. Hence a0 is odd. Hence we see that h = 1 or h = 2 are the only possibilities left. Therefore gcd (a, b) = 2 d gcd (a0 , b0 ) = hd2 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 2t , but not 2t+1 divides d. These two cases can occur: (iii a): g = d2 and h = 1 = gcd (a0 , b0 ) = gcd ((p0 − q 0 )(p0 + q 0 ), 2p0 q 0 ). Hence p0 ± q 0 are both odd, and hence one number among p0 and q 0 is odd, but the other one is even. Hence a0 is odd, but 4 divides b0 . Similarly, 4−t a is odd and 4−t b is even, since these are odd multiples of a0 and b0 . (iii b): g = 2d2 and h = gcd (a0 , b0 ) = gcd ((p0 − q 0 )(p0 + q 0 ), 2p0 q 0 ) = 2. Hence p0 ± q 0 are both even, and—different from case (a)—p0 and q 0 are both odd. Hence a0 is divisible by 4, but b0 ≡ 2 mod 4. Hence, 2−2t−1 a is even and 2−2t−1 b is odd, since these are odd multiples of a0 and b0 . Proof of sufficiency theorem 3.4 . We assume that the integers a and b satisfy the four conditions (i) through (iii). Let g = gcd (a, b) = hd2 with h = 1 or h = 2 and as above a0 := a b c , b0 := 2 , c0 := 2 2 d d d In cases (a) and (b) one gets (iii a) if g = d2 , then gcd (a0 , b0 ) = 1, and a0 is odd, b0 is even. Hence c0 is odd, and c0 ±a0 are both integers. 2 (iii b) if g = 2d2 , then gcd (a0 , b0 ) = 2, a0 2 is even and b0 2 is odd. Hence c0 2 is odd. As explained in Proposition 3.2, one obtains a square root of a complex number by the formula r r 0 + a0 √ c c 0 − a0 (3.27) a0 + ib0 = + i sign (b) 2 2 0 0 0 In both cases (iii a) and (iii b), c ±a and b2 are integers. Furthermore a2 + b2 = c2 2 implies 0 2 b c 0 + a0 c 0 − a0 = · 2 2 2 The two factors are relatively prime: 0 c + a0 c 0 − a0 gcd , =1 2 2 46 0 0 is a divisor of h = gcd (a0 , b0 ), which is Indeed, the greatest common divisor of c ±a 2 0 0 either 1 or 2. But is the case (b) where h = 2, both numbers c ±a turn out to be odd. 2 It is a (too easy) exercise to show that uniqueness of the prime factorization implies Lemma 3.2. If the product of two relatively prime factors is a perfect square, then both factors are perfect squares. Hence we conclude that both roots r c 0 + a0 0 (E:doit) p = 2 r 0 and q = sign (b) c 0 − a0 2 √ are integers. In other word, we have√shown that a0 + ib0 = p0 + iq 0 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 3.23. Generalize the Lemma 3.2 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) ≤ 2, indeed only the following possibilities occur for perfect squares a + ib = (p + iq)2 : p odd even odd 3.8 q even odd odd a mod 4 1 3 0 b mod 4 0 0 2 c mod 4 1 1 2 h 1 1 2 Roots of unity 10 roots: Problem 3.24. Find all different powers—exactly, expressed using square √ −1 + i 3 (a) ω = . 2 p √ √ 5 − 1 + i 10 + 2 5 (b) z = . 4 1+i (c) w = √ . 2 47 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: √ √ −1 − i 3 −1 + i 3 2 (a) , ω = = ω , ω3 = 1 ω= 2 2 p p √ √ √ √ 5 − 1 + i 10 + 2 5 − 5 − 1 + i 10 − 2 5 2 (b) z= , z = , 4 4 z3 = z2 , z4 = z −1 + i 1+i (c) , w4 = −1 , w = √ , w2 = i , w3 = √ 2 2 5 6 3 w = w , w = −i , w7 = 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 3.24, we get a product of argument 15◦ . 1 3 ω −1 w3 = e2πi[− 3 + 8 ] = e 24 = cos 15◦ + i sin 15◦ 2πi (b) We can calculate this product in terms of the exact square roots, obtained in Problem 3.24. √ √ √ 1 + 3 + i( 3 − 1) −1 − i 3 −1 + i −1 3 √ · √ = ω w = 2 2 2 2 (c) Finally, we compare the real- and imaginary parts obtained form (a) and (b), and use the common denominator 4. √ √ 2+ 6 ◦ cos 15 = √ 4√ 6− 2 sin 15◦ = 4 (d) In this case, we get the table immediately. 48 k 15◦ 15◦ 30◦ 45◦ 60◦ 75◦ 90◦ ◦ cos√k 15 √ ◦ sin√k 15 √ 1 0 2+ 6 4√ 3 √2 2 2 1 √ √2 6− 2 4 6− 2 4 1 √2 2 √2 3 √ √2 6+ 2 4 10 Problem 3.25. Calculate cos 15◦ + i sin 15◦ directly by formulas r r θ 1 + cos θ θ 1 − cos θ cos = , sin = 2 2 2 2 and confirm that you get the same values. Answer. ◦ 30 = cos 2 r 1+ cos 30◦ 2 s = p √ √ 2+ 3 2+ 3 = 4 2 √ ◦ Similarly, one gets sin 15 = √ 2− 3 . 2 By squaring both sides, one checks that indeed p p √ √ √ √ √ √ 2+ 3 6+ 2 2− 3 6− 2 = , = 2 4 2 4 10 Problem 3.26. The following steps can be elaborated to obtain an exact trigonometric table of cos 9◦ k and sin 9◦ k for all k = 1, . . . , 40. (a) Find natural numbers p, q such that − p5 + q 8 = 1 . 40 (b) With z, w from Problem 3.24, calculate the argument θ in z −p wq = e2πiθ (c) Hint: calculate √ p √ 5 + 1 − i 10 − 2 5 1 + i · √ 4 2 (d) Calculate z −p wq with the exact square roots, obtained in Problem 3.24. (e) Use Euler’s 2πi e 40 = cos (3.28) 2π 2π + i sin 40 40 and get exact expressions for cos 9◦ and sin 9◦ . Use the common denominator 8 in your results. Answer. (a) The natural numbers p, q have to satisfy −8p + 5q = 1. The smallest solution is p = 3, q = 5. (b) With z, w from Problem 3.24, 3 5 z −p wq = e2πi[− 5 + 8 ] = e 40 49 2πi (d) With the same z, w from Problem 3.24, using z 5 = 1 and w2 = i, we can simplify and get p √ √ 5 + 1 − i 10 − 2 5 1+i · √ z −3 w5 = −z 2 w = 4 2 p p √ √ √ √ √ √ 10 + 2 + 2 5 − 5 10 + 2 − 2 5 − 5 +i = 8 8 (e) Comparing real- and imaginary part from parts (b) and (c) yields p √ √ √ 10 + 2 + 2 5− 5 ◦ cos 9 = (3.29) 8 p √ √ √ 10 + 2 − 2 5 − 5 (3.30) sin 9◦ = 8 10 Problem 3.27. Give the exact expression for the perimeter of a regular 20gon, inscribed into the unit circle. Compare with 2π. How large is the deviation? Answer. The circumference of a regular 20-gon is q √ √ √ ◦ 40 sin 9 = 5 10 + 2 − 2 5 − 5 ≈ 6.257378602 Bur 2π ≈ 6.283185307 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 12p − 5q = ±1. The smallest solution is p = 3, q = 7. Dividing by 60 yields 7 1 3 − = 5 12 60 (b) With the twelfth and fifth roots of unity τ and z we get 3 7 z 3 τ −7 = e2πi[ 5 − 12 ] = e 60 = cos 6◦ + i sin 6◦ 2πi (d) We can calculate this product with the exact values of the square roots, simplify and get p √ √ √ 5 + 1 + i 10 − 2 5 3−i z 3 τ −7 = −z 3 τ −1 = · 4 2 p √ √ √ √ √ p √ 15 + 3 + 10 − 2 5 − 5 − 1 + 3 10 − 2 5 = +i 8 8 50 (e) Comparing real- and imaginary part from parts (b) and (c) yields p √ √ √ 15 + 3 + 10 − 2 5 ◦ (3.31) cos 6 = 8 √ √ p √ − 5 − 1 + 3 10 − 2 5 sin 6◦ = (3.32) 8 10 Problem 3.28. Give the exact expression for the perimeter of a regular 30gon, inscribed into the unit circle. Compare with 2π. How large is the deviation? Answer. The circumference of a regular 30-gon is q √ √ √ 15 ◦ − 5 − 1 + 3 10 − 2 5 ≈ 6.271707796 60 sin 6 = 2 But 2π ≈ 6.283185307—so just one decimal point coincides. 10 Problem 3.29. Prove that √ p √ √ 5 + 1 + 3 10 − 2 5 ◦ (3.33) cos 24 = 8p √ √ √ 15 + 3 − 10 − 2 5 (3.34) sin 24◦ = 8 Here is a table assembling the information we have gathered so far. θ 6◦ 9◦ 15◦ 18◦ 24◦ 30◦ 36◦ 45◦ cos θ √ √ √ 15+ 3+ 10−2 5 √ √ 8√ √ 10+ 2+2 5− 5 √ √ 8 2+ 6 q4 √ 5+ 5 8 √ √ √ √ 5+1+ 3 10−2 5 8 √ 3 2 √ 5+1 √4 2 2 √ sin θ √ √ √ √ − 5−1+ 3 10−2 5 √ √8 √ √ 10+ 2−2 5− 5 8 √ √ 6− 2 4 √ 5−1 4 √ √ √ √ 15+ 3− 10−2 5 8 1 q √2 5− 5 8√ 2 2 10 Problem 3.30. 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 1 1 1 − − . With ω, z and w from Problem 3.24, get the argument θ in 3 5 8 ωz −1 w−1 = e2πiθ = cos 2πθ + i sin 2πθ 51 (b) Calculate ωz −1 w−1 with the exact square roots, obtained in Problem 3.24. Get a readable result, do not distribute every term. (c) Get exact expressions for cos 3◦ and sin 3◦ . Solution. (a) Since 1 1 1 1 − − = , we get the argument 3 5 8 120 1 1 1 ωz −1 w−1 = e2πi [ 3 − 5 − 8 ] = cos 2π 2π + i sin 120 120 (b) With the exact square roots from Problem 3.24, " p √ # √ √ −1 + i 3 1 − i 5 − 1 − i 10 + 2 5 −1 −1 √ ωz w = ωwz = 2 4 2 p √ √ √ √ 3 − 1 + i( 3 + 1) 5 − 1 − i 10 + 2 5 √ = 4 2 2 p p √ √ √ √ √ √ √ √ ( 3 + 1)( 5 − 1) − ( 3 − 1) 10 + 2 5 ( 3 − 1)( 5 − 1) + ( 3 + 1) 10 + 2 5 √ √ +i = 8 2 8 2 (e) We get exact expressions p √ √ √ √ √ 2( 3 − 1)( 5 − 1) + 2( 3 + 1) 5 + 5 cos 3◦ = 16 p √ √ √ √ √ 2( 3 + 1)( 5 − 1) − 2( 3 − 1) 5+ 5 ◦ sin 3 = 16 10 Problem 3.31. Calculate the circumference of a regular 60-gon. Which approximation of 2π do we get? How many decimal points are correct? Answer. The circumference of an 60-gon is q √ √ √ 15 √ √ ◦ 120 sin 3 = 2( 3 + 1)( 5 − 1) − 2( 3 − 1) 5 + 5 ≈ 6.280314749 2 2π ≈ 6.283185307 One gets two correct decimal points. 