aa r X i v : . [ m a t h . N T ] J un Multiperfect Numbers in CertainQuadratic Rings
Colin Defant
Department of MathematicsUniversity of FloridaUnited Statescdefant@ufl.edu
Abstract
Using an extension of the abundancy index to imaginary quadraticrings that are unique factorization domains, we investigate what we call n -powerfully t -perfect numbers in these rings. This definition serves toextend the concept of multiperfect numbers that have been defined andstudied in the integers. At the end of the paper, as well as at variouspoints throughout the paper, we point to some potential areas for furtherresearch. Throughout this paper, we will let N denote the set of positive integers, and wewill let P = { , , , . . . } denote the set of (integer) prime numbers.The arithmetic functions σ k are defined, for every integer k , by σ k ( n ) = X c | nc> c k . For each integer k = 0, σ k is multiplicative and satisfies This work was supported by National Science Foundation grant no. 1262930. Colin Defant18434 Hancock Bluff Rd.Dade City, FL 33523 Mathematics Subject Classification : Primary 11R11; Secondary 11N80.
Keywords:
Abundancy index, quadratic ring, perfect number, multiperfect number. k ( p α ) = p k ( α +1) − p k − p and positive integers α . Theabundancy index of a positive integer n is defined by I ( n ) = σ ( n ) n . Some ofthe most interesting questions related to the abundancy index are those dealingwith perfect and multiperfect numbers.A positive integer n is said to be t -perfect if I ( n ) = t for a positive integer t ≥
2, and 2-perfect numbers, which have been studied since the ancient Greeks,are called perfect numbers. It is known that there is a one-to-one correspondencebetween even perfect numbers and Mersenne primes, so it is, therefore, unknownwhether or not there are infinitely many even perfect numbers. Although no oddperfect numbers are currently known to exist, a long list of criteria, sometimesknown as Sylvester’s Web of Conditions, places demands on the properties thatany odd perfect number would need to satisfy.For any square-free integer d , let O Q ( √ d ) be the quadratic integer ring givenby O Q ( √ d ) = Z [ √ d ] , if d ≡ Z [ √ d ] , if d ≡ , . Throughout the remainder of this paper, we will work in the rings O Q ( √ d ) for different specific or arbitrary values of d . We will use the symbol “ | ” tomean “divides” in the ring O Q ( √ d ) in which we are working. Whenever weare working in a ring other than Z , we will make sure to emphasize when wewish to state that one integer divides another in Z . For example, if we areworking in Z [ i ], the ring of Gaussian integers, we might say that 1 + i | i and that 2 | Z . We will also refer to primes in O Q ( √ d ) as “primes,”whereas we will refer to (positive) primes in Z as “integer primes.” For aninteger prime p and a nonzero integer n , we will let υ p ( n ) denote the largestinteger k such that p k | n in Z . For a prime π and a nonzero number x ∈ O Q ( √ d ) ,2e will let ρ π ( x ) denote the largest integer k such that π k | x . Furthermore, wewill henceforth focus exclusively on values of d for which O Q ( √ d ) is a uniquefactorization domain and d <
0. In other words, d ∈ K , where we will define K to be the set {− , − , − , − , − , − , − , − , − } . The set K is knownto be the complete set of negative values of d for which O Q ( √ d ) is a uniquefactorization domain [4].For now, let us work in a general ring O Q ( √ d ) such that d ∈ K . For an element a + b √ d ∈ O Q ( √ d ) with a, b ∈ Q , we define the conjugate by a + b √ d = a − b √ d .We also define the norm of an element z by N ( z ) = zz and the absolute valueof z by | z | = p N ( z ). We assume familiarity with the properties of these object,which are treated in Keith Conrad’s online notes [1]. For x, y ∈ O Q ( √ d ) , we saythat x and y are associated, denoted x ∼ y , if and only if x = uy for some unit u in the ring O Q ( √ d ) . Furthermore, we will make repeated use of the followingwell-known facts. Fact 1.1.
