Perfect Numbers in Quadratic Rings of Integers

Transcription

1 Julius-Maximilians-Universität Würzburg Fakultät für Mathematik und Informatik Institut für Mathematik Master Thesis Perfect Numbers in Quadratic Rings of Integers Author: Patrick Giron Supervisor: Prof. Dr. Jörn Steuding October 3, 017

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3 Contents Abstract 1 1 Perfect Numbers - an overview Essentials 6 3 Perfect Numbers in the Ring of Gaussian Integers 9 4 Perfect Numbers in the Ring of Eisenstein Integers 0 5 Perfect Numbers in the Ring of Kleinian Integers 33 6 Conclusion 38 iii

4 Abstract In 1961, Spira dened a sum-of-divisors function for the Gaussian integers and generalized the term of perfectness. Based on this paper, McDaniel showed an analogue to the Euler-Euclid-Theorem. These ideas encouraged Hunt, Parker and Rushall as well as Mena to transfer the concepts to the ring of Eisenstein integers. In this paper, we follow these authors and their ideas. First, we give a short overview of the history of perfect numbers. We dene the necessary terms and show the most important results (Chapter 1). Then, we are concerned with the essentials of algebra and number theory, which we need to work on the rings of integers of imaginary quadratic elds (Chapter ). The next section is devoted to Spira's and McDaniel's papers about perfect numbers in the ring of Gaussian integers. Additionally, we follow Ward, who found a proof for the form of odd perfect Gaussian integers (Chapter 3). Subsequently, we elaborate parallels in the ring of Eisenstein integers by following Hunt, Parker and Rushall, respectively Mena. Afterwards, we state a theorem for the form of odd perfect Eisenstein integers (Chapter 4). Finally, we attempt to nd similarities in the ring of Kleinian integers (Chapter 5). 1

5 1 Perfect Numbers - an overview The ancients believed that there is a magical and mystical power in numbers. Perfect numbers fascinated mathematicians from ancient times down to the present day. The Pythagoreans had the motto "All is Number" and they recognized that the only integers less than 10,000 that equal the sum of their proper divisors are 6, 8, 496, and = = = = The Greek mathematicians called these numbers ideal, complete or perfect. Denition 1.1. An integer n is perfect if it is equal to the sum of its positive divisors excluding n itself. Early denitions were not in terms of divisors but in terms of aliquot parts, which are the proper quotients of the number [1]. In modern language one can use a useful and well-known number-theoretical function: Denition 1.. The sum-of-divisors function σ(n) denotes the sum of positive divisors of n. Hence, σ(n) = d n d. Thus, perfect numbers are the positive integer solutions to the equation σ(n) = n. The following proposition shows important properties of the sum-of-divisors function. Proposition 1.3. Let n be any integer and let p be prime. Then the following holds: (i) σ(p) = p + 1 (ii) σ(p n ) = pn+1 1 p 1 (iii) If m and n are relatively prime, then σ(m n) = σ(m) σ(n). Proof. (i) is clear, since p is prime and the only positive integers that divide p are p itself and 1. (ii) By geometric sum formula σ(p n ) = n j=0 pj = pn+1 1. p 1 (iii) Since m and n are relatively prime, there exists an bijection between the pairs of divisors d m, d n and the divisors dd mn. Thus, σ(m) σ(n) = ( d )( d) = dd = σ(m n). d m d n dd mn

6 The probably rst recorded result about perfect numbers can be found in Euclid's "Elements Book IX", which was written around 300 BCE. Theorem 1.4. If k 1 is prime for an integer k > 0, then n = k 1 ( k 1) is perfect. Proof. Suppose k 1 is prime. By the properties of σ we know σ(n) = σ( k 1)σ( k 1 ). Note that σ( k 1) = k and σ( k 1 ) = k 1 = 1 k 1. We obtain that σ(n) = n and n is a perfect number. Around 100 CE Nicomachus wrote his renowned text "Introductio Arithmetica" where he classied numbers based on the concept of perfect numbers. He stated certain assumptions about perfect numbers without any attempt of proof, therefore, it is not surprising that some of them are wrong or still unproved. Unfortunately, they were taken as fact for many centuries. Perfect numbers also played a role in the early religious discourse of Christianity. In the Bible, it is written that God took six days to create the Earth and it was said that he chose the lunar cycle to be 8 days long, because both numbers 6 and 8 are perfect. "Saint Augustine ( ) writes in his famous text The City of God :- Six is a number perfect in itself, and not because God created all things in six days; rather, the converse is true. God created all things in six days because the number is perfect..." (cf. [1]). Another group that was interested in perfect numbers were the Arab mathematicians. Thabit ibn Quarra considered numbers of the form n p, where p is prime, and he investigated when they can be perfect. A partial converse to Euclid's result was proved by Ibn al-haytham. In his treatise, Ibn Fallus ( ) listed ten numbers which he claimed to be perfect. Not all of the numbers were perfect, but the list contained the rst seven perfect numbers. In European Mathematics of that time no one was aware of the results of the Arabs. The fth smallest perfect number was not discovered until the 15th century. Famous mathematicians of the Renaissance like Descartes, Fermat and Mersenne were concerned with perfect numbers. In 1644, Mersenne published his "Cogitata physica mathematica", where he gave a rst list of special prime numbers, which are of particular importance for the structure of perfect numbers. Denition 1.5. Numbers of the form n 1 are called Mersenne numbers. If n 1 is prime, then we say it is a Mersenne prime. Euclid's result shows an interesting connection between perfect numbers and Mersenne primes. Denition 1.6. Let n N and T n = n(n+1). We call T n the n-th triangular number. Corollary 1.7. If n is a Mersenne prime, then the n-th triangular number is perfect. It was not until the 18th century, however, that the next signicant contribution was made. In 173, Euler found the eighth smallest perfect number 30 ( 31 1) and in two unpublished manuscripts he proved the converse of Euclid's Theorem. 3

7 Theorem 1.8. Every even perfect number n is of the form n = k 1 ( k 1), where k > 1 is an integer and k 1 is prime. Proof. Suppose n is even and perfect. We can write n = k 1 m with an integer k > 1 and an odd integer m. By the multiplicativity of σ and since k 1 and m are coprime, we get σ(n) = σ( k 1 m) = σ( k 1 )σ(m). We also know σ(n) = n = k m and thus obtain the equation k m = ( k 1)σ(m), (1) so k 1 divides k m, hence k 1 divides m, say m = ( k 1) m. By (1) it follows that k m = σ(m). As m and m both divide m we obtain k m = σ(m) m + m = k m. So m must be prime, and m equals 1. We have shown the converse in the theorem above and hence every even perfect number is of the form n = k 1 ( k 1), where k 1 is prime. Euler did not only consider even perfect numbers, but he also showed of which form an odd perfect number has to be []. Theorem 1.9. An odd perfect number n has to be of the form n = q 4k+1 m, where q 1 (mod 4) is prime, k N and m N is odd and coprime to q. Proof. Let n be an odd perfect number with prime factorization n = p e 0 0 p e 1 1 p e p er r. Consequently, σ(n) = n. Since the p i are pairwise distinct for i = 0, 1,,..., r, we obtain σ(n) = r i=0 σ(p e i i ) = n. () This implies σ(n) (mod 4), hence there is exactly one i 0 such that σ(p e i 0 i 0 ) is even. If there is more than one such i 0, then it contradicts (). Without loss of generality let i 0 = 0, thus σ(p e 0 0 ) (mod 4). For the remaining p e i i we have to distinguish two cases. First, we consider p i 1 (mod 4). Then σ(p e i i ) = 1 + p i + p i + p 3 i p e i i { 0 (mod 4) if e i odd, 1 (mod 4) if e i even. = 1 + ( 1) + ( 1) + ( 1) ( 1) e i 4