10 Problem 3.32. Gather the entire table with cos k3◦ and sin k3◦ for all k = 1 . . . 30. Go on as far as you get. 52 ω k z −k w−k = e 2πik 120 #k " √ "√ p √ √ #k 3 − 1 + i( 3 + 1) 5 − 1 − i 10 + 2 5 √ = 4 2 2 # " √ " √ p √ # 3 + i − 5 − 1 − i 10 − 2 5 − ω 2 z −2 w−2 = e = 2 4 p √ √ √ √ p √ √ 15 + 3 + 10 − 2 5 − i 5 − i + i 3 10 − 2 5 = 8 4πi 120 p √ # "√ √ # 1−i 3 5 − 1 + i 10 + 2 5 ω z w = e = −ωz = 2 4 p √ p √ √ √ √ √ 5 − 1 + 3 10 + 2 5 − i 15 + i 3 + i 10 + 2 5 = 8 " 4 −4 −4 8πi 120 " √ # " √ p √ √ # − 3 − 1 + i( 3 − 1) 5 − 1 − i 10 − 2 5 − √ = ωwz −2 = ω 7 z −7 w−7 = e 4 2 2 p p √ √ √ √ √ √ √ √ √ √ 2( 3 + 1)( 5 + 1) + 2( 3 − 1) 5 − 5 2(− 3 + 1)( 5 + 1) + 2( 3 + 1) 5 − 5 = +i 16 16 14πi 120 Further values were obtained by trial and error for the signs of the roots. 53 θ 3◦ 6◦ 9◦ 12◦ 15◦ 18◦ 21◦ 24◦ 27◦ 30◦ 33◦ 36◦ 39◦ 42◦ 45◦ 48◦ 51◦ 54◦ 57◦ 60◦ 63◦ 66◦ 69◦ 72◦ 75◦ 78◦ 81◦ 84◦ 87◦ 90◦ cos θ √ √ √ √ √ √ 2( 3−1)( 5−1)+2( 3+1) 5+ 5 16 √ √ √ √ 15+ 3+ 10−2 5 √ √ 8√ √ 10+ 2+2 5− 5 √ √ 8√ √ 5−1+ 3 10+2 5 8 √ √ 2+ 6 q4 √ 5+ 5 8 √ √ √ √ √ √ 2( 3+1)( 5+1)+2( 3−1) 5− 5 16 √ √ √ √ 5+1+ 3 10−2 5 √ √ 8√ √ 10− 2+2 5+ 5 8 √ 3 2√ √ √ √ √ √ − 2( 3−1)( 5−1)+2( 3+1) 5+ 5 16 √ 5+1 4 √ √ √ √ √ √ − 2(− 3+1)( 5+1)+2( 3+1) 5− 5 √ √ √ √16 15− 3+ 10+2 5 8 √ 2 2√ √ √ √ − 5+1+ 3 10+2 5 8√ √ √ √ √ √ 2( 3+1)( 5+1)−2( 3−1) 5− 5 16 q √ 5− 5 8 √ √ √ √ √ √ 2( 3+1)( 5−1)+2( 3−1) 5+ 5 16 1 2√ √ √ √ − 10+ 2+2 5+ 5 √ √ 8√ √ 15+ 3− 10−2 5 √ √ √ √ 8 √ √ 2(− 3+1)( 5+1)+2( 3+1) 5− 5 16 √ 5−1 √4 √ 6− 2 √4 √ √ √ − 15+ 3+ 10+2 5 8√ √ √ √ 10+ 2−2 5− 5 8 √ √ √ √ − 5−1+ 3 10−2 5 8√ √ √ √ √ √ 2( 3+1)( 5−1)−2( 3−1) 5+ 5 16 sin θ √ √ √ √ √ √ 2( 3+1)( 5−1)−2( 3−1) 5+ 5 √16 √ √ √ − 5−1+ 3 10−2 5 √ √8 √ √ 10+ 2−2 5− 5 8 √ √ √ √ − 15+ 3+ 10+2 5 8 √ √ 6− 2 4 √ 5−1 4 √ √ √ √ √ √ 2(− 3+1)( 5+1)+2( 3+1) 5− 5 16 √ √ √ √ 15+ 3− 10−2 5 √ √ 8 √ √ − 10+ 2+2 5+ 5 8 1 √ √2 √ √ √ √ 2( 3+1)( 5−1)+2( 3−1) 5+ 5 16 q √ 5− 5 8 √ √ √ √ √ √ 2( 3+1)( 5+1)−2( 3−1) 5− 5 √16 √ √ √ − 5+1+ 3 10+2 5 8 √ 2 √ √ √ √2 15− 3+ 10+2 5 √ √ √ √ 8 √ √ − 2(− 3+1)( 5+1)+2( 3+1) 5− 5 16 √ 5+1 4 √ √ √ √ √ √ − 2( 3−1)( 5−1)+2( 3+1) 5+ 5 16 √ 3 √ √2 √ √ 10− 2+2 5+ 5 8 √ √ √ √ 5+1+ 3 10−2 5 √ √ 8 √ √ √ √ 2( 3+1)( 5+1)+2( 3−1) 5− 5 16 q √ 5+ 5 √ 8√ 6+ 2 4√ √ √ √ 5−1+ 3 10+2 5 √ √ 8 √ √ 10+ 2+2 5− 5 8 √ √ √ √ 15+ 3+ 10−2 5 √ √ √ √ 8 √ √ 2( 3−1)( 5−1)+2( 3+1) 5+ 5 16 1 0 54 Part II Algebra 1 Polynomials 1.1 Rational zeros of polynomials Proposition 1.1. Assume x is a real or complex zero of a monic 5 integer polynomial xr + ar−1 xr−1 + · · · + a1 x + a0 = 0 where ar−1 , . . . a1 , a0 are integers. If x is rational, then it is even an integer, and a divisor of the constant term a0 , too. Proof. Assume x = yields m n where m and n are relatively prime integers. Multiplying by nr mr mr−1 m + a + · · · + a + a0 = 0 r−1 1 nr nr−1 n mr + ar−1 mr−1 n + · · · + a1 mnr−1 + a0 nr = 0 mr = − ar−1 mr−1 + · · · + a1 mnr−2 + a0 nr−1 n 6 Now we argue the same way as in the proof of Proposition 1.9: Assume that any prime p divides n. By the last line, the same prime divides mr . Hence, by Euclid’s Lemma . Hence n = 1, and p divides m. This cannot occur after cancellation of the fraction m 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 mr + ar−1 mr−1 + · · · + a1 m + a0 = 0 m mr−1 + ar−1 mr−2 + · · · + a1 = −a0 Hence any zero m has to be a divisor of the constant coefficient a0 . Proposition 1.2. Assume x is a real or complex zero of any polynomial ar xr + ar−1 xr−1 + · · · + a1 x + a0 = 0 where ar 6= 0, ar−1 , . . . a1 , a0 are integers. If the root x = m is rational with m and n n relatively prime, then m is a divisor of a0 and n is a divisor of ar . 5 6 A polynomial is called monic if the coefficient of the leading term xr is one. Hence n is a divisor of mr and hence n = 1, since n and m are relatively prime. 55 Proof. ar mr + ar−1 mr−1 n + · · · + a1 mnr−1 + a0 nr = 0 ar mr = − ar−1 mr−1 + · · · + a1 mnr−2 + a0 nr−1 n − ar mr−1 + ar−1 mr−2 n + · · · + a1 nr−1 m = a0 nr The second line implies that n is a divisor of ar mr . Since gcd (n, m) = 1, the Euclidean Property stated in Proposition 1.4 implies that n is a divisor of ar . The last line implies that m is a divisor of a0 nr . Since gcd (n, m) = 1, the Euclidean Property stated in Proposition 1.4 implies that m is a divisor of a0 . Rule: The rational solutions of an integer polynomial ar xr + · · · + a0 = 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 ar X + a0 = 0. 1.2 Gauss’ Lemma Proposition 1.3 (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 (1.1) ar xr + · · · + a0 = (nx − m)(br−1 xr−1 + · · · + b0 ) If additionally gcd (ar , . . . , a0 ) = 1, then gcd (br−1 , . . . , b0 ) = 1, too. Proposition 1.4 (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 xr + · · · + a0 = (x − m)(xr−1 + · · · + b0 ) Proof. We may assume from the start that gcd (ar , . . . , a0 ) = 1. The factoring of equa, . . . , b0 = BB0 . We choose the tion (1.1) holds with rational coefficients br−1 = Br−1 B smallest possible denominator B. In that case, the assumption that gcd (ar , . . . , a0 ) = 1 implies that gcd (Br−1 , . . . , B0 ) = 1, too. 7 After multiplying with the least common denominator B of the b0 , . . . br−1 , comparison of coefficients yields Ba0 Ba1 Ba2 Bar−1 Bar = −mB0 = nB0 − mB1 = nB1 − mB2 = nBr−2 − mBr−1 = nBr−1 7 Indeed, a common prime divisor of all Br−1 , . . . , B0 could not divide B and hence would be a common divisor of all ar , . . . , a0 , what we just have ruled out. 56 Assume towards a contradiction that a prime p divides all Ba0 , . . . Bar . Either p does not divide n or p does not divide m. Take the second case. Successively, we conclude that p divides Ba0 = −mB0 , B0 , Ba1 − nB0 = −mB1 , B1 , Ba2 −nB1 = −mB2 , B2 , . . . , Br−1 , hence all B0 , . . . Br−1 —contrary to the assumption. Hence we conclude gcd (Ba0 , . . . Bar ) = 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 Bar = −nBr−1 , Br−1 , Bar−1 +mBr−1 = nBr−2 , Br−2 , . . . , Ba1 + mB1 = nB0 , B0 . In both cases, the conclusion gcd (Ba0 , . . . Bar ) = 1 holds. Hence B = 1, and in the first place, the rational numbers b0 , . . . br−1 need to be integers. Finally, we see that gcd (br−1 , . . . , b0 ) = 1. Indeed, since B = 1, a prime p dividing all coefficients br−1 , . . . , b0 would divides all coefficients ar , . . . , a0 , too, contrary to the assumption. Theorem 1.1 (Gauss’ Lemma). An integer polynomial which factors over the rational numbers into factors of lower degree, already factored into integer polynomials of lower degree. The latter factoring is obtained from the former one by adjusting integer factors. Especially, a monic integer polynomial that factors over the rational numbers, even factors over the integers into monic integer polynomials. Proof. Suppose that an integer polynomial of degree r ≥ 2 can be factored into rational polynomials of degree s ≥ 1 and t ≥ 1. Multiplying with the least common denominators, we get the integer formula (1.2) B (ar xr + · · · + a0 ) = A (bs xs + · · · + b0 ) ct xt + · · · + c0 where ar xr + · · · + a0 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 (1.3) gcd (A, B) = 1 gcd (ar , . . . , a0 ) = 1 gcd (bs , . . . , b0 ) = 1 gcd (ct , . . . , c0 ) = 1 Question. Show that A = 1. Answer. Assume any prime number p divides A, and derive a contradiction. By formula (1.2) this would imply that p divides all numbers Bar , . . . Ba0 . Since gcd (ar , . . . , a0 ) = 1, we conclude that p divides B. But this contradicts the assumption gcd (A, B) = 1. Lemma 1.1. If a prime p divides all coefficients of the product of two polynomials with integer coefficients (1.4) (bs xs + · · · + b0 ) ct xt + · · · + c0 , then p divides either gcd (bs , . . . , b0 ) or gcd (ct , . . . , c0 ). 57 Reason for the Lemma. Multiplying the terms of the product (1.4), we see that p divides all the integers b0 c 0 b0 c 1 + b1 c 0 b0 c 2 + b1 c 1 + b2 c 0 b0 ck + b1 ck−1 + · · · + bk c0 for all k ≥ 0. 