Let d ∈ K . If p is an integer prime, then exactly one of the followingis true. • p is also a prime in O Q ( √ d ) . In this case, we say that p is inert in O Q ( √ d ) . • p ∼ π and π ∼ π for some prime π ∈ O Q ( √ d ) . In this case, we say p ramifies (or p is ramified) in O Q ( √ d ) . • p = ππ and π π for some prime π ∈ O Q ( √ d ) . In this case, we say p splits (or p is split) in O Q ( √ d ) . Fact 1.2.
Let d ∈ K . If π ∈ O Q ( √ d ) is a prime, then exactly one of the followingis true. • π ∼ q and N ( π ) = q for some inert integer prime q . • π ∼ π and N ( π ) = p for some ramified integer prime p . π π and N ( π ) = N ( π ) = p for some split integer prime p . Fact 1.3. If d ∈ K , q is an integer prime that is inert in O Q ( √ d ) , and x ∈O Q ( √ d ) \{ } , then υ q ( N ( x )) is even and ρ q ( x ) = υ q ( N ( x )) . Fact 1.4.
Let p be an odd integer prime. Then p ramifies in O Q ( √ d ) if and onlyif p | d in Z . If p ∤ d in Z , then p splits in O Q ( √ d ) if and only if d is a quadraticresidue modulo p . Note that this implies that p is inert in O Q ( √ d ) if and onlyif p ∤ d in Z and d is a quadratic nonresidue modulo p . Also, the integer prime ramifies in O Q ( √− and O Q ( √− , splits in O Q ( √− , and is inert in O Q ( √ d ) for all d ∈ K \{− , − , − } . Fact 1.5.
Let O ∗ Q ( √ d ) be the set of units in the ring O Q ( √ d ) . Then O ∗ Q ( √− = {± , ± i } , O ∗ Q ( √− = (cid:26) ± , ± √− , ± − √− (cid:27) , and O ∗ Q ( √ d ) = {± } whenever d ∈ K \{− , − } . For a nonzero complex number z , let arg( z ) denote the argument, or angle,of z . We convene to write arg( z ) ∈ [0 , π ) for all z ∈ C . For each d ∈ K , wedefine the set A ( d ) by A ( d ) = { z ∈ O Q ( √ d ) \{ } : 0 ≤ arg( z ) < π } , if d = − { z ∈ O Q ( √ d ) \{ } : 0 ≤ arg( z ) < π } , if d = − { z ∈ O Q ( √ d ) \{ } : 0 ≤ arg( z ) < π } , otherwise . Thus, every nonzero element of O Q ( √ d ) can be written uniquely as a unit timesa product of primes in A ( d ). Also, every z ∈ O Q ( √ d ) \{ } is associated to aunique element of A ( d ). The author has defined analogues of the arithmeticfunctions σ k in quadratic rings O Q ( √ d ) with d ∈ K [2], and we will state theimportant definitions and properties for the sake of completeness. Definition 1.1.
Let d ∈ K , and let n ∈ Z . Define the function4 n : O Q ( √ d ) \{ } → [1 , ∞ ) by δ n ( z ) = X x | zx ∈ A ( d ) | x | n . Remark 1.1.
We note that, for each x in the summation in the above definition,we may cavalierly replace x with one of its associates. This is because associatednumbers have the same absolute value. In other words, the only reason for thecriterion x ∈ A ( d ) in the summation that appears in Definition 1.1 is to forbidus from counting associated divisors as distinct terms in the summation, but wemay choose to use any of the associated divisors as long as we only choose one.This should not be confused with how we count conjugate divisors (we treat2 + i and 2 − i as distinct divisors of 5 in Z [ i ] because 2 + i − i ). Remark 1.2.
We mention that the function δ n is different in each ring O Q ( √ d ) .Perhaps it would be more precise to write δ n ( z, d ), but we will omit the lattercomponent for convenience. We note that we will also use this convention withfunctions such as I n (which we will define soon).We will say that a function f : O Q ( √ d ) \{ } → R is multiplicative if f ( xy ) = f ( x ) f ( y ) whenever x and y are relatively prime (have no nonunit common di-visors). The author has shown that, for any integer n , δ n is multiplicative [2]. Definition 1.2.