8 So p 0 has to be 1 (mod 4), and if p i 3 (mod 4), e i has to be even since σ(n) 0 (mod 4). Secondly, we consider p i 1 (mod 4). Then ei σ(p e i i ) 1 e i + 1 (mod 4). j=0 This implies e 0 1 (mod 4), because p 0 1 (mod 4). From () follows that e i has to be even for i > 0. This shows that n is of the form n = q 4k+1 p β 1 1 p β p β 3 3 p βr r = q 4k+1 m, where q 1 (mod 4). To the present day no odd perfect number has been found, but exhaustive computer researches reveal that an odd perfect number has to exceed [3]. A more detailed view of the history of perfect numbers can be found on [1]. 5

9 Essentials The gentle reader should have background knowledge in algebra and algebraic number theory (most of the results that we use can be found in textbooks like [4],[5] and [6]). Nevertheless, we will take a closer look at some basic denitions we need to generalize the term of perfect numbers. In order to understand objects in more complex rings than Z, we have to generalize some objects and processes from Z to any ring R. Denition.1. Let α be an element in some ring R. If there exist some β R such that α β = 1, then we call α a unit in R. Denition.. Let R be a ring and α, β R. If there exists a γ R such that α γ = β, then we say α divides β and we write α β. Denition.3. We say α in a Ring R is prime if, whenever α βγ for some β, γ R, then α divides at least one of the factors β or γ and α is not unit. In this thesis we work on special eld extension over Q of degree. Denition.4. A number eld K of the form K = Q[ d] where d Q is not a square in Q is called quadratic eld. If d < 0, we say Q[ d] is an imaginary quadratic eld. If d > 0, we say Q[ d] is a real quadratic eld. In order to do number theory, we have to carry over the idea of integers to other number elds. Denition.5. Let α C be a root of a monic polynomial in Z[t], then we say α is an algebraic integer. Denition.6. Let K be a number eld. We dene the ring of integers of K to be the set O K = B K, where B is the set of algebraic integers. The ring of integers of quadratic elds can be of two dierent forms. Theorem.7. Let d be a squarefree integer with d 1. Let K = Q[ d]. We have if d 1 (mod 4), then O K = {a + b d : a, b Z} if d 1 (mod 4), then O K = { 1 (a + b d) : a, b Z, a b (mod )} Proof. First, we show that for any eld K = Q( d) and each x O K, there exist a, b Z such that x = 1(a + b d). Since {1, d} is a basis of the quadratic eld K, we may write x = 1(a + b d) with a, b Q. Suppose b = 0, then x = a Q. However, x is an algebraic integer, which implies x Z and hence a Z. Now we assume b 0. In this case x is not a rational number. Thus, it has a minimal polynomial of degree two. In particular, f(t) = (t 1 (a + b d))(t 1 (a b d)) = t at (a db ) 6

10 is the minimal polynomial. The coecients of f are (rational) integers, since x is algebraic, therefore a Z and 1 4 (a db ) Z. This implies db Z because d is squarefree. We still need to show b Z. Suppose b Q and b = m with m and n are coprime n integers, n > 1. We have dm = kn for some k Z. Let p n be a (rational) prime. Then p dm, but as p and m are coprime p d, which contradicts our assumption. We have shown that x O K is of the form x = 1(a + b d) for some a, b Z. Finally, we discuss in which cases this expression is algebraic. Suppose a and b are both even, then a = m and b = n for some m, n Z. Thus, x = m + n d is an algebraic integer, since B is a ring. Suppose a and b are both odd, say a = m + 1 and b = n + 1. Then x = m + n d + 1 (1 + d). So x is algebraic if and only if 1 (1 + d). The minimal polynomial of 1 (1 + d) is t t 1 4 (1 d), hence 1 (1 + d) is an algebraic integer if and only if d 1 (mod 4). Suppose a = m is even and b = n + 1 is odd. Then x = m + n d + 1 d. But 1 d B, since the minimal polynomial of 1 d is t 1 d which does not have 4 (rational) integer coecients. Suppose a = m+1 is odd and b = n is even. Then x = m + n d + 1, but 1 B. We obtain the assertion by combining the cases above. A useful tool for our studies will be the norm-function. Since we work on the integers of imaginary quadratic elds, we can specify it. Denition.8. Let K = Q( d) with d < 0 squarefree. norm-function N : O K Z is dened as Given any α O K, the N(α) = αα. It holds some useful properties. Lemma.9. Let O K be the ring of integers for an imaginary quadratic eld K. For all α O K we have N(α) 0. In particular, N(α) = 0 if and only if α = 0. Proof. Let α = a + i b, then N(α) = αα = (a + i b)(a i b) = a + b 0, since a, b Q. If α = 0, it is clear that N(α) = 0. Now let N(α) = 0 and hence αα = 0. This implies α = 0, since O K is an integral domain. Lemma.10. Let O K be the ring of integers for an imaginary quadratic eld K. If α, β O K, then N(α β) = N(α) N(β). 7

11 Proof. N(α β) = αβαβ = αβαβ = ααββ = N(α) N(β) The following propositions help us to structure and analyze integers of imaginary quadratic elds. Proposition.11. Let O K be the ring of integers for an imaginary quadratic eld K. An element α O K is a unit if and only if N(α) = 1. Proof. Let α be a unit in O K, then there exists β O K such that αβ = 1 and N(α)N(β) = 1 and since N(α), N(β) are positive integers, N(α) = 1. Conversely, if N(α) = 1, then αᾱ = 1 and ᾱ O K implies that α is a unit. Proposition.1. Let O K be the ring of integers for an imaginary quadratic eld K. An element α in O K is prime if N(α) is a rational prime. Proof. Let α O K. Assume N(α) is prime and suppose α = β γ for some β, γ O K. By the multiplicativity of the norm-function we obtain N(α) = N(β γ) = N(β)N(γ). This is a contradiction to our assumption that N(α) is prime. Further we limit ourselves to integer rings that full an important condition: unique factorization. Denition.13. An integral domain in which every non-zero element possesses a unique factorization into irreducible elements is called unique factorization domain. Remark.14. In most algebra textbooks, i.e. [4], one can nd a proof of the following inclusions: euclidean domain principal ideal domain unique factorization domain. 8