8 We see that p divides either b0 or c0 . Now take the case that p divides b0 . The case that p divides c0 can be deals with similarly. Because of the assumption gcd (bs , . . . , b0 ) = 1, there exists an index 0 ≤ σ < s such that p divides b0 , b1 , . . . bσ but not bσ+1 . But we have seen that p divides all the sums from above. Successively, we put k = σ + 1, σ + 2 . . . and cancel the terms already known to be divisible by p. Hence we see that p divides the sums bσ+1 c0 bσ+1 c1 + bσ+2 c0 bσ+1 c2 + bσ+2 c1 + bσ+3 c0 bσ+1 cj + bσ+2 cj−1 + · · · + bσ+j+1 c0 and so on for all j ≥ 0. Hence we conclude that p divides c0 , c1 , c2 , . . . cj for all j ≥ 0. Question. Use the Lemma to show that B = 1. Answer. Assume any prime number p divides B, and derive a contradiction. By formula (1.2), and since A = 1, this would imply that p divides all coefficients of the product (1.4). Hence by the Lemma, either p divides all coefficients bs , . . . , b0 or all coefficients ct , . . . , c0 . This has been ruled out by assumption (1.3). Hence we conclude that B = 1. Now, since A = B = 1, the factoring (1.2) is indeed a factoring of the integer polynomial ar xr + · · · + a0 into integer polynomials. Since the integer polynomial ar xr + · · · + a0 was obtained from the given integer polynomial by dividing through with the greatest common divisor of the coefficients, we have factored the originally given polynomial, too. 10 Problem 1.1. Find all symmetric monic polynomials x4 + ax2 + bx2 + ax + 1, that factor over the rationals and have (i) Four rational zeros. (ii) Exactly two rational zeros. (iii) No rational zeros. 8 For simplicity, we put the coefficients of a polynomial equal zero if we exceed its degree. 58 Do the cases with one or three zeros occur? Why not? Solution. The only possible rational zeros are +1 and −1. If x = 1 is a zero, it is a double zero since P (x) = x4 P (x−1 ) P 0 (x) = 4x3 P (x−1 ) − x2 P 0 (x−1 ) Hence P (x) = 0 and x2 = 1 imply P 0 (1) = −P 0 (1) = 0. (i) Four rational zeros. Hence there are five choices with all zeros in the set {+1, −1}. Three of them turn out to be symmetric: (x + 1)4 , (x2 − 1)2 , (x − 1)4 (ii) Exactly two rational zeros. The cases with a double zero 1 are (x − 1)2 (x2 + (a + 2)x + 1) = x4 + ax3 − 2(a + 1)x2 + ax + 1 where a is any integer a 6= 0, a 6= −4. The cases with a double zero −1 are (x + 1)2 (x2 + (a − 2)x + 1) = x4 + ax3 + 2(a − 1)x2 + ax + 1 where a is any integer a 6= 0, a 6= +4. Too, we see that no solutions with a zero of multiplicity three exist. (iii) No rational zeros. The polynomial factors into two irreducible monic quadratics: Q and R. The symmetry implies P (x) = Q(x)R(x) = x2 Q(x−1 )x2 R(x−1 ) = x4 P (x−1 ) Uniqueness of factorization into irreducible factors leads to four possibilities: (a) x2 Q(x−1 ) = Q(x) and x2 R(x−1 ) = R(x). (b) x2 Q(x−1 ) = −Q(x) and x2 R(x−1 ) = −R(x). (c) x2 Q(x−1 ) = R(x) and x2 R(x−1 ) = Q(x). (d) x2 Q(x−1 ) = −R(x) and x2 R(x−1 ) = −Q(x). Here are the results: (a) From x2 Q(x−1 ) = Q(x) = x2 + cx + 1 and x2 R(x−1 ) = R(x) = x2 + dx + 1, we get Q(x)R(x) = (x2 +cx+1)(x2 +dx+1) = x4 +(c+d)x3 +(cd+2)x2 +(c+d)x+1 with any integer c, d 6= ±2. 59 (b) x2 Q(x−1 ) = −Q(x) and x2 R(x−1 ) = −R(x) gives only the solution (x2 − 1)2 that we already have. (c) x2 Q(x−1 ) = R(x) and x2 R(x−1 ) = Q(x) gives only the special case of (a) with c = d, but nothing new. (d) −x2 Q(x−1 ) = R(x) = x2 − cx − 1 and −x2 R(x−1 ) = Q(x) = x2 + cx − 1 gives the extra solutions Q(x)R(x) = (x2 + cx − 1)(x2 − cx − 1) = x4 − (c2 + 2)x2 + 1 These are new solution for any c 6= 0. 1.3 Eisenstein’s irreducibility criterium Proposition 1.5 (The Eisenstein criterium). For a polynomial P to be irreducible in the ring Z[z], it is sufficient that there exists a prime number p such that 1. the leading coefficient of the polynomial is not divisible by p; 2. all other coefficients except the leading one are divisible by p; 3. the constant coefficient is not divisible by p2 . Proof. Assume toward a contradiction that a reducible polynomial would satisfy the criterium. We would get the integer factorization (1.5) a0 + · · · + ar xr = (b0 + · · · + bs xs ) c0 + · · · + ct xt , where r = s + t and s, t, ≥ 1. Multiplying the terms of the product yields a0 a1 a2 ak =b0 c0 =b0 c1 + b1 c0 =b0 c2 + b1 c1 + b2 c0 =b0 ck + b1 ck−1 + · · · + bk c0 for all k ≥ 0. It is assumed that the prime p, but not p2 divides a0 = b0 c0 . Hence p divides exactly one of the two numbers b0 and c0 . We may assume that p divides c0 , but does not divide b0 . Furthermore, by assumption, p divides a0 , a1 , . . . ar−1 , but not ar . Recursively, we conclude that p divides b0 c1 = a1 − b1 c0 and hence c1 , next b0 c2 = a2 − b1 c1 − b2 c0 and hence c2 , . . . , cr−1 . Hence p divides all coefficients of the factor ct xt +· · ·+c0 , since t ≤ r−1. This implies that p divides all coefficients of the original polynomial ar xr + · · · + a0 , contradicting the assumption that the prime p does not divide the leading coefficient ar . From this contradiction, we see that no factoring of the polynomial into polynomials of lower degree is possible, and hence it is irreducible. 60 q √ 10 Problem 1.2. Find the irreducible polynomial with zero z1 = 5−2 5 . Find all its algebraic conjugates, and show that the number is a totally real algebraic integer. Solution of Problem 1.2. Simple arithmetic shows that (2z 2 − 5)2 − 5 = 0 and hence P (z) := z 4 − 5z 2 + 5 = 0 is the monic polynomial in the ring Z[z] with zero z1 . Hence z1 is an algebraic integer. The Eisenstein criterium applies with p = 5. Hence its zeros are exactly the algebraic conjugates of z. Obviously, these zeros are s s s s √ √ √ √ 5+ 5 5+ 5 5− 5 5− 5 , − , , − 2 2 2 2 which all four turn out to be real. Lemma 1.2. For any prime p, the binomial coefficients p for k = 1 . . . p − 1 k are divisible by p, but not by p2 . Proposition 1.6. For any prime p, the polynomial Φp (z) = zp − 1 z−1 is irreducible over the integers. Increasing powers. The assertion is true for p = 2. We assume now that p is an odd prime. We substitute z = 1 + x and use the Eisenstein criterium to show that the resulting polynomial P (x) = Φp (1 + x) is irreducible. The binomial formula implies p (1 + x)p − 1 X p k−1 P (x) = = x x k k=1 p p 2 p =p+ x+ x + ··· + xp−3 + pxp−2 + xp−1 2 3 p−2 Because of Lemma 1.2, we see that all three assumptions for the Eisenstein criterium (Proposition 1.5) are satisfied. Hence the polynomial P and hence Φp are irreducible over the integers. Lemma 1.3. For any odd prime p, the binomial coefficients 2 p for k = 1 . . . p − 1 k are divisible by p. 61 Proposition 1.7. For any prime p, the polynomial 2 zp − 1 Φ (z) = p z −1 p2 is irreducible over the integers. Proof. In the case p = 2, we get Φ4 = 1 + z 2 . which we can check to be irreducible. We assume now that p is an odd prime. We substitute z = 1 + x and use the Eisenstein criterium to show that the resulting polynomial p(p−1) Φp2 (1 + x) = X bi xi = p + · · · + xp(p−1) i=0 is irreducible. The definition and the binomial formula imply 2 zp − 1 zp − 1 = Φp2 (z) · z−1 z−1 p2 (1 + x)p − 1 (1 + x) − 1 = Φp2 (1 + x) · x x p(p−1) p2 2 p X p X X p xl−1 = bi x i · xk−1 l k i=0 l=1 k=1 We have obtained the integer factorization a0 + · · · + ar xr = (b0 + · · · + bs xs ) c0 + · · · + ct xt , which is really possible in this context since it has been constructed above. The degrees are now r = p2 − 1, s = deg Φp2 = p(p − 1) and t = deg Φp = p − 1. Multiplying the polynomials and comparing the coefficients yields once more a0 a1 a2 ak =b0 c0 =b0 c1 + b1 c0 =b0 c2 + b1 c1 + b2 c0 =b0 ck + b1 ck−1 + · · · + bk c0 for all k ≥ 0. We know from Lemma 1.2 and 1.3 that the prime p divides all the coefficients a0 , . . . , ap2 −1 , but ap2 = 1. That is enough to check inductively that p divides all coefficients b0 , . . . , bp2 −p−1 . But the leading coefficient of Φp2 (1 + x) is bp2 −p = 1. We see that the polynomial Φp2 (1+x) satisfies all three assumptions for the Eisenstein criterium (Proposition 1.5). Hence Φp2 (1 + x) and hence Φp2 (x) are irreducible over the integers. 