An Extension of the Abundancy Index to Certain Quadratic Rings
aa r X i v : . [ m a t h . N T ] J un An Extension of the Abundancy Indexto Certain Quadratic Rings
Colin Defant
Department of MathematicsUniversity of FloridaUnited Statescdefant@ufl.edu
Abstract
We begin by introducing an extension of the traditional abundancy in-dex to imaginary quadratic rings with unique factorization. After showingthat many of the properties of the traditional abundancy index continue tohold in our extended form, we investigate what we call n -powerfully soli-tary numbers in these rings. This definition serves to extend the conceptof solitary numbers, which have been defined and studied in the integers.We end with some open questions and a conjecture. Throughout this paper, we will let N denote the set of positive integers, and wewill let N denote the set of nonnegative integers.The arithmetic functions σ k are defined, for every integer k , by σ k ( n ) = X c | nc> c k , and it is conventional to write σ = σ . It is well-known that, for 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, solitary number, friendly number. k = 0, σ k is multiplicative and satisfies σ k ( p α ) = p k ( α +1) − p k − p and positive integers α . The abundancy index of a positiveinteger n is defined by I ( n ) = σ ( n ) n . Using the formulas σ ( n ) = Y p α k n p α +1 − p − σ − ( n ) = Y p α k n p − ( α +1) − p − − I = σ − . Some of themost common questions associated with the abundancy index are those relatedto friendly numbers.Two or more distinct positive integers are said to be friends (with each other)if they have the same abundancy index. For example, I (6) = I (28) = I (496) =2, so 6, 28, and 496 are friends. A positive integer that has at least one friendis said to be friendly, and a positive integer that has no friends is said to besolitary. Clearly, 1 is solitary as I ( n ) > I (1) for any positive integer n > I = σ − , that every primepower is solitary. In the next section, we extend the notions of the abundancyindex and friendliness to imaginary quadratic integer rings that are also uniquefactorization domains. Observing the infinitude of possible such generalizations,we note four important properties of the traditional abundancy index that wewish to preserve (possibly with slight modifications). • The range of the function I is a subset of the interval [1 , ∞ ). • If n and n are relatively prime positive integers, then I ( n n ) = I ( n ) I ( n ). • If n and n are positive integers such that n | n , then I ( n ) ≤ I ( n ),with equality if and only if n = n . • All prime powers are solitary.For any square-free integer d , let O Q ( √ d ) be the quadratic integer ring given2y 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,” whereaswe will refer to (positive) primes in Z as “integer primes.” 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 [3].For now, let us work in a 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 ). From now on, we will assume familiarity with theseobjects and their properties (for example, z z = z z and N ( z ) ∈ N ), whichare treated in Keith Conrad’s online notes [1]. For x, y ∈ O Q ( √ d ) , we say that 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.
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 , we definethe 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 . 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, which we will call B ( z ), of A ( d ). We are now ready to defineanalogues of the arithmetic functions σ k . Definition 2.1.
Let d ∈ K , and let n ∈ Z . Define the function δ n : O Q ( √ d ) \{ } → [1 , ∞ ) by δ n ( z ) = X x | zx ∈ A ( d ) | x | n . Remark 2.1.
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 2.1 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 2.2.
We note that, by choosing different values of d , the functions δ n change dramatically. For example, δ (3) = 10 when we work in the ring O Q ( √− , but δ (3) = 16 when we work in the ring O Q ( √− . Perhaps it wouldbe more precise to write δ n ( z, d ), but we will omit the latter component forconvenience. We note that we will also use this convention with functions suchas 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). Theorem 2.1.
Let d ∈ K , and let f, g : O Q ( √ d ) \{ } → R be multiplicative func- ions such that f ( u ) = g ( u ) = 1 for all units u ∈ O ∗ Q ( √ d ) . Define F : O Q ( √ d ) \{ } → R by F ( z ) = X x,y ∈ A ( d ) xy ∼ z f ( x ) g ( y ) . Then F is multiplicative.Proof. Suppose z , z ∈ O Q ( √ d ) \{ } and gcd( z , z ) = 1. For any x, y ∈ A ( d )satisfying xy ∼ z z , we may write x = x x , y = y y so that x y ∼ z and x y ∼ z . To make the choice of x , x , y , y unique, we require x , y ∈ A ( d ).Conversely, if we choose x , x , y , y ∈ O Q ( √ d ) \{ } such that x , y ∈ A ( d ), x y ∼ z , x y ∼ z , and x x , y y ∈ A ( d ), then we may write x = x x and y = y y so that xy ∼ z z . To simplify notation, write B ( x ) = x , B ( y ) = y , and let C be the set of all ordered quadruples ( x , x , y , y ) suchthat x , x , y , y ∈ O Q ( √ d ) \{ } , x , y ∈ A ( d ), x y ∼ z , x y ∼ z , and x x , y y ∈ A ( d ). We have established a bijection between C and the set ofordered pairs ( x, y ) satisfying x, y ∈ A ( d ) and xy ∼ z z . Therefore, F ( z z ) = X x,y ∈ A ( d ) xy ∼ z z f ( x ) g ( y ) = X ( x ,x ,y ,y ) ∈ C f ( x x ) g ( y y )= X ( x ,x ,y ,y ) ∈ C f ( x ) f ( x ) g ( y ) g ( y )= X ( x ,x ,y ,y ) ∈ C f ( x ) f ( B ( x )) g ( y ) g ( B ( y ))= X x ,y ∈ A ( d ) x y ∼ z f ( x ) g ( y ) X x ,y ∈ A ( d ) x y ∼ z f ( x ) g ( y ) = F ( z ) F ( z ) . Corollary 2.1.