12 3 Perfect Numbers in the Ring of Gaussian Integers In two papers in 189 and 1831, Gauss formulated the law of biquadratic reciprocity [7]. There he introduced what came to be known as the Gaussian integers, which are the elements of the set Z[i] = {a + b i : a, b Z}. Since 1 1 (mod 4), this is the ring of integers of Q[i]. The Gaussian integers form a unique factorization domain, as it is a euclidean ring with the norm-function N : Z[i] Z with N(a + b i) = a + b as euclidean function. The units are ±1, ±i. We call all elements in a ring, which dier only via multiplication by a unit, associates. Thus, for every element in Z[i], there are four dierent associates. By denition associates have equal norm. Figure 1: Gaussian integers Example 3.1. The Gaussian integer 1+i is a prime in Z[i]. We apply Proposition.1: N(1 + i) = =, which is prime in Z. This is the prime of minimal norm value in Z[i], up to multiplication with a unit, since the only integer solutions of a +b = are a = ±1, b = ±1. Figure 1 shows a visualization of Z[i], where units are red, Gaussian primes are blue, and the remaining Gaussian integers are black. In the following proposition we characterize the primes of Z[i]. Proposition 3.. The prime elements of Z[i] are 9

13 (i) 1 + i and its associates, (ii) the rational primes, which are congruent to 3 modulo 4 and their associates, (iii) π and π and their associates, where N(π) = N(π) = p N and p is a rational prime congruent to 1 modulo 4. Proof. See [5, p. 19] or [6, p. 73]. This proposition points out one of the fundamental computational dierences between Z and Z[i]. The primes that are congruent to 1 modulo 4 split over Z[i], for example 13 = ( + 3i)( 3i) and 5 = ( + i)( i). These are exactly the primes which are the sum of two squares. Also the rational prime has a non-trivial factorization in the ring of Gaussian integers, = (1+i)(1 i). This fact leads to another generalization we have to make for studying perfect numbers in the Gaussian integers: Evenness. Denition 3.3. An element α Z[i] is even if it is divisible by 1 + i. Any element which is not divisible by 1 + i is said to be odd. The laws of parity also hold for the Gaussian integers. In order to prove this fact, one has to check the 16 dierent cases, which occur from the fact that a Gaussian integer a + bi is odd if and only if a and b have dierent real parity (a b (mod )). We follow Spira [8] and McDaniel [9], who came up with a generalization of perfect numbers to the Gaussian integers and their characterization. In 1961, Spira introduced a complex sum-of-divisors function σ for Gaussian integers. First, we dene our sum-of-divisors σ for powers of primes, then we extend it multiplicatively. Such a function depends on the choice of associates. A reasonable property is N(σ(π n )) N(π n ), such that the associates lie in the right half-plane. For uniqueness we will pick the associates in the rst quadrant, which is thought of containing the positive real semiaxis, but not the imaginary semiaxis (Re(π) > 0, Im(π) 0). To get a well-dened function we set σ(ɛ) = 1 for all units ɛ of Z[i]. We wish to achieve multiplicativity, in particular σ(α β) = σ(α) σ(β) for gcd(α, β) = 1, and independence of the choice of associates. Therefore, we set σ(π k ) = σ((iπ) k ) = σ(( π) k ) = σ(( iπ) k ) = a certain number dened below for Gaussian primes π. Since every element η in Z[i] has a unique factorization in prime elements, we set σ(η) = n i=1 σ(πk i ) where η = n i=1 (πk i ). To summarize it: Denition 3.4. Let η Z[i] with η = ɛ n i=1 πk i i, where ɛ is a unit of Z[i], k i N and each π i is a Gaussian prime lying in the rst quadrant, then σ : Z[i] Z[i] is a well-dened function with σ(η) = n i=0 π k i+1 i 1 π i 1. We call σ the complex sum-of-divisors function or complex σ-function. 10

14 If η only splits in powers of rational primes congruent to 3 modulo 4, the sum-of-divisors functions coincide. Nonetheless, for the real σ-function we have σ() = 3, but for the complex σ-function we get σ() = + 3i. a\b i + i 5 + 5i + 4i 6 + 8i + 6i + 3i 3 + i 5i 3 + 3i + 10i 3 + 5i i i 4 + i 8 + 4i 6 + 5i 9 + 7i 8 + 8i i 5 + i i 1 + 6i 8 + i 5 + 5i i i 3 + 9i 6 + i 10i 6 + 4i 0i 6 + 6i i 7 + i i 1 + 4i + 16i 7 + 5i 0i i 8 + i i 1i + 14i i Table 1: σ(a + bi) One could say that a Gaussian integer α is perfect if σ(α) = α. However, now that we have dened evenness for complex integers, we also should reconsider the denition of perfect numbers. Denition 3.5. For α Z[i], we say α is perfect if and only if σ(α) = (1 + i)α. The Mersenne primes play an important role for the form of even perfect numbers in Z, hence it is reasonable to carry over the concept of Mersenne numbers to the Gaussian integers. Denition 3.6. For k Z, we call M k = i[(1 + i) k 1] Z[i] complex Mersenne number, and if M k is prime, we say M k is a complex Mersenne prime. Using these denitions we obtain an analogous result to Euclid's Theorem (1.4). Theorem 3.7. Let M p be a complex Mersenne prime and let p be a rational prime congruent to 1 modulo 8, then (1 + i) p 1 M p is perfect. Proof. We have M p = i((1 + i) p 1) and σ((1 + i) p 1 M p ) = σ((1 + i) p 1 )σ( i((1 + i) p 1)) = i((1 + i) p 1)σ( i)σ((1 + i) p 1) = i((1 + i) p 1)(1 + i) p = (1 + i)(1 + i) p 1 M p. There are seventeen Gaussian Mersenne primes for p < 50, in particular p =, 3, 5, 7, 11, 19, 9, 47, 73, 79, 113, 151, 157, 163, 167, 39, and 41. However, only for three of them p 1 (mod 8), namely p = 73, 113, 41. Spira [8] and McDaniel [9] observed that the characterization of perfect Gaussian integers above, which seems both reasonable and adequate, is somewhat limited. They both could not nd a proof for the converse of Euclid's Theorem. By using the norm-function Spira had another idea to dene perfect numbers in Z[i]. 11

15 Denition 3.8. For α Z[i] and N(σ(α)) = N(α), we say α is norm-perfect. We say α is primitive if it has no (norm-)perfect number as proper divisor. McDaniel [9] characterizes the even primitive norm-perfect numbers by using the following inequalities. Theorem 3.9. Let z C with N(z) 5 and z = x + iy, x, y R, and x 1. Then, for n N, ( ) N(1 + z z n ) > N(z n x 1.4 ) 1 + N(z) and furthermore, if y x 1, then Proof. For n = 1, we obtain N(1 + z z n ) > N(z n ) ( 1 + x ). N(z) N(1 + z) = (1 + x) + y = 1 + x + x + y = N(z) ( 1 + x + 1 ). N(z) For n =, using the inequality N(z n+1 1) N(z n )N(z 1) valid for z C with Re z 1 and equality holding if and only if z = 1 (see [10, p. 140]), we obtain N(1 + z + z ) = N(z)N(z z) ) = N(z) (N(z) + x x + x + 1 y x + y ( ) N(z ) 1 + x 1, for arbitrary y, N(z) > ( ) N(z ) 1 + x+1, for y x 1. N(z) Henceforth, we write z = re iθ and we get ( ) z N(1 + z z n n+1 1 ) = N z 1 If n 3, this term is greater than = (zn+1 1)(z n+1 1) (z 1)(z 1) = N(zn+1 ) + 1 (z n+1 + z n+1 ) r x + 1 = N(z n ) r + r n r 1 n cos((n + 1)θ). r x + 1 N(z n ) r r r x