62 Example p = 3, p2 = 9. We claim that Φ9 (1 + x) = 6 X bi x i i=0 is irreducible. The definition and the binomial formula imply z9 − 1 z3 − 1 = Φ9 (z) · z−1 z−1 9 (1 + x) − 1 (1 + x)3 − 1 = Φ9 (1 + x) · x x 9 6 X 9 X xl−1 = bi xi · 3 + 3x + x2 l i=0 l=1 We have obtained an integer factorization. Multiplying the polynomials and comparing the coefficients yields once more 9 9 9 = b0 · 3 , = b 0 · 3 + b1 · 3 , = b0 · 1 + b1 · 3 + b2 · 3 1 2 3 9 9 9 = b1 · 1 + b2 · 3 + b3 · 3 , = b2 · 1 + b3 · 3 + b4 · 3 , = b3 · 1 + b4 · 3 + b5 · 3 4 5 6 9 9 9 = b4 · 1 + b5 · 3 + b6 · 3 , = b5 · 1 + b6 · 3 , = b6 · 1 7 8 9 We get the constant coefficients a0 = 9, b0 = 3 and c0 = 3 and the leading coefficient b6 = 1. As in the general case, we to use the inductive argument to conclude that 3 is a divisor of the coefficients b0 , . . . , b5 . That is all we need. All assumptions in the Eisenstein criterium are satisfied and hence Φ9 (1 + x) and Φ9 (x) are irreducible over the integers. Remark. The explicit result Φ9 (1 + x) = 3 + 9x + 18x2 + 21x3 + 15x4 + 6x5 + x6 is not needed. 1.4 Descartes’ Rule of Signs Definition 1.1 (Degree of a polynomial). We say that the polynomial P ∈ C[z] has degree n provided that an 6= 0 is the highest nonzero coefficient, and write deg(P ) = n. The constant nonzero polynomial has degree zero. The zero polynomial has degree ∞ or undefined. 9 Remark. For any two polynomials deg(P Q) = deg(P ) + deg(Q). 9 —depending on the taste of the author. 63 Proposition 1.8 (Descartes’ rule of signs). Let P (x) = an xn + an−1 xn−1 + · · · + a0 be a polynomial with real coefficients. The number of real positive zeros of P (x) is less or equal the number of sign changes in the sequence an , an−1 , . . . , a0 ; with the zero terms deleted. The assertion holds even counting zeros with their multiplicities. Furthermore, the difference between the number of sign changes and the number of positive zeros, counted with multiplicities, is even. 10 Problem 1.3. Produce at least five polynomials of degree at least three, four, five—with zeros you have chosen conveniently. Multiply out and check Descartes’ rule of signs. Look for covering different possibilities, multiple zeros, and so on. Examples. (i) (x − 1)4 = x4 − 4x3 + 6x2 − 4x + 1 has 4 sign changes and four positive roots. (ii) (x − 1)4 − 1 = x4 − 4x3 + 6x2 − 4x has 3 sign changes and one positive roots x1 = 2. The other roots are 0, 1 ± i. (iii) (x − 1)4 + 1 = x4 − 4x3 + 6x2 − 4x + 2 has 4 sign changes and no positive root. √ . The roots are 1 + (±1±i) 2 (iv) (x − 1)4 + 64 = x4 − 4x3 + 6x2 − 4x + 65 has 4 sign changes and no positive root. The roots are −1 ± 2i, 3 ± 2i. Proof of Descartes’ rule of signs. Prove the assertion by induction on the degree n. The assertion is true for n = 0 and n = 1. For induction assumption, assume that the assertion holds for all polynomials with degree less than n. Given is a polynomial of degree n. We can assume that an > 0. and let c count the number of sign changes in the sequence an , an−1 , . . . , a0 . Let 0 < x1 ≤ x2 ≤ · · · ≤ xN be the increasing list of positive zeros of the polynomial P , counting multiplicities. Too, we write N = N (P, (0, ∞)) for the count of zeros of P . In the case that a0 = 0, the polynomial x−1 P (x) has degree n − 1, less than n, but the same number of positive zeros and sign changes. Hence the assertion follows from the induction assumption. Assume a0 6= 0 and let ai 6= 0 be next nonzero coefficient. We need to distinguish the cases a0 ai > 0 and a0 ai < 0. (a) case a0 ai > 0. The divided derivative Q(x) := x−i+1 P 0 (x) has the same number c of sign changes in the sequence nan , (n − 1)an−1 , . . . , iai of its coefficients. Furthermore, Q(x) has at least one zero in each of the intervals (1.6) (0, x1 ), (x1 , x2 ), . . . , (xN −1 , xN ) 64 hence at least N positive zeros. Hence N (P, (0, ∞)) ≤ N (Q, (0, ∞)). Note that a zero of P 0 and hence Q exists in the open interval (0, x1 ), since |P | restricted to [0, x1 ] does not achieve its maximum at 0, but in the open interval (0, x1 ). In the case of a multiple zero a > 0, the interval (a, a) has to be replaced by the single element set {a}. By this interpretation of the list (1.6), we see that the assertion remains true counting multiplicities. Since deg(Q) < n, the induction assumption yields N (P, (0, ∞)) ≤ N (Q, (0, ∞)) ≤ c, as to be shown. (b) case a0 ai < 0. The divided derivative Q(x) := x−i+1 P 0 (x) has only c − 1 of sign changes in the sequence of its coefficients. Furthermore, Q(x) has at least one zero in each of the intervals (x1 , x2 ), . . . , (xN −1 , xN ) hence at least N − 1 positive zeros. Hence N (P, (0, ∞)) ≤ N (Q, (0, ∞)) + 1. Note that the interval (0, x1 ) has to be left out in this case, since now |P | restricted to [0, x1 ] can achieve its maximum at 0. With the interpretation of any interval (a, a) with a > 0 as the single element set {a}, we see the assertion remains true counting multiplicities. Since deg(Q) < n, the induction assumption yields N (Q, (0, ∞)) ≤ c − 1 N (P, (0, ∞)) ≤ N (Q, (0, ∞)) + 1 ≤ c − 1 + 1 = c as to be shown. It remains to check the last assertion about the even difference. 10 The assertion clearly holds for degree zero and one. Note that in any interval (a, b) between two consecutive zeros P (a) = P (b) of a polynomials, the derivative P 0 has an odd number of zeros. This yields an even number of extra zeros of P 0 in (0, ∞) beyond the ones claimed above. Take the case (b): the difference N (Q, (0, ∞))+1−N (P, (0, ∞)) is even, Too, by the induction assumption, the difference c − 1 − N (Q, (0, ∞)) is even. Hence the difference c − N (P, (0, ∞)) is even, too. In the case (a), the interval (0, x1 ) has to be included. Therefore, this time, the difference N (Q, (0, ∞)) − N (P, (0, ∞)) is even. Too, by the induction assumption, the difference c − N (Q, (0, ∞)) is even. Hence, in the end, the difference c − N (P, (0, ∞)) turns out to be even in all cases. For any polynomial P (x) = an xn + an−1 xn−1 + · · · + a0 with real coefficients, we define c+ : the number of sign changes in the sequence an , an−1 , . . . , a0 —with zero terms deleted; N+ : the number N (P, (0, ∞)) of positive zeros of P , counting their multiplicities; 10 This assertion has been added by Gauss. 65 c− : the number of sign changes in the sequence (−1)n an , (−1)n−1 an−1 , . . . , a0 —with zero terms deleted; N− : the number N (P, (−∞, 0)) of negative zeros of P , counting their multiplicities; c0 : the multiplicity—also denoted by N (P, {0})— of z = 0 as a zero of polynomial P . Hence ai = 0 for i < c0 , but ai 6= 0 for i = c0 . Lemma 1.4 (Sum of sign changes). For any sequence an , an−1 , . . . , a0 with an 6= 0, the sum c+ + c0 + c− is less or equal to n, and the difference n − (c+ + c0 + c− ) ≥ 0 is even. Furthermore, if n = c+ + c0 + c− , there cannot be more than one zero coefficient between two nonzero coefficients; and for such the zero coefficient, the two nonzero neighbors have different signs. Proof. Prove the assertion by induction on the degree n. The assertion is true for n ≤ 1. For induction assumption, assume that the assertion holds for all sequences ak , ak−1 , . . . , a0 with k < n and ak 6= 0. Given any sequence an , an−1 , . . . , a0 with an 6= 0. In the case an is the only nonzero term, the assertion holds. Otherwise, let k < n be the maximal index with ak 6= 0. The two sequences (±1)n an , (±1)n−1 an−1 , . . . , a0 have the added signs changes c+ + c0 + c− — these are more than the added signs changes c0+ + c0 + c0− of the two shorter sequences (±1)k ak , (±1)k−1 ak−1 , . . . , a0 . The difference are the additional sign changes occurring between (±1)n an and (±1)k ak which are ( 1 if n − k is odd c+ + c− − c0+ − c0− = 0 or 2 if n − k is even By induction assumption, the difference k − (c0+ + c0 + c0− ) ≥ 0 is even. Hence the difference n − (c+ + c0 + c− ) ≥ 0 is even, too. Furthermore, if n = c+ + c0 + c− , then k = c0+ + c0 + c0− , too. ( 1=n−k if n − k is odd c+ + c− − c0+ − c0− = 2=n−k if n − k is even Thus there cannot be more than one zero coefficient between an 6= 0 and ak 6= 0. If n − k = 2, then an ak < 0. Corollary. A polynomial with real coefficients of degree n has at most n real zeros, counting multiplicities. Corollary. For a real polynomial of degree n with n real zeros, Descartes’ rule of signs gives the exact number of positive and negative zeros. Furthermore, there cannot be more than one zero coefficient between two nonzero coefficients. For such the zero coefficient, the two nonzero neighbors have different signs. 