For each positive integer n , define the function I n : O Q ( √ d ) \{ } → [1 , ∞ ) by I n ( z ) = δ n ( z ) | z | n . For a positive integer t ≥
2, we saythat a number z ∈ O Q ( √ d ) \{ } is n -powerfully t -perfect in O Q ( √ d ) if I n ( z ) = t ,and, if t = 2, we simply say that z is n -powerfully perfect in O Q ( √ d ) . Whenever n = 1, we will omit the adjective “1-powerfully.”As an example, we will let d = − O Q ( √ d ) = Z [ i ]. Let us compute I (9 + 3 i ). We have 9 + 3 i = 3(1 + i )(2 − i ), so δ (9 + 3 i ) = N (1) + N (3) + N (1 +5 ) + N (2 − i ) + N (3(1 + i )) + N (3(2 − i )) + N ((1 + i )(2 − i )) + N (3(1 + i )(2 − i )) =1+9+2+5+18+45+10+90 = 180. Then I (9 + 3 i ) = 180 N (3(1 + i )(2 − i )) = 2,so 9 + 3 i is 2-powerfully perfect in O Q ( √− .We omit the (fairly simplistic) proof of the following theorem because it isincluded in [2]. Theorem 1.1.
Let n ∈ N , d ∈ K , and z , z , π ∈ O Q ( √ d ) \{ } with π a prime.Then, if we are working in the ring O Q ( √ d ) , the following statements are true.(a) The range of I n is a subset of the interval [1 , ∞ ) , and I n ( z ) = 1 if andonly if z is a unit in O Q ( √ d ) . If n is even, then I n ( z ) ∈ Q .(b) I n is multiplicative.(c) I n ( z ) = δ − n ( z ) .(d) If z | z , then I n ( z ) ≤ I n ( z ) , with equality if and only if z ∼ z . Henceforth, we will focus on the existence of n -powerfully t -perfect numbersfor n = 2. n -powerfully t -perfect Numbers for n = 2 We begin this section with a theorem (after two short lemmata) that dramati-cally limits the number of possibilities that we may consider when dealing with n -powerfully t -perfect numbers. Lemma 2.1.
Let d ∈ K , and let n ∈ N . If z ∈ O Q ( √ d ) \{ } and n ≥ , then I n ( z ) < ζ (cid:16) n (cid:17) , where ζ denotes the Riemann zeta function. roof. Let Ψ( z ) be the set of primes in A ( d ) that divide z , and let Φ be the setof primes in A ( d ). By parts ( b ) and ( c ) of Theorem 1.1, as well as Fact 1.2, wemay write I n ( z ) = Y π ∈ Ψ( z ) ρ π ( z ) X j =0 | π j | n = Y π ∈ Ψ( z ) | π |∈ N ρ π ( z ) X j =0 | π j | n Y π ∈ Ψ( z ) | π |6∈ N π ∼ π ρ π ( z ) X j =0 | π j | n Y π ∈ Ψ( z ) | π |6∈ N π π ρ π ( z ) X j =0 | π j | n < Y π ∈ Φ | π |∈ N ∞ X j =0 | π j | n Y π ∈ Φ | π |6∈ N π ∼ π ∞ X j =0 | π j | n Y π ∈ Φ | π |6∈ N π π ∞ X j =0 | π j | n = Y q ∈ P q is inert ∞ X j =0 q jn Y p ∈ P p ramifies ∞ X j =0 √ p jn Y p ∈ P p splits ∞ X j =0 √ p jn < Y q ∈ P q is inert ∞ X j =0 √ q jn Y p ∈ P p ramifies ∞ X j =0 √ p jn Y p ∈ P p splits ∞ X j =0 √ p jn = Y p ∈ P ∞ X j =0 √ p jn = ζ (cid:16) n (cid:17) . We state the following lemma without proof, though the proof may be foundas a corollary of Lemma 3.4 in [2].
Lemma 2.2.