For any integer n , δ n is multiplicative. roof. Noting that δ n ( w ) = δ n ( w ) whenever w ∼ w , we may let f, g : O Q ( √ d ) \{ } → R be the functions defined by f ( z ) = | z | n and g ( z ) = 1 forall z ∈ O Q ( √ d ) \{ } . Then the desired result follows immediately from Theorem2.1. Definition 2.2.
For each positive integer n , define the function I n : O Q ( √ d ) \{ } → [1 , ∞ ) by I n ( z ) = δ n ( z ) | z | n . We say that two or more numbers z , z , . . . , z r ∈ O Q ( √ d ) \{ } are n -powerfully friendly (or n -powerful friends ) in O Q ( √ d ) if I n ( z j ) = I n ( z k ) and | z j | 6 = | z k | for all distinct j, k ∈ { , , . . . , r } .Any z ∈ O Q ( √ d ) \{ } that has no n -powerful friends in O Q ( √ d ) is said to be n -powerfully solitary in O Q ( √ d ) . Remark 2.3.
Whenever n = 1, we will omit the adjective “1-powerfully” inthe preceding definitions.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 + i ) + 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.Although I (3 + 9 i ) is also equal to 2, 3 + 9 i and 9 + 3 i are not 2-powerful friendsin Z [ i ] because | i | = | i | . We now establish some important propertiesof the functions I n . Theorem 2.2.
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 .(e) If z ∼ π k for a nonnegative integer k , then z is n -powerfully solitary in O Q ( √ d ) .Proof. The first sentence in part ( a ) is fairly clear, and the second sentencebecomes equally clear if one uses the fact that | z | n ∈ N whenever n is even. Toprove part ( b ), suppose that z and z are relatively prime elements of O Q ( √ d ) .Then, by Corollary 2.1, I n ( z z ) = δ n ( z z ) | z z | n = δ n ( z ) δ n ( z ) | z | n | z | n = I n ( z ) I n ( z ). Inorder to prove part ( c ), it suffices, due to the truth of part ( b ), to prove that I n ( π α ) = δ − n ( π α ) for any prime π and nonnegative integer α . To do so is fairlyroutine, as I n ( π α ) = δ n ( π α ) | π α | n = P αj =0 | π j | n | π α | n = α X j =0 | π j − α | n = α X j =0 | π α − j | − n = α X l =0 | π l | − n = δ − n ( π α ) . The truth of statement ( d ) follows from part ( c ) because, if z | z , then I n ( z ) = δ − n ( z ) = X x | z x ∈ A ( d ) | x | − n = X x | z x ∈ A ( d ) | x | − n + X x | z x ∤ z x ∈ A ( d ) | x | − n = I n ( z ) + X x | z x ∤ z x ∈ A ( d ) | x | − n . Finally, for part ( e ), we provide a proof for the case when n is even. Wepostpone the proof for the case in which n is odd until the next section. Let π be a prime in O Q ( √ d ) , and suppose that z ∼ π k for a nonnegative integer k . If k = 0, then z is a unit and the result follows from part ( a ). Therefore, assume k >
0. Assume, for the sake of finding a contradiction, that I n ( z ) = I n ( z )and | z | 6 = | z | for some z ∈ O Q ( √ d ) \{ } . Under this assumption, we have8 z | n δ n ( z ) = | z | n δ n ( z ). Either N ( π ) = p is an integer prime or N ( π ) = q ,where q is an integer prime.First, suppose N ( π ) = p is an integer prime. Then the statement | z | n δ n ( z ) = | z | n δ n ( z ) is equivalent to N ( z ) n/ δ n ( π k ) = p kn/ δ n ( z ). Notingthat N ( z ) n/ , δ n ( π k ), and δ n ( z ) are integers (because n is even) and that p ∤ δ n ( π k ) = 1+ p n/ + · · · + p kn/ in Z , we find p kn/ | N ( z ) n/ in Z . This impliesthat p k | N ( z ) in Z , and we conclude that there exist nonnegative integers t , t satisfying π t π t | z and t + t = k . If π ∼ π , then we have π k | z , from whichpart ( d ) yields the desired contradiction. Otherwise, π and π are relativelyprime, so we may use parts ( b ) and ( d ) to write I n ( z ) ≥ I n ( π t ) I n ( π t ) = 1 + p n/ + · · · + p t n/ p t n/ p n/ + · · · + p t n/ p t n/ = (1 + p n/ + · · · + p t n/ )(1 + p n/ + · · · + p t n/ ) p kn/ ≥ p n/ + · · · + p kn/ p kn/ = I n ( π k ) = I n ( z ) . This implies that I n ( z ) = I n ( π t π t ), from which part ( d ) tells us that z ∼ π t π t . Therefore, | z | = | π t π t | = √ p t + t = √ p k = | π k | = | z | , which weassumed was false.Now, suppose that N ( π ) = q , where q is an integer prime ( q is inert). Thenthe statement | z | n δ n ( z ) = | z | n δ n ( z ) is equivalent to N ( z ) n/ δ n ( π k ) = q kn δ n ( z ). As before, N ( z ) n/ , δ n ( π k ), and δ n ( z ) are in-tegers, and q ∤ δ n ( π k ) = 1 + q n + · · · + q kn in Z . Therefore, q kn | N ( z ) n/ in Z , so q k | N ( z ) in Z . As q is inert, this implies that q k | z , so z | z (note that z ∼ π k ∼ q k ). Therefore, part ( d ) provides the final contradiction, and theproof is complete. 