66 Corollary. The even or odd coefficients of an even or odd real polynomial with only real zeros have alternating signs and are nonzero—except for the lowest ones. Proof. Put together, the rule of Descartes 1.8 and the Lemma 1.4 imply N (P, (0, ∞)) + N (P, {0}) + N (P, (−∞)) ≤ c+ + c0 + c− ≤ n (1.7) In the case of n real zeros, equality holds everywhere, and hence N (P, (0, ∞)) = c+ , N (P, (−∞)) = c− , and c+ + c0 + c− = n Because of the last equality, as shown in the Lemma, the list of coefficients cannot have more than one zero coefficient between two nonzero coefficients, and in that case the two nonzero neighbors have different signs. 2 Tschebychev polynomials Definition 2.1 (Tschebychev polynomials). The Tschebychev polynomials of first and second kind are defined by Tn (cos t) = cos nt and Un (cos t) = sin(n + 1)t sin t for integers n = 0, 1, 2, . . . . Proposition 2.1. Both polynomials Tn and Un satisfy the same recursion formula Tn+1 = 2xTn − Tn−1 Un+1 = 2xUn − Un−1 with the initial data T0 = U0 = 1 and T1 = x but U1 = 2x, respectively. Both Tn and Un are integer polynomials of degree n. They are even for even n, odd for odd n. Moreover Tn (1) = 1 and Un (1) = n + 1 for all n. The Tn satisfy the composition formula Tn ◦ Tm = Tnm . The polynomials 2Tn (v/2) are integer polynomials of the variable v. n 0 1 2 3 4 5 6 7 8 9 Tn 1 x −1 + 2x2 −3x + 4x3 1 − 8x2 + 8x4 5x − 20x3 + 16x5 −1 + 18x2 − 48x4 + 32x6 −7x + 56x3 − 112x5 + 64x7 1 − 32x2 + 160x4 − 256x6 + 128x8 9x − 120x3 + 432x5 − 576x7 + 256x9 67 Un 1 2x −1 + 4x2 −4x + 8x3 1 − 12x2 + 16x4 6x − 32x3 + 32x5 −1 + 24x2 − 80x4 + 64x6 −8x + 80x3 − 192x5 + 128x7 1 − 40x2 + 240x4 − 448x6 + 256x8 10x − 160x3 + 672x5 − 1024x7 + 512x9 Lemma 2.1. Assume that n = 2m + 1 is odd. Let z = x + iy = eit and v = 2x = 2 cos t. (2.1) n z −1 = z−1 n−1 X " z k = (x + iy)m 1 + 2 k=0 m X # " cos kt = (x + iy)m 1 + 2 k=1 m X # Tk (cos t) k=1 (2.2) m n−1 Y 2π k 2πi k zn − 1 Y m x − cos = z − exp = (2x + 2iy) = (x + iy)m R(2x) z−1 n n k=1 k=0 The polynomial (2.3) m m Y X 2π k R2m+1 (2x) = 2 x − cos =1+2 Tk (x) 2m + 1 k=1 k=1 m is a monic integer polynomial of the variable v = 2x of degree m = (n − 1)/2. Proof. m m n−1 X X X zn − 1 k −m k (z k + z −k ) =: R2m+1 (2x) z = 1 + = z z = (z − 1)z m k=1 k=−m k=0 To prove that these are integer polynomials of the variable v = 2x = z +z −1 , inductively for all m, we use the binomial formulas for (z + z −1 )m . The example with n = 5, m = 3 has already been calculated in the formula (??) from the section about the heptagon. Too, we can substitute (z k + z −k )/2 = cos kt and introduce the Tschebychev polynomials defined via Tk (cos t) = cos kt: m m m m X X X X zn − 1 z k + z −k k = z = 1 + 2 = 1 + 2 cos kt = 1 + 2 Tk (cos t) (z − 1)z m k=−m 2 k=1 k=1 k=1 with variable cos t = (z +z −1 )/2 = v/2. We see, again by induction on m, that 2Tm (v/2) and hence Tm are integer polynomials. Combining the factors from conjugate complex roots yields (2.4) Y n−1 m zn − 1 Y 2πi k 2πi k 2πi k = z − exp = z − exp z − exp − z−1 n n n k=0 k=1 m m Y Y z + z −1 2π k 2π k 2 m = z − 2z cos + 1 = (2z) − cos n 2 n k=1 k=1 With x = (z + z −1 )/2 and v = 2x, the formula 2.2 is confirmed. We need to factor the polynomial K ∈ Z[z] over the integers. 68 Proposition 2.2. Assume that n is odd. The polynomial K(z) ∈ Z[z] is irreducible over the integers if and only if the polynomial R is irreducible over the integers. Proof. Assume that there is an integer factoring K = P Q. The conjugate complex of any root eiτ of P is a root e−iτ of P , too. Since n is odd, the real root −1 does not occur, and the real root 1 has been divided out in the beginning. Hence the degree of deg P = 2s is even and the factors with conjugate complex roots can be combined as in formula (2.4). Thus we obtain an integer factoring of the polynomial R(v). The converse is even easier to confirm. The item k ≤ 8 of the sequence of formulas (2.3) are given. m m Y X 2π k 2 Tk (x) x − cos =1+2 n k=1 k=1 2π 2 x − cos = 1 + 2T1 = 1 + 2x 3 4π 2π x − cos = 1 + 2T1 + 2T2 = −1 + 2x + 4x2 4 x − cos 5 5 3 3 Y X 2π k Tk = −1 − 4x + 4x2 + 8x3 8 x − cos =1+2 7 k=1 k=1 4 4 Y X 2π k 16 x − cos Tk = 1 − 4x − 12x2 + 8x3 + 16x4 =1+2 9 k=1 k=1 m m m Y X 2π k 2 x − cos =1+2 Tk (x) n k=1 k=1 5 Y 2π k 32 x − cos = 1 + 6x − 12x2 − 32x3 + 16x4 + 32x5 11 k=1 6 Y 2π k = −1 + 6x + 24x2 − 32x3 − 80x4 + 32x5 + 64x6 64 x − cos 13 k=1 7 Y 2π k 128 x − cos = −1 − 8x + 24x2 + 80x3 − 80x4 − 192x5 + 64x6 + 128x7 15 k=1 8 Y 2π k = 1 − 8x − 40x2 + 80x3 + 240x4 − 192x5 − 448x6 + 128x7 + 256x8 256 x − cos 17 k=1 m 69 One sees that the substitution v = 2x eliminates many powers of two. 2π (2.5) v − 2 cos =1+v 3 2 Y 2π k v − 2 cos (2.6) = −1 + v + v 2 5 k=1 4 Y 2π k (2.7) v − 2cos = 1 − 2v − 3v 2 + v 3 + v 4 9 k=1 7 Y 2π k (2.8) v − 2 cos = −1 − 4v + 6v 2 + 10v 3 − 5v 4 − 6v 5 + v 6 + v 7 15 k=1 8 Y 2π k v − 2 cos (2.9) = 1 − 4v − 10v 2 + 10v 3 + 15v 4 − 6v 5 − 7v 6 + v 7 + v 8 17 k=1 The polynomial R9 has the divisor R3 , as expected. The irreducible factors are: R9 = 1 − 2v − 3v 2 + v 3 + v 4 = (1 + v) · (1 − 3v + v 3 ) Moreover, we see that R15 has the divisor R3 · R5 . This is expected since R15 =???????????????????? R3 · R5 The irreducible factors of R15 are: −1 − 4v + 6v 2 + 10v 3 − 5v 4 − 6v 5 + v 6 + v 7 = (1 + v) · (−1 − 3v + 9v 2 + v 3 − 6v 4 + v 6 ) = (1 + v) · (−1 + v + v 2 ) · (1 + 4v − 4v 2 − v 3 + v 4 ) The Five Quadrable Lunes Brian Shelburne Faculty, Wittenberg University Abstract Number: 5. A lune, the crescent shaped area formed by the intersection of two circles, is quadrable if it is possible to construct a square with the same area using only straight-edge and compass. Hippocrates of Chios (ca. 430 BCE) is credited with discovering three such Lunes; two more were discovered in the 18th century. In the 20th century Tschebatorev and Dorodnov proved there were only five. This talk will examine the equation used to derive the five cases and present as examples the two cases which are not normally seen. 70 3 3.1 Polynomials over the Complex Numbers A proof of the Fundamental Theorem of Algebra Given is any polynomial P (z) = n X ak z k k=0 with complex coefficients ak ∈ C. We want to estimate its zeros. 10 Problem 3.1. Prove for any zero of a monic polynomial: n P (z) = z + n−1 X ak z k k=0 " |z| ≤ R0 := max 1, n−1 X # |ak | k=0 Answer. The assertion clearly holds for all zeros |z| ≤ 1. On the other hand, |z| > 1 implies ! n−1 n n X X X |ak | |an−k ||z|n−1 = |z|n−1 |z| − |an−k ||z|n−k ≥ |z|n − |P (z)| ≥ |z|n − k=1 k=1 k=0 Now P (z) = 0 and |z| > 1 imply that the right-hand side is non-positive, and hence |z| ≤ n−1 X |ak | ≤ R0 k=0 10 3.2. Choose any R > R0 . Show that the monic polynomials P (z) = P Problem k z n + n−1 a z and Q(z) = z n have the same number of zeros with absolute value k=0 k |z| ≤ R. Solution. We use the Theorem of Rouché for the functions P and Q in the disk B(R, 0)— of radius R around 0. On the boundary ∂B(R, 0), we get |z| = R > 1, and hence n−1 n−1 X X k |P (z) − Q(z)| = ak z ≤ |ak | ≤ R0 < R < Rn = |Q(z)| k=0 k=0 Thus all assumptions of Rouché’s Theorem are satisfied. Hence the assertion follows. 71 Clearly, the result implies even that the polynomials P and Q have the same number of zeros with absolute value |z| ≤ R0 —in the closed disk B(R0 , 0) = {z ∈ C : |z| ≤ R0 }. By the last Problem (3.1), all zeros z of polynomial P lie in the disk B(R0 , 0). On the other hand, the polynomial Q(z) = z n obviously has n zeros, counting multiplicities, and they all lie in the disk B(R0 , 0). Hence we get the celebrated result: Main Theorem (The Fundamental Theorem of Algebra). Every polynomial of degree n has exactly n complex zeros, counting their multiplicities. 