Let us fix d ∈ K and work in the ring O Q ( √ d ) . Let n be an oddpositive integer, and let z ∈ O Q ( √ d ) \{ } . If I n ( z ) is rational, then all primesdividing z are associated to inert integer primes. Theorem 2.1.
Let d ∈ K . For any integers n ≥ and t ≥ , there areno n -powerfully t -perfect numbers in O Q ( √ d ) because I n ( z ) < whenever z ∈ Q ( √ d ) \{ } and I n ( z ) ∈ Q .Proof. Let z ∈ O Q ( √ d ) \{ } be such that I n ( z ) ∈ Q . If n ≥
5, the proof followsfrom Lemma 2.1 because, in that case, we have I n ( z ) < ζ (cid:18) (cid:19) ≈ . < n = 3. Because n is an odd positive integer and I n ( z ) is rational,Lemma 2.2 tells us that any prime dividing z must be associated to an inertinteger prime. Therefore, I n ( z ) = Y π | zπ ∈ A ( d ) ρ π ( z ) X j =0 | π j | < Y q ∈ P q is inert ∞ X j =0 q j < Y q ∈ P ∞ X j =0 q j = ζ (3) < . The only case left to consider is the case n = 4. As an intermediate step inthe proof of Lemma 2.1, we arrived at the inequality I n ( z ) < Y q ∈ P q is inert ∞ X j =0 q jn Y p ∈ P p ramifies ∞ X j =0 √ p jn Y p ∈ P p splits ∞ X j =0 √ p jn . (1)Substituting n = 4, we have I ( z ) < Y q ∈ P q is inert ∞ X j =0 q j Y p ∈ P p ramifies ∞ X j =0 p j Y p ∈ P p splits ∞ X j =0 p j . Now, suppose that the ring O Q ( √ d ) in which we are working is one in which theinteger prime 2 is inert. Then I ( z ) < ∞ X j =0 j Y q ∈ P q is inertq =2 ∞ X j =0 q j Y p ∈ P p ramifies ∞ X j =0 p j Y p ∈ P p splits ∞ X j =0 p j ∞ X j =0 j Y q ∈ P q is inertq =2 ∞ X j =0 q j Y p ∈ P p ramifies ∞ X j =0 p j Y p ∈ P p splits ∞ X j =0 p j = ∞ X j =0 j Y p ∈ P p =2 ∞ X j =0 p j = ζ (2) ∞ X j =0 j − ∞ X j =0 j = π · · π < . Now, recall from Fact 1.4 that the only d ∈ K for which 2 is not inert in O Q ( √ d ) are d = − d = −
2, and d = −
7. If d = −
1, then 2 ramifies and 3 is inert.Therefore, if we write H = ∞ X j =0 j ∞ X j =0 j , then I ( z ) < H Y q ∈ P q is inertq =3 ∞ X j =0 q j Y p ∈ P p ramifiesp =2 ∞ X j =0 p j Y p ∈ P p splits ∞ X j =0 p j ≤ H Y p ∈ P p , } ∞ X j =0 p j = ζ (2) ∞ X j =0 j − ∞ X j =0 j − ∞ X j =0 j = π · · · π < . Similarly, if d = −
2, then 2 ramifies and 5 is inert. Therefore, we may replaceall of the 3’s in the above chain of inequalities with 5’s to arrive at I ( z ) < ζ (2) ∞ X j =0 j − ∞ X j =0 j − ∞ X j =0 j = π · · · π < . Finally, we consider the case d = −
7. In this case, 3 and 5 are both inert. If we9rite J = ∞ X j =0 j ∞ X j =0 j , then I ( z ) < J Y q ∈ P q is inertq , } ∞ X j =0 q j Y p ∈ P p ramifies ∞ X j =0 p j Y p ∈ P p splits ∞ X j =0 p j ≤ J Y p ∈ P p , } ∞ X j =0 p j = J ζ (2) ∞ X j =0 j − ∞ X j =0 j − = 4 π < . This completes the final case.
Theorem 2.2.