16 For N(z) 5, we can improve these inequalities. By using r 5, we compute that ( ) 1 +, N(z n ) r r r x + 1 N(zn ) x 1.4 x + y and for y x 1, we obtain that ( N(z n ) r r r x + 1 N(zn ) 1 + x ). x + y Corollary Let π be an odd Gaussian prime. Then, for k N, ( ) σ(π k ) N(π) + x 1.4 N >, π k N(π) where ɛπ = x + iy is the rst-quadrant associate. If y x 1, then ( ) σ(π k ) N(π) + x N >. π k N(π) Proof. Note that with the assumptions above and apply Theorem 3.9. σ(π t ) = 1 + (ɛπ) (ɛπ) t Spira [8] found a helpful result to show the structure of even perfect Gaussian integers. Theorem For an odd α Z[i], the Gaussian integer (1 + i) k 1 α can be perfect only if k 0, ±1 (mod 8). k mod 8 Re h h 0 h h h 0 h h = 1 k Im 0 h h h 0 h h h Table : Real and imaginary parts of (1 + i) k Proof. Let (1 + i) k 1 α be perfect, then we have σ((1 + i) k 1 α) = i((1 + i) k 1)σ(α) = (1 + i) k α. But N((1 + i) k 1) > N((1 + i) k ) if and only if k, 3, 4, 5, 6 (mod 8), and always N(σ(α)) N(α), thus N( i((1+i) k 1)σ(α)) > N((1+i) k α) for k, 3, 4, 5, 6 (mod 8), which contradicts perfectness. 13

17 Consequently, (1 + i) k 1 α can be norm-perfect only if k 0, ±1 (mod 8). From now on we write A k = N(M k ) and note that M k = i( k/ 1) and A k = k k+ + 1 if k 0 (mod 8), M k = ± k 1 ( k 1 1)i and A k = k k if k ±1 (mod 8). Lemma 3.1. Let η be norm-perfect, π be an odd prime divisor of η, and ɛπ be the rst-quadrant associate of π. (i) If k 0, ±1 (mod 8) and k 11, then N(π) > A 1 3 k, (ii) if k ±1 (mod 8), then N(π) > 0.3( k+1 1), (iii) if k ±1 (mod 8) and Re(ɛπ) 1, then N(π) > A 1 k. Proof. Let a Z such that π a ( η and ) π a+1 η. By using the multiplicativity of the complex σ-function and since N σ(δ b ) > 1 for any prime power δ b, we obtain δ b 1 = N(σ(η)) N(η) N(σ((1 + i)k 1 ))N(σ(π a )). k N(π a ) N(π) + c By Corollary 3.10, this is greater than A k where c = 0.6 if Re(ɛπ) = 1 and k N(π) c =.6 if Re(ɛπ) > 1. Now, solving for N(π) shows that N(π) > ca k k A k. (3) (i) Let k 0, ±1 (mod 8), then A k k k+ + 1 as seen above. Thus, from (3), N(π) 0.6 k k+ + 1 k+ 1 which is greater than k 3 > A 1 3 k for k 11. > 0.6( k 1), (ii) and (iii) Let k ±1 (mod 8), then A k = k k > ( k+1 1), thus, from (3), N(π) > c( k+1 1) ( k+1 1) = c ( k+1 1) > 0.3( k+1 1), and if Re(ɛπ) 1, then c k+1 ( 1) > 1.3( k+1 1) > A 1 k. 14

18 Lemma Let η = (1 + i) k 1 µ be norm-perfect, µ be odd, k ±1 (mod 8) and σ((1 + i) k 1 = ɛπϱ, where ɛ is a unit, and π and ϱ are rst-quadrant primes. If π is of the form a + bi and ϱ is of the form c + di, then at most one of the rational integers a, b, c and d is equal to 1. Proof. The case k = 1 is trivial. For k = 7, we obtain σ((1 + i) 6 ) = (8 + 7i), and N(σ((1 + i) 6 )) = = 113 is prime, hence σ((1 + i) 6 ) is prime. For k = 9, we have σ((1 + i) 8 ) = ( + 3i)(1 + 6i) and N( + 3i) = + 3 = 13 < 481 = A 1 9, which is why η is not norm-perfect by Lemma 3.1 (iii). Now we consider k 15. The complex sum-of-divisors for (1 + i) k 1 is an odd number, hence neither a and b nor c and d can both be 1. Suppose one coecient of each pair, (a, b) and (c, d), is equal to 1. Without loss of generality suppose N(π) N(ϱ). Then N(π) = 1 + g, where g is a or b, and N(ϱ) = 1 + h, where h is c or d. Since η is norm-perfect, π η or π η, which implies By Lemma 3.1 (ii), ηη = N(η) = N(σ(η)) = N(σ((1 + i) k 1 µ)) = N(ɛπϱσ(µ)) = ππn(ϱσ(µ)). 1 + g = N(π) = N(π) > 0.3( 8 1) > 76, which implies g 9. Let π = re iθ and ϱ = se iϑ, then 0 < θ < 7 or 83 < θ < 90. Furthermore, it holds that 0 < ϑ < 7 or 83 < ϑ < 90 since N(ϱ) N(π). However, for k 15, the argument of σ((1 + i) k 1 ) is within one degree of 45 if k 1 (mod 8) or within one degree of 135 if k 1 (mod 8). This contradicts our previous estimate since the argument of σ((1 + i) k 1 ) θ + ϑ (mod 90). Lemma If η = (1+i) k 1 µ is norm-perfect, with µ being an odd Gaussian integer, then M k = σ((1 + i) k 1 ) is prime and k is a rational prime congruent to ±1 modulo 8. Proof. We choose π Z[i] to be the rst-quadrant prime factor of σ((1 + i) k 1 ) of least norm. As seen in the proof above, either π η or π η. By Theorem 3.11, k 0, ±1 (mod 8). If k is a multiple of eight, say k = 8m, then A k = A 8m = ( 4m 1). Assume m = 1, then 3, which remains prime in Z[i], divides A k. Thus, 3 l η for some l 1. By using Corollary 3.10, we obtain ( ) σ(3 l ) N > = 15 3 l 3 9, 15