3.2 Meditation on Descartes’ rule of signs Which insight does one gain by combining the Fundamental Theorem of algebra 3.1 with Descartes’ rule of signs? From Descartes’ rule of signs and Lemma 1.7, we get for the number of complex zeros of a real polynomial P : (3.1) N (P, C \ R) = n − N (P, (0, ∞)) − N (P, {0}) − N (P, (−∞)) n − c+ − c0 − c− c+ − N+ c− − N− + + =2 2 2 2 where all three fractions are nonnegative integers. Corollary. A polynomial with real coefficients of degree n has at least n − c+ − c0 − c− complex zeros. They occur in pairs of conjugate complex ones. 10 Problem 3.3 (The sleepless Descartes). Give eight or more examples of polynomials which show that all three fractions in equation (3.1) can be zero or nonzero in all combinations. Polynomials of degree up to six are needed! P only real zeros x2 + 1 (x + 2)((x − 1)2 + 1) (x − 2)((x + 1)2 + 1) x2 (x − 1)2 + 1 x2 (x + 1)2 + 1 (x2 − 1)2 + 1 (x2 + 1)(x4 − 2x2 + 2) distributed n − c+ − c0 − c− 0 x2 + 1 2 3 x − 2x + 4 0 x3 − 2x − 4 0 4 3 2 x − 2x + x + 1 2 4 3 2 x + 2x + x + 1 2 x4 − 2x2 + 2 0 6 4 x −x +2 2 c+ − N+ 0 0 2 0 2 0 2 2 2 c− − N− 0 0 0 2 0 2 2 2 Definition 3.1. For any integers a and b, the notation ”b ≥ a” means b ≥ a and b − a is even. Similar notation is used with other comparison signs. Similarly, for any integer m ≥ 1, the notation means b ≤ a and b − a is divisible by m. 72 m ”b ≤ a” In the following I want to count the zeros of a real polynomial in the right- and left half plane. H+ : Let H+ := N (P, {x + iy : x > 0} denote the number of zeros in the open right half plane; H− : Let H− := N (P, {x + iy : x < 0} denote the number of zeros in the open right half plane; H0 : denote the number of zeros on the imaginary axis. 10 Problem 3.4. Prove that a real polynomial with only imaginary roots is always even or odd. Furthermore, there are no sign changes in its sequence of coefficients, and all even respectively odd coefficients between the lowest nonzero one and the leading one are nonzero. Proof. Since P (z) = P (z) for all complex z, the zeros occur in conjugate complex pairs. Because all roots are assumed to be purely imaginary, the roots occur in pairs of zi and zi = −zi , with the same multiplicity. Hence the polynomial is even or odd. P (x) = xp [ap + ap+2 x2 + · · · + an xn−p ] = xp Q(x2 ) (3.2) where n − p is even and deg Q = (n − p)/2. The polynomial Q has only real negative zeros. Hence by Corollary 1.4 to the rule of Descartes, its sequence of coefficients has no sign changes. Proposition 3.1. An odd or even real polynomial (3.2) has at least as many roots in the right half plane as there are sign changes in the sequence ap , ap+2 , ap+4 , . . . an of its coefficients. On the other hand, the number of roots on the positive imaginary axis is at most the number q− of sign changes in the alternated sequence ap , −ap+2 , ap+4 , · · · ± an of its coefficients. In both assertions, the differences are even. Put together, we even get (3.3) 2 2 2 2 c+ ≤ m − q− ≤ N (P, {z : Re z > 0}) ≤ N (P, {z : Re z ≥ 0}) ≤ c0 + m + q− ≤ n − c+ 4 (3.4) N (P, {z : Re z = 0}) ≤ c0 + 2q− Proof. Let n = c0 + 2m and p = c0√and R(z) := z −p P (z). Since the polynomial R(z) is even, the polynomial Q(z) := R( z) is well defined, and obviously Q(0) = R(0) 6= 0 and deg Q = m. 73 The mapping z → w = z 2 is a bijection of the right half plane {x + iy : x > 0} onto the slitted plane C \ (−∞, 0]. Hence the zeros of polynomial R(z) in the right half plane are mapped to the zeros of w 7→ Q(w) = R(z) in the slitted plane. We use—for polynomial Q—the Fundamental Theorem of Algebra, and Descartes’ rule of signs, and Lemma 1.4 on the sum of sign changes. Hence we conclude N (P, {z : Re z > 0}) = N (R, {z : Re z > 0}) = N (Q, C \ (−∞, 0]}) 2 2 = m − N (Q, (0, ∞)) ≥ m − q− ≥ q+ = c+ The mapping z → w = z 2 is a bijection of the positive imaginary axis to the negative real axis. Again counting zeros of polynomial z → R(z) and zeros of w 7→ Q(w) = R(z), we get 2 N (P, {iy : y > 0}) = N (R, {iy : y > 0}) = N (Q, (−∞, 0)}) ≤ q− 4 N (P, {z : Re z = 0} = c0 + 2N (Q, (−∞, 0)}) ≤ c0 + 2q− 10 Problem 3.5. Is it true or not. For a real polynomial, the number of real or complex zeros with positive real part is less or equal the number of sign changes in the sequence an , an−1 , . . . , a0 ; with the zero terms deleted. Remark. This assertion is wrong. Here are examples, which show: The number of real or complex zeros of P (x) with positive real part can be less, equal or even greater the number of sign changes in the sequence an , an−1 , . . . , a0 ; with the zero terms deleted. In each case, the difference is even. P (x + 1)4 + 64 (x − 1)4 + 64 (x − 1)4 − 1 (x + 2)((x − 1)2 + 1) distributed zeros x + 4x + 6x + 4x + 65 1 ± 2i, −3 ± 2i x4 − 4x3 + 6x2 − 4x + 65 3 ± 2i, −1 ± 2i x4 − 4x3 + 6x2 − 4x 0, 2, 1 ± i 3 x − 2x + 4 1 ± i, −2 4 3 2 H+ 2 2 3 2 c+ 0 4 3 2 10 Problem 3.6. Is it true or not. For a real polynomial, the number of real or complex zeros with positive real part is at least half the number of sign changes in the sequence an , an−1 , . . . , a0 ; with the zero terms deleted. Remark. This assertion is wrong. Here are examples, which show: 74 The number of real or complex zeros of P (x) with positive real part can be less, equal, or greater than half the number of sign changes in the sequence an , an−1 , . . . , a0 ; with the zero terms deleted. P (x − 3)((x + 1)2 + 9) (x − 1)(x2 + 1) (x − 1)4 + 64 (x + 1)4 + 64 distributed zeros x3 − x2 + 4x − 30 3, −3 ± i x3 − x2 + x − 1 1, ±i 4 3 2 x − 4x + 6x − 4x + 65 3 ± 2i, −1 ± 2i x4 + 4x3 + 6x2 + 4x + 65 1 ± 2i, −3 ± 2i H+ 1 1 2 2 c+ 3 3 4 0 10 Problem 3.7. List all examples mentioned up to now—and some more interesting ones. 75 P P distributed only real zeros zeros only imaginary zeros x2 + 1 x2 + 1 even or odd polynomial ±i (x2 + 1)2 − 2 x4 + 2x2 − 1 (x2 + 1)(x4 − 6x2 + 25) x6 − 5x4 + 19x2 + 25 (x2 − 1)2 + 1 x4 − 2x2 + 2 (x2 + 1)(x4 − 2x2 + 2) x6 − x4 + 2 (x − 3)((x + 1)2 + 9) x3 − x2 + 4x − 30 (x − 1)(x2 + 1) x3 − x2 + x − 1 (x − 1)(x − 2)(x2 + 1)3 x8 − 3x7 + 5x6 − 9x5 + 9x4 − 9x3 + 7x2 − 3x + 2 (x + 2)((x − 1)2 + 1) x3 − 2x + 4 ((x − 2)2 + 2)((x + 1)2 + 2) x4 − 2x3 + x2 + 18 (x + 1)4 + 64 x4 + 4x3 + 6x2 + 4x + 65 (x − 1)4 − 1 x4 − 4x3 + 6x2 − 4x (x − 1)4 + 1 x4 − 4x3 + 6x2 − 4x + 2 (x − 1)4 + 4 x4 − 4x3 + 6x2 − 4x + 5 ((x − 1)4 + 4)2 x8 − 8x7 + 28x6 − 56x5 + 78x4 − 88x3 + 76x2 − 40x + 18 x2 (x − 2)2 + 1 x4 − 4x3 + 4x2 + 1 76 ± p√ p√ 2 − 1, ±i 2+1 ±2 ± i, ±i √ ± 1±i √ ±i, ± 1 ± i 3, −3 ± i 1, ±i 1, 2, (±i)3 1 ± i, −2 √ √ 2 ± i 2, −1 ± i 2 1 ± 2i, −3 ± 2i 2, 1 ± i, 0 1+ (±1±i) √ 2 2 ± i, ±i (2 ± i)2 , (±i)2 1± √ 1±i H+ c+ c+ c+ 0 0 0 0 ≥ c+ c+ 1 1 2 2 2 2 2 2 1 3 1 3 2 8 2 2 2 2 2 0 3 3 4 4 2 4 4 8 2 2 H− c− c+ c− 0 0 0 0 ≥ c+ c+ 1 1 0 2 2 2 2 2 2 0 0 0 0 0 1 1 2 0 2 4 0 0 0 0 0 0 4 0 2 0 Lemma 3.1. The number of positive imaginary zeros of a real polynomial is at most the number of the sign changes q− (odd) and q− (even) in the nonempty alternating sequences a0 , −a2 , a4 , −a6 , . . . a1 , −a3 , a5 , −a7 , . . . of all nonzero coefficients of one parity, either odd or even. Proof. Let P (x) + σP (−x) = xp Qσ (x2 ) with parity σ = ±1. If P (iy) = 0 with y > 0, then P (−iy) = 0, and hence Qσ (−y 2 ) = 0. Hence Descartes rule implies N (P, (0, i∞)) ≤ N (Q, (−∞, 0)) ≤ q− (σ) as long as Q is not identically zero. This happens for σ = −1 and P even, as well as for σ = 1 and P odd. Otherwise, one can use both parities. In general, one gets two inequalities for both parities. 10 Problem 3.8. Is it true or not. For a real polynomial, the number of real or complex zeros with positive or negative real part is at least twice the minimum of the number of sign changes in the two sequence (±1)n an , (±1)n−1 an−1 , . . . , a0 ; with the zero terms deleted. Conjecture 1 (Sleepy Descartes Conjecture). The nonnegative integer numbers c0 , c+ , c− , N+ , N− , H+ , H− characterizing a real polynomial of degree deg P can take any values compatible with the restrictions 2 2 (3.5) N+ ≤ c+ , , N− ≤ c− (3.6) N+ ≤ H+ , N− ≤ H− , (3.7) c+ + c− + c0 ≤ deg P , (3.8) H+ + H− + c0 ≤ deg P (3.9) 2 min(c+ , c− ) ≤ H+ + H− 2 2 2 2 2 Proof of the sleepy Descartes conjecture. 