Let d ∈ K . A number z ∈ O Q ( √ d ) \{ } satisfies I ( z ) = b ∈ Q if and only if it is associated to an integer whose (traditional) abundancy indexis b and whose prime factors (in Z ) are all inert in O Q ( √ d ) .Proof. Suppose z ∈ O Q ( √ d ) \{ } satisfies I ( z ) = b ∈ Q . If b = 1, then thedesired result is clear because z ∼
1. Therefore, we may assume b >
1. Lemma2.2 tells us that all primes dividing z are associated to inert integer primes,which implies that z is associated to an integer whose prime factors (in Z ) areall inert in O Q ( √ d ) . We may, therefore, write z ∼ r for some r ∈ N . As allprimes dividing r are associated to inert integer primes, we have I ( r ) = I ( r ),where I is the traditional abundancy index defined over N . Therefore, I ( r ) = I ( r ) = I ( z ) = b .Conversely, if z ∼ r , where r is an integer whose (traditional) abundancyindex is a rational number b and whose prime factors (in Z ) are all inert in O Q ( √ d ) , then I ( z ) = I ( r ) = I ( r ) = b ∈ Q .10 orollary 2.1. Let d ∈ K . A number z ∈ O Q ( √ d ) \{ } is a t -perfect numberin O Q ( √ d ) if and only if it is associated to an integer that is t -perfect in Z andwhose prime factors (in Z ) are all inert in O Q ( √ d ) .Proof. Simply set b = t in Theorem 2.2.Let us now restrict the scope of our exploration to perfect numbers in a ring O Q ( √ d ) ( d ∈ K ). That is, we will search for n -powerfully t -perfect numbers with n = 1 and t = 2. We will repeatedly make use of Corollary 2.1 and the followingtwo well-known facts about perfect numbers in Z [3, 5]. Fact 2.1.
A positive integer r is an even perfect number (in Z ) if and only if r = 2 p − (2 p − for some Mersenne prime p − . Fact 2.2. If r is an odd integer that is perfect in Z (assuming such a numberexists), then r = p k m , where p is an integer prime and k, m ∈ N . Furthermore, p ≡ k ≡ , m > , and p ∤ m in Z . The expression p k m is known asthe Eulerian form of the odd perfect number r . Theorem 2.3.
There are no perfect numbers in O Q ( √− , the ring of Gaussianintegers.Proof. Suppose z is perfect in O Q ( √− . Then, by Corollary 2.1, z ∼ r forsome positive integer r that is perfect in Z . Furthermore, all integer primesthat divide r in Z must be inert in O Q ( √− . We know (by Fact 1.4) that aninteger prime is inert in O Q ( √− if and only if it is congruent to 3 modulo 4, sowe conclude that all integer primes that divide r in Z must be congruent to 3modulo 4. Thus, r is odd, so Fact 2.2 tells us that there exists an integer prime p that divides r in Z and is congruent to 1 modulo 4. This is a contradiction,and the desired result follows. 11 heorem 2.4. There are no perfect numbers in O Q ( √− , the ring of Eisensteinintegers.Proof. Suppose z is perfect in O Q ( √− . Then, by Corollary 2.1, z ∼ r forsome positive integer r that is perfect in Z . Furthermore, all integer primesthat divide r in Z must be inert in O Q ( √− . We know (by Fact 1.4) that aninteger prime is inert in O Q ( √− if and only if it is congruent to 2 modulo 3,so we conclude that all integer primes that divide r in Z must be congruent to2 modulo 3. If r is even, then we may write r = 2 p − (2 p − p − p − r in Z ,so we conclude 2 p − ≡ r is odd, sowe may write r in the Eulerian form r = p k m . From the fact that r is perfectin Z , we have 2 p k m = σ ( p k m ) = σ ( p k ) σ ( m ) = σ ( m ) k X j =0 p j . Now, Fact 2.2tells us that k is odd, so p + 1 | P kj =0 p j in Z . As p ≡ | k X j =0 p j | p k m = 2 r in Z . Therefore, 3 | r in Z , which is a contradiction becauseall the integer primes that divide r in Z are congruent to 2 modulo 3.The method used to prove Theorems 2.3 and 2.4 may be used to explorethe properties that perfect numbers in the other seven rings (correspondingto d ∈ {− , − , − , − , − , − , − } ) must possess, but the caseworkcan become tedious very quickly. For this reason, we will only briefly exploreproperties of hypothetical perfect numbers in O Q ( √− .If z is perfect in O Q ( √− (a ring in which 2 ramifies), then z ∼ r , where r is an odd integer that is perfect in Z and has Eulerian form r = p k m . Inaddition, − p ≡ p must be inert), we find that p ≡ m = m m with m , m ∈ N so that all integer primes dividing12 in Z are congruent to 5 modulo 8 and all integer primes dividing m in Z arecongruent to 7 modulo 8. Let m = s Y j =1 q α j j be the canonical prime factorizationof m in Z , and let L = |{ j ∈ { , , . . . , s } : α j is odd }| . Using this notation,we may state and prove the following theorem. Theorem 2.5.