19 and by an easy calculation ( ) σ((1 + i) 7 (1+i) ) 8 1 (1+i) 1 N = N = 5 (1 + i) 8 (1 + i) 8 56, hence 1 = N ( ) σ(η) = N (1 + i)η ( ) σ((1 + i) 7 )σ(3 l ) (1 + i)(1 + i) 7 3 l > 1. On the contrary, if l > 1, then N(π) l 1 = k 8 1 < A 1 3 k, which contradicts Lemma 3.1 (i). Now, without loss of generality we can assume that k ±1 (mod 8). The assertion is true if k = 7, since σ((1 + i) 6 ) = (8 + 7i) is prime. If k = 9, η is not norm-perfect as seen in the proof of Lemma The case k 15 remains. We should remember that π is the prime factor of least norm of σ((1 + i) k 1 ). Now, by Lemma 3.1 (i), N(π) > A 1 3 k, thus σ((1+i) k 1 ) has at most two (not necessarily distinct) prime factors. Suppose σ((1 + i) k 1 ) is not a prime and σ((1 + i) k 1 ) = ɛπϱ, where ɛ is a unit and ϱ is the second rst-quadrant prime factor. Lemma 3.1 (iii) implies that Re(π) = 1, because if Re(π) 1, then N(π) > A 1 k and σ((1 + i) k 1 ) has to be prime. By Lemma 3.13, we get Re(ϱ) 1 since π is the prime factor of least norm, N(ϱ) > A 1 k. We know that A k = k k+1 + 1, so N(ϱ) > 0.7( k+1 1), and by Lemma 3.1 (ii), N(π) > 0.3( k+1 1). By using these inequalities and Corollary 3.10, we obtain for some c, d Z ( ) ( ) ( ) σ(η) σ((1 + i) k 1 π c ϱ d ) σ((1 + i) k 1 )σ(π c )σ(ϱ d ) 1 = N N N (1 + i)η (1 + i) k π c ϱ d (1 + i) k π c ϱ d (N(π) + 0.6)(N(ϱ) +.6) > N(πϱ) k N(π)N(ϱ) > k + 0. k k > 1. Thus, σ((1 + i) k 1 ) is prime. We still have to show that k is prime. Suppose k = qs, where q is a rational prime and q s. But then ((1+i) q 1) divides M k = σ((1+i) k 1 ), which contradicts the primality of σ((1 + i) k 1 ). This proves our claim. We proceed by putting Lemma 3.14 to good use in proving the following theorem. Theorem If η Z[i] is an even norm-perfect number, then there exists a Gaussian Mersenne prime M p such that, for some t N and an odd δ Z[i], either or (i) η = (1 + i) p 1 M t pδ, with p 1 (mod 8), (ii) η = (1 + i) p 1 M p t δ, with p 1 (mod 8). 16

20 Proof. We start with the fact that σ(η)σ(η) = ηη. This argument holds because η is norm-perfect. In Lemma 3.14, we have shown that η is of the form (1 + i) p 1 µ, where σ((1 + i) p 1 ) = M p is a Gaussian Mersenne prime and p is a rational prime congruent to ±1 modulo 8. Since M p σ(η), either M p η or M p η. Now let p 1 (mod 8), then M p = p 1 + ( p 1 )i. Suppose ((1 + i) p 1 t M p ) η for some t N. Since Im(M p ) = Re(M p ) 1, we can use Corollary 3.10 and we obtain ( ) ( ) σ(η) σ((1 + i) p 1 ) N N N η (1 + i) p 1 > N(M p ) N(M p) + p+1 p 1 N(M p ) ( σ(m p t ) M p t ) = p + 1 p 1 >. This contradicts the norm-perfectness of η, which implies that M p η. Analogously, we get M p η if p 1 (mod 8). Gathering what we have proved so far, we can nally characterize the even primitive norm-perfect numbers. Theorem Let η Z[i] be an even norm-perfect number, then either η = ɛ(1 + i) p 1 M p, where M p is a Gaussian Mersenne prime with p 1 (mod 8) and ɛ is a unit, or η = ɛ(1 + i) p 1 M p, where M p is a Gaussian Mersenne prime with p 1 (mod 8) and ɛ is a unit. Conversely, for an arbitrary Gaussian Mersenne prime M p and a unit ɛ, we have either or is a primitive norm-perfect number. η = ɛ(1 + i) p 1 M p if p 1 (mod 8), η = ɛ(1 + i) p 1 M p if p 1 (mod 8) Proof. If η is a primitive norm-perfect number, then by Theorem 3.15, η is divisible by (1+i) p 1 M p or (1+i) p 1 M p, where M p is a Gaussian Mersenne prime with p 1 (mod 8) respectively p 1 (mod 8). Hence, η is equal to a unit ɛ times one of these divisors. For the converse we suppose η = ɛ(1 + i) p 1 M p if p 1 (mod 8) and η = ɛ(1+) p 1 M p if p 1 (mod 8), then N(σ(η)) = N(σ((1 + i) p 1 )σ(m p ) = N(M p p 1 (1 + i)) hence η is primitive and norm-perfect. = N( p 1 Mp ) = N((1 + i) p 1 M p ) = N(η), 17

21 Since we have characterized the even norm-perfect numbers, we can use our results to show an analogue to Euler's Theorem (1.8). Corollary An element η of Z[i] is an even primitive perfect number if and only if there exists a rational prime p 1 (mod 8) such that η = (1 + i) p 1 M p, where M p is a Gaussian Mersenne prime. Proof. In Theorem 3.7, we have shown that all η Z[i] of the form (1 + i) p 1 M p, where M p is a Gaussian Mersenne prime, are primitive and perfect. The necessity of η being of this form is a consequence of Theorem 3.15 and Theorem As every perfect number η is norm-perfect, by Lemma 3.14 we know that η = (1+i) p 1 µ where µ Z[i] is odd and p is a rational prime congruent to ±1 modulo 8. We have σ(η) = σ((1+i) p 1 µ) = M p σ(µ) and σ(η) = (1 + i)η, and hence M p η. This implies that p 1 (mod 8), since η could not be norm-perfect if p 1 (mod 8) and M p η, as seen in the proof of Theorem By Theorem 3.15, η is divisible by (1 + i) p 1 M p, and σ(ɛ(1 + i) p 1 M p ) = (1 + i) p M p, thus η = (1 + i) p 1 M p. Another interesting result was discovered by Matthew Ward [11]. He found an analogue for Euler's characterization of odd perfect numbers in the complex case. Before we observe this case, we have to show the following lemma. Lemma Let π Z[i] be an odd prime, then σ(π m ) is even if and only if m is odd. Proof. First, we claim that for any η j Z[i] we have N( j η j) j N(η j) (mod ). Using the basic parity rules, which we have seen above, we will prove this claim. It is valid that N( j η j) 0 (mod ) if and only if j η j is even in the Gaussian sense. The second sum is even if and only if there is an even number of odd η j 's, or in other words, if there is an even number of η j 's with the property N(η j ) 1 (mod ). This proves our rst claim. Now, let π Z[i] be an odd prime. We examine N(σ(π m )) = N(1 + π π m ) N(1) + N(π) N(π m ) (mod ). Clearly, this sum is congruent to 0 modulo if and only if m is odd since each summand is odd. Similar to Euler's characterization of odd perfect numbers in the rational case, we get the following result in the complex case. Theorem Let α Z[i] be an odd norm-perfect number, then α = π k γ where k is an rational integer congruent to 1 modulo, γ Z[i] is odd, π Z[i] is prime, and gcd(π, γ) = ɛ. Proof. Since α is odd, we have α = n i=1 πk i i, where no π i is even. Thus, no π i is an associate of (1 + i). Furthermore, N(σ(α)) = N(α) because of the norm-perfectness 18