3.3 Estimation of zeros with Rouché’s Theorem 10 Problem 3.9. Assume that (3.10) |an−1 | > n−2 X |ak | + |an | k=0 P k Use the Theorem of Rouché to show that the polynomial P (z) = n−1 k=0 ak z has exactly n − 1 zeros of absolute value less than one. Under the additional assumption an 6= 0, the polynomial has exactly one zero with absolute value greater than one. 77 Solution. We use the Theorem of Rouché for the functions P and Q(z) = an−1 z n−1 in the unit disk B(1, 0) = {z ∈ C : |z| ≤ 1}. Indeed, on the boundary of this disk, we get |z| = 1, and n−2 X X |P (z) − Q(z)| = ak z k ≤ |ak | + |an | < |an−1 | = |Q(z)| k6=n−1 k=0 All assumptions of Rouché’s Theorem are satisfied, and the first assertion follows. The second assertion can for example be shown by applying the Theorem of Rouché to the functions f (z) = z n P (z −1 ) = a0 z n + · · · + an−1 z + an —also called the mirror polynomial of P — and the function g(z) = an−1 z, again in the unit disk B(1, 0). Since again |f (z) − g(z)| < |g(z)| on the boundary ∂B(1, 0), both functions f and g have exactly one zero inside the unit disk. Under the assumption an 6= 0, we get z 0 6= 0 from f (z 0 ) = 0. Hence P (z 0−1 ) = z 0−n f (z 0 ) = 0 and |z 0−1 | > 1 yields the n-th zero outside the unit disk. 3.4 Perron’s irreducibility criterium Proposition 3.2 (The Perron criterium). For a monic polynomial to be irreducible in the ring Z[z], it is sufficient that all its constant coefficient a0 6= 0 and the second highest coefficient dominates (3.11) |an−1 | > n−2 X |ak | + 1 k=0 Corollary. An an integer monic polynomial with a0 6= 0 and exactly one root of absolute value greater than one and no root of absolute value equal to one, is irreducible. The Proposition follows from the Corollary. Indeed, by Problem (3.2), the polynomial has n − 1 zeros of absolute value in 0 < |z| < 1 and one zero of absolute value larger one. Proof of the Corollary. Assume toward a contradiction that there would exist a rational factorization. By Gauss’ Lemma, we would get a factorization even into monic integer polynomials P (z) = B(z)C(z) with deg(B) = s ≤ n − 1 and deg(C) = t ≤ n − 1. Furthermore, we can choose C to be the irreducible factor with the unique zero of absolute value larger one. Hence B(z) would have only zeros z1 , z2 , . . . zs of absolute value in (0, 1). Since P is assumed to be monic, B and C are both monic, too. Furthermore a0 = b0 c0 , and Viëta’s formula implies (3.12) 0 6= |b0 | = |z1 | · |z2 | · · · |zs | < 1 78 which cannot be true for any integer polynomial. The only way out of this contradiction is that B = 1 and deg(B) = 0. Hence the polynomial P is irreducible. 10 Problem 3.10. Try to generalize Perron’s criterium to polynomials which are not monic. 11 3.5 The limiting case This subsection investigates zeros and reducibility for the polynomial P (z) = in the limiting case (3.13) |an−1 | = n−2 X Pn k=0 ak z k |ak | + |an | k=0 Here is a main result. Proposition 3.3. Assume the equality (3.13), and additionally 0 6= |an | ≤ |a0 | for n ≥ 3, or 0 6= |an | < |a0 | for n =P 1 or 2. Then the polynomial P (z) = nk=0 ak z k has exactly one zero of absolute value greater than one and at most one simple zero of absolute value equal one. 10 Problem 3.11. Find an integer polynomial satisfying equality (3.13), and additionally |an | > |a0 | = 6 0 and n ≥ 3, which has no zero of absolute value greater than one. Can it be even reducible? 10 Problem 3.12. Assume the equality (3.13). Use the logarithmic residue Theorem to show that the polynomial P has at most n − 1 zeros of absolute value less one—even counting multiplicities. Why can we not claim existence of exactly n − 1 zeros of absolute values less one? Show there is at most one zero with absolute value greater one. Solution of Problem 3.12. Let ε > 0 be small enough such that the circle ∂B(1 − ε, 0) incloses all zeros P (z) = 0 with |z| < 1. Choose α > 0 small enough such that α|Q(z)| < |P (z)| on ∂B(1−ε, 0). The Theorem of Rouché 4.2 implies N (P, B(1−ε, 0)) = N (P + αQ, B(1 − ε, 0))— equal number of zeros for the functions P and P + αQ inside the disk B(1 − ε, 0). Since the coefficients of the polynomial P + αQ satisfy the strict inequality (3.9), we know from the solution of Problem 3.9 that this polynomial has n − 1 zeros of absolute value less than one. Hence N (P + αQ, B(1 − ε, 0)) ≤ n − 1. Together, we conclude that N (P, B(1 − ε, 0)) = N (P + αQ, B(1 − ε, 0)) ≤ n − 1, as claimed. We cannot claim existence of exactly n − 1 zeros of P with absolute values less 11 I did not have any success. 79 one. Indeed, in the limit α → 0, some of the zeros of P + αQ can move from inside to the boundary ∂B(1, 0) of the unit disk. We show there is at most one zero with absolute value greater one. Let ε > 0 be small enough and R > 1 large enough such that annulus A := B(R, 0) \ B(1 + ε, 0) contains all zeros P (z) = 0 with |z| > 1. Choose α > 0 small enough such that α|Q(z)| < |P (z)| on its boundary ∂A. The Theorem of Rouché 4.2 and the result of Problem 3.9 imply N (P, A) = N (P + αQ, A)) ≤ 1, as claimed. 10 Problem 3.13. Assume once more the equality (3.13), and additionally Pn a0 6= 0k and an 6= 0. Show by elementary considerations that the polynomial P (z) = k=0 ak z has at most one zero of absolute value one. Indeed, if all coefficients ak are real, this zero can possibly only be either 1 or −1. Proof. Assume there exists a zero with |z| = 1. The estimate X X X n−1 k |ak | = |an−1 | |ak z k | = |an−1 | = |an−1 z | = ak z ≤ k6=n−1 k6=n−1 k6=n−1 contains all equalities. Equality in the triangle inequality implies that all summands are nonnegative multiples of any nonzero term. Hence |ak | a0 for all k 6= n − 1 |a0 | X a0 X ak z k = − an−1 z n−1 = − |ak | |a 0| k6=n−1 k6=n−1 a 0 an−1 z n−1 = − |an−1 | |a0 | a0 |an | an z n = |a0 | ak z k = which rules out—as possible zero—all but one value (3.14) z1 = − |an | an−1 an |an−1 | In the case of real coefficients, this leaves only either +1 or −1 for a possible zero on the unit circle. Rather obviously (z − 1)2 = z 2 − 2z + 1 is a polynomial the coefficients of which satisfy equality (3.13), but has a double zero. 10 Problem 3.14. Assume once more the equality (3.13), and additionally 0 6= |an | ≤ |a0 | and n ≥ 3, or 0 6= |an | < |a0 | for n = 1 orP2. Use the previous Problems 3.12 n k and 3.13, and show that the polynomial P (z) = k=0 ak z has exactly one zero of absolute value greater than one. 80 Solution of Problem 3.14. Assume—towards a contradiction—|z| ≤ 1 for all zeros. Viëta’s formula, and the assumption would imply |a0 | = |an | |z1 | · |z2 | · · · |zn | ≤ |an | ≤ |a0 | Hence there is equality everywhere. All zeros would have absolute values one. By Problem 3.13, they need all to be equal to the only one possible value z1 , given by formula (3.14). Hence P (z) = an (z − z1 )n n ak = an (−z1 )k for all k = 0, . . . , n k n X 2|an−1 | = 2n|an | = |ak | = |an |2n k=0 Hence n = 1 or n = 2 and P (z) = an (z − z1 )n , which we ruled out by assumption. With these steps, we have proved Proposition 3.3. The Corollary 3.4 immediately implies Proposition 3.4. A monic integer polynomial P ∈ Z[z] satisfying the weak inequality (3.15) |an−1 | = n−2 X |ak | + 1 k=0 and additionally 0 6= a0 can either (i) be irreducible, (ii) or be equal to P = (x + 1)2 or P = (x − 1)2 , (iii) or have a factorization P = QR with Q = x + 1 or Q = x − 1, the second factor R being irreducible. 10 Problem 3.15. Assume that the polynomial P ∈ R[z] satisfies equality (3.13), and additionally that ak > 0 for all k = 0, 1, . . . , n. Let n ≥ 2. How many positive zeros does P have. What follows from Descartes’ rule of signs about the negative zeros? Solution. The sequences of coefficients a0 , a1 , . . . an has 2 sign changes. By the rule of Descartes number of positive zeros can only be 0 or 2. Since P (1) = 0, there are exactly two positive zeros, counting multiplicity: N (P, (0, ∞)) = 2. The sequences of coefficients a0 , −a1 , . . . (−1)n an has n − 2 sign changes. By the rule of Descartes number of negative zeros can only be N (P, (−∞, 0)) ≤ n − 2, where the missing ones give pairs of conjugate complex zeros. 