Let all notation be as in the preceding paragraph. If k ≡ , then L is odd. If k ≡ , then L is even.Proof. First, as p , k , and m are all odd, we have p k m = p · ( p ) k − m ≡ p (1)(1) ≡ r = 2 p k m ≡ r = σ ( p k m ) = σ ( p k ) σ ( m ) σ ( m ) = σ ( m ) k X l =0 p l ! s Y j =1 α j X l =0 q lj ! . Let q be an integer prime that divides m in Z . Then σ ( q υ q ( m ) ) = 1 + q + q + · · · + q υ q ( m ) ≡ υ q ( m ) is even and q ≡ σ ( m ) ≡ j ∈ { , , . . . , s } ,we have α j X l =0 q lj ≡ α j X l =0 l ≡ α j + 1 (mod 8). Assume k ≡ k X l =0 p l ≡ k X l =0 l ≡ ≡ r ≡ σ ( m ) k X l =0 p l ! s Y j =1 α j X l =0 q lj ! ≡ s Y j =1 (6 α j + 1) (mod 8) , which implies s Y j =1 (6 α j + 1) ≡ α j is even, 6 α j + 1 ≡ α j is odd, 6 α j + 1 ≡ L ,which is the number of integers j ∈ { , , . . . , s } such that α j is odd, must bean odd number.On the other hand, if k ≡ k X l =0 p l ≡ k X l =0 l ≡ ≡ r ≡ σ ( m ) k X l =0 p l ! s Y j =1 α j X l =0 q lj ! ≡ s Y j =1 (6 α j + 1) (mod 8) , which implies s Y j =1 (6 α j + 1) ≡ α j is even, 6 α j + 1 ≡ α j is odd, 6 α j + 1 ≡ L mustbe an even number in this case.We note that it has been conjectured (supposedly by Descartes) that thevalue of k in the Eulerian form of any hypothetical odd number that is perfectin Z must be 1. If the conjecture is true, then Theorem 2.5 implies that L mustbe odd.We note that there are definitely rings O Q ( √ d ) with d ∈ K that containperfect numbers. For example, one may show that an integer prime is inertin O Q ( √− if and only if that integer prime is congruent to 2, 6, 7, 8, or 10modulo 11. Therefore, if 2 p − p is prime, that 2 p − p − (2 p −
1) is perfect in O Q ( √− . For example,28, 8128, 2 − (2 − − (2 −
1) (this list is not exhaustive) are allperfect in O Q ( √− . We acknowledge the entirely possible generalization of the definitions presentedhere to the other quadratic integer rings. In particular, generalizing the abun-dancy index to unique factorization domains O Q ( √ d ) with d > O Q ( √ d ) with d ∈ K ,we may ask some interesting questions. For example, in a given ring O Q ( √ d ) with d ∈ K , one may wish to examine the properties of t -perfect numbers for t > The author would like to thank Professor Pete Johnson for inviting him to the2014 REU Program in Algebra and Discrete Mathematics at Auburn University.The author would also like to thank the unknown referee for his or her carefulreading.