22 of α. Moreover, N(α) Z is odd, hence N(α) is the sum of two squares. This implies N(α) 1 (mod 4), so N(σ(α)) = N(α) (mod 4). We consider N(σ(α)) = n i=0 N(σ(π k i i )) = N(σ(π k 1 1 )) N(σ(π kn n )). Without loss of generality we choose π i such that N(σ(π k 1 1 )) (mod 4) and all other factors are congruent to 1 modulo 4. This holds since no term can be congruent to 0 modulo 4, as otherwise the whole product would be congruent to 0 modulo 4. Further, no N(σ(π k i i )) can be congruent to 3 modulo 4 since the norm is the sum of two squares. From Lemma 3.18 we know that k 1 has to be odd and each k i has to be even for 1 < k n. Thus, if α Z[i] is an odd norm-perfect number, it is of the form α = π k γ where k is odd and gcd(π, γ) = 1. In conclusion, the perfect Gaussian integers have to be of the form above because they are all norm-perfect. Corollary 3.0. Let α be an odd perfect Gaussian integer, then α = π k γ, where k is a rational integer congruent to 1 modulo, γ Z[i] is odd, π Z[i] is prime, and gcd(π, γ) = ɛ. An odd perfect Gaussian integer has not been found yet, but there exist odd normperfect Gaussian integers. A curiosity of this class of numbers is the number + i and its associates. They are odd norm-perfect numbers, which are also primes in the ring of Gaussian integers. Nonetheless, it is easy to show that this integer and its associates are the only norm-perfect primes. For a Gaussian prime π, we have σ(π) = 1 + π, where π lies in the rst-quadrant. Furthermore, if π is norm-perfect, then N(σ(π)) = N(π). For π = a + bi, this leads to (a + 1) + b = (a + b ), which is equivalent to (a 1) + b =. The only integer solutions of this equation are (, 1), (, 1), (0, 1) and (0, 1). Only one of these is a rst-quadrant prime, namely + i, hence + i and its associates are the only odd norm-perfect primes. 19

23 4 Perfect Numbers in the Ring of Eisenstein Integers The Gaussian integers as well as the ring of integers, which we study in this section, can be referred to Carl Friedrich Gauss. He studied polynomials of the form x n 1 and showed that a polynomial of this form has exactly n solutions in the complex plane. These solutions are e πik/n for 1 k n and they are called n-th roots of unity. We then write x n 1 = n (x e πik/n ). k=1 These polynomials will not be irreducible in general, so we restrict this product to factors such that k and n are relatively prime. We obtain the minimal polynomial of e πik/n for gcd(k, n) = 1, Φ n (x) = (x e πik/n ), 1 k n gcd(k,n)=1 which is known as n-th cyclotomic polynomial. These polynomials occur in several questions of algebraic number theory. In this paper, we consider the third cyclotomic polynomial Φ 3 (x) = 1 + x + x and focus on a specic root of it, which will be denoted as ω = e πik/n = The ring of Eisenstein integers Z[ω] is named after a student of Gauss. By Theorem.7, it is the ring of integers of Q[ 3] since 3 1 (mod 4). Eisenstein and Jacobi used this integer ring to formulate the cubic reciprocity law [7]. For the norm-function we have N(a + bω) = a ab + b for a, b Z. The units are ±1, ±ω, ±ω, meaning that for every element in Z[ω], there are six dierent associates. Example 4.1. The Eisenstein integer + ω is a prime in Z[ω]. We apply Proposition.1 and we obtain N( + ω) = = 3, which is prime in Z. This is the prime of minimal norm value in Z[ω] since a ab + b = has no integer solutions, which indicates there is no prime element with norm. The following proposition gives a characterization of the primes in Z[ω]. Proposition 4.. An α Z[ω] is an Eisenstein prime if and only if one of the following conditions hold: (i) α is the product of a unit and a rational prime congruent to modulo 3. (ii) N(α) is a rational prime. Proof. See [6, p. 69]. 0

24 Figure : Eisenstein integers Figure shows a visual depiction of the Eisenstein integers, where units are red, primes are blue, and the remaining integers are black. In Z[ω] the rational prime remains prime, but we have a prime of smaller norm, leading to another generalization of evenness: Denition 4.3. An element α Z[ω] is even if it is divisible by + ω. Any element that is not divisible by + ω is said to be odd. Figure 3: Elements α in Z[ω] with N(σ (α)) N(α) Like Spira for the Gaussian integers, the authors of [] extended the sum-of-divisors function σ on Z[ω]. As before we want our extension σ to full the property N(σ (α)) N(α). 1

25 The ring of Eisenstein integers also features unique factorization, which is why for any Eisenstein integer η we can unambiguously write η = ɛ n i=1 πk i i, where ɛ is a unit, each π i is an Eisenstein prime lying in the shaded region of Figure 3, and all k i are positive integers. For the remainder of this section, we suppose that all prime factorizations are of the form above. We assume the shaded region in Figure 3 to contain the Eisenstein integers on the lower boundary (the real axis) and not the upper boundary (dashed blue line). This area contains the Eisenstein integers a + bω if and only if a > b 0. Denition 4.4. For η = ɛ n i=1 πk i i σ : Z[ω] Z[ω] is dened as in Z[ω], the Eisenstein sum-of-divisors function σ (η) = n i=1 π k i+1 i 1 π i 1. Note that σ is multiplicative over Z[ω] and, if η Z, the sum-of-divisors functions σ and σ coincide. We have seen that Mersenne primes play a key role for characterizing even perfect numbers in Z, as well as in Z[i]. Therefore, we should dene them in a similar way for Z[ω]. Denition 4.5. Let k Z, we call M k = ω[( + ω) k 1] Z[ω] Eisenstein Mersenne number, and, if M k is prime, we say M k is an Eisenstein Mersenne prime. The Eisenstein Mersenne primes seem to be rarer than the real ones. There are only seven Eisenstein Mersenne primes with an exponent p 160, namely n =, 5, 7, 11, 17, 19 and 79, but in the same range we have twelve rational Mersenne primes. Example 4.6. We consider the Eisenstein Mersenne numbers for k = and k = 3. (i) Let k = : Now we compute the norm of M : M = ω[( + ω) 1] = ω(3 + 4ω + ω ) = 3ω 4ω ω 3 = 3 + ω. N(3 + ω) = = = 7. Thus, the norm of M is prime in Z, and therefore, M = ω[( + ω) 1] is an Eisenstein Mersenne prime.

26 (ii) Let k = 3: M 3 = ω[( + ω) 3 1] = ω[(8 + 1ω + 6ω + ω 3 ) 1] = ω( + 6ω) = ω 6ω = 6 + 4ω. It is easy to verify that M 3 can be factored in a non-trivial way over Z[ω] since both coecients are even rational integers. The norm-function also veries this: N(6 + 4ω) = = = 8. Hence, M 3 = ω[( + ω) 3 1] is not prime in Z[ω]. Since we have generalized evenness for Z[ω], we are now able to dene perfectness in a natural way. Denition 4.7. Let α Z[ω], we say α is (Eisenstein) perfect if and only if σ (α) = ( + ω)α. As for the Gaussian integers, we hope to nd an analogue to Euclid's Theorem (1.4). A key step to structure even perfect numbers is the following equation: σ(m p ) = (even prime) p for a Mersenne prime in the appropriate ring of integers. In the ring of Eisenstein integers this identity only holds if the Mersenne prime is the correct associate. In order to check for which p Z (+ω) p lies in the shaded region, we rewrite ( + ω) p in the form a + bω where a, b Z. We only have to consider p modulo 1 thanks to the periodic behaviour of a and b. p mod 1 a 3 h 3 h 3 h 3 h 0 3 h 3 h 3 h 3 h 3 h 0 3 h h = 1p b 0 3 h 3 h 3 h 3 h 3 h 0 3 h 3 h 3 h 3 h 3 h Table 3: Values of a and b such that ( + ω) p = a + bω This shows that we get the appropriate associate of M p if p 1 (mod 1). Theorem 4.8. If M p is an Eisenstein Mersenne prime and if p 1 (mod 1), then ( + ω) p 1 M p is Eisenstein perfect. 3