10 Problem 3.16. Assume that the polynomial P ∈ R[z] satisfies equality (3.13), (z) and additionally that an > 0, and P (1) = 0. Divide Pz−1 and calculate P 0 (1). 81 Proof. All ak ≥ 0 for k 6= n − 1, but an−1 < 0. Hence R(z) := n−2 X k X P (z) = an z n−1 − aj z k z−1 k=0 j=0 R(1) = an − n−2 X k X n−2 X aj = an − (n − 1 − k)ak k=0 j=0 = n X k=0 kak = P 0 (1) k=0 P Putting rk := kj=0 aj for k = 0, . . . , n − 2, and rn−1 := an —and replacing n − 1 by n yields the following result: Corollary. Any integer polynomial n R(z) = rn z − n−1 X rk z k with 1 ≤ rn ≤ r0 ≤ r1 ≤ r2 ≤ · · · ≤ rn−1 k=0 has exactly one zero of value greater than one, and n − 1 zeros of absolute value less than one. Furthermore, if rn = 1, then R is irreducible. Proof. The result is clearly true for n = 1. Let n ≥ 2, and define P (z) = (z − 1)R(z). The polynomial P ∈ Z[z] satisfies 1 ≤ an+1 ≤ a0 and has degree deg(P ) = n + 1 ≥ 3. It is left to the reader to check that equality (3.13) holds, now with n replaced by n + 1. Hence Proposition 3.3 applies. As a conclusion, we see once more that P has exactly one zero of absolute value greater than one, and one simple zero of absolute value equal one. Hence the polynomial R has one zero of absolute value greater than one, and n − 1 zeros of absolute value less than one. By Corollary 3.4, R is irreducible, provided that it is monic. 4 4.1 The Residue Theorem and its Consequences The residue theorem and the logarithmic residue theorem Main Theorem (The residue theorem). Let Ω ⊆ C be any domain and γ be a closed piecewise smooth path in Ω. Let f be a function that is analytic in Ω except for isolated singularities. Assume that the path γ is contractible to a point inside Ω, and assume that the function f has no singularities on the path γ. Then I X 1 (4.1) f (z) dz = n(γ, aj )Res (f, aj ) 2πi γ j 82 where n(γ, aj ) are the winding numbers of the path γ around the singularities aj of the function f and Res (f, aj ) are the residues at them. Corollary (The residue theorem in the simple form). Let γ be a simple closed piecewise smooth curve, and let Ω ⊆ C be a domain that contains both γ and the interior domain D surrounded by γ. Let f be a function that is analytic in Ω except for isolated singularities. Assume that the function f has no singularities on the path γ. Then I X 1 f (z) dz = Res (f, aj ) (4.2) 2πi γ j where the sum runs over the singularities aj of the function f in the interior domain D of γ. Definition 4.1 (Multiplicity of singularity). For any meromorphic function f with either an isolated singularity or zero at z0 , the multiplicity µ(f, z0 ) is defined to be the multiplicity of the zero, or the negative multiplicity of the pole, respectively. In other words, the Laurent expansion of f at z0 has a leading term a(z − z0 )µ(f,z0 ) with a 6= 0. Theorem 4.1 (The argument principle (logarithmic residue theorem)). Let Ω ⊆ C be any domain and γ be a closed piecewise smooth path in Ω. Let f be a nonzero meromorphic function in Ω. Assume that the path γ is contractible to a point inside Ω, and assume that the function f has no zeros or poles on the path γ. Then I 0 X X f (z) 1 dz = n(γ, aj )µ(f, aj ) − n(γ, pk )µ(g, bk ) (4.3) 2πi γ f (z) j k Here n(γ, aj ) is the winding number of the path γ around the singularity aj , whereas µ(f, aj ) is the multiplicity of the singularity or zero. Remark. In the two sums, zeros and poles are counted according to their multiplicities, which is negative for the poles. Reason for the logarithmic residue theorem. If f has a pole or zero at z0 , the Laurent expansions in a neighborhood with small |z − z0 | are f (z) = a(z − z0 )µ + b(z − z0 )µ+1 + . . . f 0 (z) = aµ(z − z0 )µ−1 + b(µ + 1)(z − z0 )µ + . . . f 0 (z) µ b = + + ... f (z) z − z0 a 0 Hence the residue Res ( ff , z0 ) = µ = µ(f, z0 ) is just the multiplicity of the zero or pole of f . Now the result follows from the residue Theorem. 83 Corollary (The logarithmic residue theorem in its simple form). Let γ be a simple closed piecewise smooth curve, and let Ω ⊆ C be a domain that contains both γ and the interior domain D surrounded by γ. Let f be a meromorphic function in Ω. Assume that the function f has no zeros or poles on the path γ. Then the integral I 0 f (z) 1 dz = N − P (4.4) 2πi γ f (z) equals the number N of zeros of f minus the number P of poles in the interior domain of γ. Both zeros and poles have to be counted according to their multiplicities. 4.2 Comparing two functions Proposition 4.1 (Comparison at the zeros of two functions). Let Ω ⊆ C be any domain and γ be a closed piecewise smooth path in Ω. Let f, g be any meromorphic, and h be any holomorphic functions in Ω. Assume that the path γ is contractible to a point inside Ω, and assume that the functions f and g have no zeros or singularities on the path γ. With the quotient function F := fg , we get (4.5) 1 2πi I h(z) γ X X F 0 (z) dz = h(aj )n(γ, aj )µ(f, aj ) − h(bk )n(γ, bk )µ(g, bk ) F (z) j k Here n(γ, aj ) is the winding number of the path γ around the singularity aj , whereas µ(f, aj ) is the multiplicity of the singularity or zero. Corollary (Comparison at the zeros in the interior domain). Let γ be a simple closed piecewise smooth curve, and let Ω ⊆ C be a domain that contains both γ and the interior domain D surrounded by γ. Let f and g be two holomorphic functions in Ω, and assume that f and g have no zeros on the path γ. Put F := fg . (4.6) 1 2πi I γ F 0 (z) dz = N (f, D) − N (g, D) F (z) is the number N (f, D) of zeros of f in the interior domain of γ, counted according to their multiplicities—minus N (g, D), the number of zeros of the denominator g. Main Theorem (Theorem of Rouché). Let γ be a simple closed piecewise smooth curve, and let Ω ⊆ C be a domain that contains both γ and the interior domain D surrounded by γ. Let the two holomorphic functions f and g be defined in Ω, and assume that they satisfy the strict inequality |f (z) − g(z)| < |f (z)| + |g(z)| for all z ∈ γ on the path. Then the equality N (f, D) = N (g, D) holds for the number of zeros of f , and g, respectively, in the interior domain of γ, counted according to their multiplicities. 84 Remark. The more traditional form of the theorem makes the stronger assumption |f (z) − g(z)| < |f (z)| on the path z ∈ γ. Theorem 4.2 (Theorem of Rouché in general form). Let Ω ⊆ C be any domain and γ be a closed piecewise smooth path in Ω. Assume that the path γ is contractible to a point inside Ω. Let the two meromorphic functions f and g be defined in Ω, and assume that they have no poles and satisfy the strict inequality |f (z) − g(z)| < |f (z)| + |g(z)| for all z ∈ γ on the path. Then the equality X X (4.7) n(γ, aj )µ(f, aj ) = n(γ, bk )µ(g, bk ) j k holds. Here n(γ, aj ) is the winding number of the path γ around the singularity aj , whereas µ(f, aj ) is the multiplicity of the singularity or zero. The proof depends on the following more general Proposition. Proposition 4.2. As in the general Theorem of Rouché 4.2, let Ω be any domain, f, g be any nonzero meromorphic functions in Ω with no poles or zeros on γ. Let γ be a closed piecewise smooth path which is in Ω contractible to a point. Put F := fg and assume that the path F (γ) is contractible to a point in the domain C \ {0}. In other words, we assume that the winding number n(F (γ), 0) = 0. Then the same equality (4.7), as claimed in the general Theorem of Rouché 4.2 holds. Completing the proof for the Theorem of Rouché. On the path γ, the quotient F := f satisfies |F (z) − 1| < |F (z)| + 1. Hence F (z) ∈ C \ (−∞, 0], which is a simply g connected domain. Hence the image path F (γ) is contractible to a point inside the domain C \ {0} ⊃ C \ (−∞, 0] ⊃ F (γ). Hence the winding number is n(F (γ), 0) = 0. Hence the substitution rule yields I 0 I 1 F (z) 1 dw N (f, D) − N (g, D) = dz = =0 2πi γ F (z) 2πi F (γ) w The equality N (f, D) = N (g, D) shows that f and g have the same number of zeros in the interior domain of γ, counting their multiplicities. 85