27 Proof. The following applies: M p = ω[( + ω) p 1] and σ (( + ω) p 1 M p ) = σ (( + ω) p 1 )σ (M p ) = ω(( + ω) p 1)σ ( ω[( + ω) p 1]) = ω(( + ω) p 1)σ ( ω)σ (( + ω) p 1) = ω(( + ω) p 1)( + ω) p = ( + ω)( + ω) p 1 M p, since the associate of ( + ω) p for p 1 (mod 1), which lies in the shaded region of Figure 3, is itself. Hence, σ (( + ω) p 1) = ( + ω) p. As mentioned in the previous chapter, this particular denition of perfectness is somewhat limited. To characterize the even Eisenstein perfect numbers we have to use the norm-function again. Denition 4.9. Let α Z[ω] and N(σ (α)) = 3 N(α), then we say α is (Eisenstein) norm-perfect. In order to show the structure of even Eisenstein perfect numbers we follow Roja [1], who has expanded the work of Spira [8] and McDaniel [9] to the ring of Eisenstein integers. Lemma Let M p be an arbitrary Eisenstein Mersenne prime and ɛ be a unit. If p 1 (mod 1), then η = ɛ(+ω) p 1 M p is a norm-perfect number. If p 1 (mod 1), then η = ɛ( + ω) p 1 M p is a norm-perfect number. Proof. For p 1 (mod 1), M p is not in the shaded region of Figure 3, but M p = (1 + ω)m p = ( + ω) p 1. If η = ɛ( + ω) p 1 M p, then N(σ (η)) = N(σ (ɛ( + ω) p 1 M p )) = N(σ (ɛ)σ (( + ω) p 1 )σ (M p )) = N(M p (1 + M p )) = N(( + ω) p M p ) = 3N(( + ω) p 1 M p ) = 3N(η). For p 1 (mod 1), M p = ωmp = ( + ω) p 1 is in the shaded region of Figure 3. Thus, for η = ɛ( + ω) p 1 M p, we obtain N(σ (η)) = N(σ (ɛ( + ω) p 1 M p )) = N(σ (ɛ)σ (( + ω) p 1 )σ (M p )) = N(M p (1 + M p )) = N(( + ω)p M p ) = 3N(( + ω) p 1 M p ) = 3N(η), since the norm of the complex conjugate of an element is equal to the norm of the element. 4

28 An arbitrary element η Z[ω] can be represented by the form η = x+iy, where x, y R. Hence, the inequalities, which we have seen in Chapter 3, still hold for Eisenstein integers. However, following Mena [1] we have to improve these inequalities. Lemma Let z C with z = x + iy, x, y R, and n N. If x 5, then 4 ( N(1 + z z n ) > N(z n ) 1 + x 1 ) N(z) and furthermore, if y x 1, then N(1 + z z n ) N(z n ) with equality holding if and only if n = 1. ( 1 + x + 1 ) N(z) Proof. We know from the proof of Theorem 3.9 that for n = 1, the inequalities are valid and that for n 3, we have Since x 5 4, N(1 + z z n ) > N(z n 1 r 4 ) r x + 1. (r + x 1)(r x + 1) = r 4 (x 1) < r 4. ( Thus, N(1 + z z n ) > N(z n ) ( Thus, N(1 + z z n ) > N(z n ) 1 + x 1 N(z) ). If further y x 1, then (r + x + 1)(r x + 1) = (r + 1) 4x = r 4 (x y ) + 1 r 4 (4x 3) r x+1 N(z) We obtain the following corollary as a consequence of the previous lemma. Corollary 4.1. Let π be an odd Eisenstein prime. Then, for k N, ( ) σ (π k ) N(π) + x 1 N >, π k N(π) where ɛπ = x + iy lies in the shaded region of Figure 3, and if y x 1, then ( ) σ (π k ) N(π) + x + 1 N π k N(π) with equality holding if and only if k = 1. With this knowledge we can prove an equivalent result to Theorem ). 5

29 Theorem Let α Z[ω] be odd, then ( + ω) k 1 α can be perfect if and only if k 0, ±1, ± (mod 1). k (mod 1) M k A k 1 0 ( k ) + 1i( k ) 1 3 k + 3 k 1 1 ( k ) + 1i( 3 3 k ) k 3 1+k 1 ( k ) + i k + 3 k 1 3 ( k ) + 1i( k ) k 1 4 ( k ) + 1i( k ) k + 3 k i( k ) k + 3 k ( 1 3 k ) + 1 i( k ) k + 3 k 1 7 ( k ) + 1i( k ) k + 3 k ( 1 3 k ) + i k + 3 k 1 9 ( k ) + 1i( 3 3 k ) k 1 10 ( 1 3 k ) + 1i( k ) 1 3 k + 3 k i( 3 3 k ) k 3 k+1 Table 4: M k and A k Proof. Let ( + ω) k 1 α be perfect, then we have σ (( + ω) k 1 α) = ω(( + ω) k 1)σ (α) = ( + ω) k α. But N((+ω) k 1) > N((+ω) k ) if and only if k 3, 4, 5, 6, 7, 8, 9 (mod 1) (see tables above), and always N(σ (α)) N(α) by Corollary 4.1. Thus, N( ω(( + ω) k 1)σ (α)) > N(( + ω) k α) for k 3, 4, 5, 6, 7, 8, 9 (mod 1), which contradicts the perfectness. The next result will bring us one step further on our way to characterize Eisenstein even norm-perfect numbers and already shows that the generalized Mersenne primes play an important role. Lemma If η = ( + ω) k 1 µ is norm-perfect, with µ being an odd Eisenstein integer, then M k or M k divide η and both are prime. Proof. Let π be a prime divisor of M k, which lies in the shaded region of Figure 3. Since η is norm-perfect, we obtain 3ηη = N(( + ω)η) = N(σ (η)) ( ) = N(M k σ Mk (µ)) = ππn π σ (µ). 6

30 This implies that π 3ηη. Since 3 = ω( + ω), ω is a unit and gcd(m k, + ω) = 1, from which follows that π 3. Hence, π ηη, respectively π η or π η. In particular, π µ or π µ since gcd(m k, + ω) = 1. For any prime π, suppose a Z such that π a µ, but π a+1 µ. By Corollary 4.1, we obtain Solving for N(π) gives us 1 = N(σ (η)) N(( + ω)η) = N(σ (( + ω) k 1 )σ (π a )) N(σ ( µ )) π a N(( + ω) k π a ) N( µ ) π a N(σ (( + ω) k 1 )σ (π a )) N(( + ω) k π a ) N(π) > > A k(n(π) + x 1) N(( + ω) k )N(π). A k (x 1) N(( + ω) k ) A k. Let π, respectively π, be the associate of π, respectively π, which lies in the shaded region of Figure 3. They are dierent from, hence Re(π ), Re(π ). Consequently, Theorem 4.13 and Table 4 imply N(π) > 3A k N(( + ω) k ) A k. This yields 3A 1 k 3(3 k 1) N(( + ω) k ) A k 3 k 1 > 1. Rearranging leads to N(π) > 3A k > A 1 N(( + ω) k k ) A = N(M k) 1. k N(π) > N(M k ). Assume that M k is composite. Say M k = ɛπ 0 π 1 π r for r N, where all π i are prime, and ɛ is a unit. Without loss of generality let π 0 be the prime of least norm. Then, by the inequality above, N(π 0 ) > N(π 1 ) N(π r ), which is a contradiction. Thus, M k is prime. Assume M k is not prime, then M k = αβ where α, β Z[ω] are not units. However, this contradicts the fact that M k is prime since M k = αβ. Theorem If η Z[ω] is an even norm-perfect number, then there exists an Eisenstein Mersenne prime M p such that, for some t N, a unit ɛ and an odd δ Z[ω], either 7

31 or (i) η = ɛ( + ω) p 1 M t pδ, with p 1 (mod 1), (ii) η = ɛ( + ω) p 1 M p t δ, with p 1 (mod 1). Proof. By Lemma 4.14, η = ɛ( + ω) p 1 M t pδ or η = ɛ( + ω) p 1 M p t δ for an odd δ Z[ω] and a unit ɛ. By Theorem 4.13 we know p = or p ±1 (mod 1), all other possibilities are divisible by, hence not prime. First, we show that for p =, η = ɛ( + ω)m t δ and η = ɛ( + ω)m t δ are not normperfect. Consider η = ɛ( + ω)m t δ, M = 3 + ω. By Lemma 4.11 and Corollary 4.1, we get N(σ (η)) N(( + ω)η) = N(σ ( + ω))n(σ (M))N(σ t (δ)) N(( + ω) N(M)N(δ) t N(σ ( + ω))n(σ (M)) t N(( + ω) )N(M) t > N(3 + ω) N(( + ω) ) A + Re(M ) 1 A = 10 9 > 1, where M is the associate of M, which lies in the shaded region of Figure 3. We are left to show that if p 1 (mod 1), then η = ɛ( + ω) p 1 M t pδ is not normperfect and if p 1 (mod 1), then η = ɛ( + ω) p 1 t M p δ is not norm-perfect. Let η = ɛ( + ω) p 1 Mpδ t with p 1 (mod 1) and Mp = ωm p be the associate in the shaded region of Figure 3. Since Im(Mp ) Re(Mp ) 1, N(σ (η)) N(( + ω)η) A p + Re /Mp ) + 1 N(( + ω) p ) = 3 p 1 > 1, hence, η is not norm-perfect. The remaining case is analogous. With these tools we can present an analogue to Euler's Theorem (1.8). Theorem Let η be an even norm-perfect number, then for some unit ɛ and Mersenne prime M p, either or (i) η = ɛ( + ω) p 1 M p, if p 1 (mod 1), (ii) η = ɛ( + ω) p 1 M p, if p 1 (mod 1). Proof. If M p is a Mersenne prime and p 1 (mod 1), then Im(M p ) Re(M p ) 1 and by Corollary 4.1, N(σ (M t p)) N(M t p) N(σ (M p ), (4) N(M p ) 8

32 with equality if and only if t = 1. Analogously, for M p Mersenne prime and p 1 (mod 1), N(σ t (M p )) t N(σ (M p ), (5) N(M p ) N(M p ) with equality if and only if t = 1. Since η is norm-perfect, by the previous theorem, η is either of the form η = ɛ( + ω) p 1 M t pδ or η = ɛ( + ω) p 1 M p t δ. Assume that η is of the former form, then the analogue to Euclid's Theorem for norm-perfect Eisenstein integers (Lemma 4.10), Corollary 4.1 and inequality (4) lead to 1 = N(σ (η)) N(( + ω)η) = N(σ (( + ω) p 1 )) N(( + ω) p ) N(σ (( + ω) p 1 )) N(( + ω) p ) N(σ (M t p)) N(M t p) N(σ (δ)) N(δ) N(σ (M p )) N(σ (δ)) N(M p ) N(δ) = N(σ (( + ω) p 1 M p )) N(σ (δ)) N(( + ω) p M p ) N(δ) = N(σ (δ)). N(δ) We get N(σ (δ)) = 1 because of the properties of the norm-function. Thus, δ is a unit in N(δ) Z[ω]. Furthermore, since δ is a unit, we have N(σ (M t p)) N(M t p) = N(σ (M p )), N(M p ) which implies that t = 1. By the same arguments, it follows that δ is a unit in the second form and also t = 1. This proves our claim. By putting the results together, we are now able to characterize the even norm-perfect numbers. Theorem Let η Z[ω] be an even norm-perfect number, then either or η = ɛ(+ω) p 1 M p, where M p is an Eisenstein Mersenne prime with p 1 (mod 1) and ɛ is a unit, η = ɛ(+ω) p 1 M p, where M p is an Eisenstein Mersenne prime with p 1 (mod 1) and ɛ is a unit. Conversely, for an arbitrary Eisenstein Mersenne prime M p with p 1 (mod 1) respectively p 1 (mod 1) and a unit ɛ, η = ɛ( + ω) p 1 M p respectively η = ɛ( + ω) p 1 M p is a norm-perfect number. 9

33 Corollary There exists no imprimitive even norm-perfect number in the ring of Eisenstein integers. After we have shown the form of even norm-perfect numbers, we can apply the result to even Eisenstein perfect numbers. Corollary An element η Z[ω] is an even perfect number if and only if there exists a rational prime p 1 (mod 1) such that η = ( + ω) p 1 M p, where M p is an Eisenstein Mersenne prime. Neither in [] nor in [1] odd norm-perfect Eisenstein numbers are investigated. But the authors of [] believe that an odd perfect Eisenstein integer has to be of the following form: Conjecture 4.0. Let α Z[ω] be odd and norm-perfect, then α = π k γ 3, where π and γ are both odd, π is prime, k (mod 3) and gcd(π, γ) = 1. We will show that this form is possible, but we can only point out a weaker result for the form of odd norm-perfect numbers. Before we start, it is useful to consider some rules of parity for the ring of Eisenstein integers. Lemma 4.1. Let α Z[ω] and α = a + bω, where a and b are rational integers. Then α is even if and only if a + b 0 (mod 3). Proof. Assume α = a + bω is even. Thus, we can write α = a + bω = ( + ω)(c + dω) for some c, d Z. Simplifying yields to α = a + bω = ( + ω)(c + dω) = c + dω + cω + dω = c + dω + cω d dω = (c d) + (c + d) ω. }{{}}{{} =a =b We have to consider the nine possible pairs of congruence classes modulo 3 for c and d. If c d 0 (mod 3), then a b 0 (mod 3) and hence a + b 0 (mod 3). If c 0 (mod 3) and d 1 (mod 3), then a (mod 3) and b 1 (mod 3). It follows that a + b 0 (mod 3). If c 0 (mod 3) and d (mod 3), then a 1 (mod 3) and b (mod 3). It follows again that a + b 0 (mod 3). If c 1 (mod 3) and d 0 (mod 3), then a (mod 3) and b 1 (mod 3), hence a + b 0 (mod 3). If c (mod 3) and b 0 (mod 3), we obtain a 1 (mod 3) and b (mod 3), and again a + b 0 (mod 3). 30