9t is much easier to deal with the functions I n when n is even than when n isodd because, when n is even, the values of δ n ( z ) and | z | n are positive integers.Therefore, we will devote the next section to developing an understanding ofthe functions I n for odd values of n . n is Odd We begin by establishing some definitions and lemmata that will later provethemselves useful. Let W be the set of all square-free positive integers, and write W = { w , w , w , . . . } so that w = 1 and w i < w j for all nonnegative integers i < j . Let F be the set of all finite linear combinations of elements of W withrational coefficients. That is, F = { a + a √ w + · · · + a m √ w m : a , a , . . . , a m ∈ Q , m ∈ N } . For any r ∈ F , the choice of the rational coefficients is unique. Moreformally, if a + a √ w + · · · + a m √ w m = b + b √ w + · · · + b m √ w m , where a , a , . . . , a m , b , b , . . . , b m ∈ Q , then a i = b i for all i ∈ { , , . . . , m } [2]. Notethat F is a subfield of the real numbers. Definition 3.1.
For r ∈ F and j ∈ N , let C j ( r ) be the unique rational coef-ficient of √ w j in the expansion of r . That is, the sequence ( C j ( r )) ∞ j =0 is theunique infinite sequence of rational numbers that has finitely many nonzeroterms and that satisfies r = ∞ X j =0 C j ( r ) √ w j .As an example, C (cid:18) − √ √ (cid:19) = 13 because w = 7. Definition 3.2.
Let p be an integer prime. For r ∈ F , we say that r has a √ p part if there exists some positive integer j such that C j ( r ) = 0 and p | w j (in Z ).We say that r does not have a √ p part if no such positive integer j exists.For example, if r = 12 + 3 √
10, then r has a √ r has a √ √
10 does not have a √ Lemma 3.1. If r , r ∈ F each do not have a √ p part for some integer prime p , then r r does not have a √ p part.Proof. Suppose p | w j for some positive integer j . Then, if we let SF ( n ) denotethe square-free part of an integer n and consider the basic algebra used tomultiply elements of F , we find that C j ( r r ) = X i ,i ∈ N SF ( w i w i )= w j C i ( r ) C i ( r ) r w i w i w j . For every pair of nonnegative integers i , i satisfying SF ( w i w i ) = w j , either p | w i or p | w i . This implies that either C i ( r ) = 0 or C i ( r ) = 0 by thehypothesis that each of r and r does not have a √ p part. Thus, C j ( r r ) = 0.As w j was an arbitrary square-free positive integer divisible by p , we concludethat r r does not have a √ p part. Lemma 3.2.
If each of r , r , . . . , r l ∈ F does not have a √ p part for someinteger prime p , then r r · · · r l does not have a √ p part.Proof. The desired result follows immediately from repeated use of Lemma 3.1.
Lemma 3.3. If r ∈ F has a √ p part and r ∈ F \{ } does not have a √ p partfor some integer prime p , then r r has a √ p part.Proof. Write r = r + k X i =1 a i √ x i , where r ∈ F does not have a √ p part and,for all distinct i, j ∈ { , , . . . , k } , we have a i ∈ Q \{ } , x i ∈ W , p | x i in Z , and x i = x j . If we write v i = x i p for all i ∈ { , , . . . , k } , then each v i is a square-freepositive integer that is not divisible by p . Therefore,11 r = r + √ p k X i =1 a i √ v i ! r = r r √ p + r r , where r = k X i =1 a i √ v i . By thehypothesis that r has a √ p part, r = 0. As each of r , r is nonzero and doesnot have a √ p part, Lemma 3.1 guarantees that r r is nonzero and does nothave a √ p part. Now, it is easy to see that this implies that √ pr r has a √ p part. Furthermore, each of r , r does not have a √ p part, so Lemma 3.1 tellsus that r r does not have a √ p part. Thus, it is clear that r r √ p + r r hasa √ p part, so the proof is complete. Lemma 3.4.
Let us fix d ∈ K and work in the ring O Q ( √ d ) . Let π be a primesuch that N ( π ) = p is an integer prime. If n is an odd positive integer and π | z for some z ∈ O Q ( √ d ) \{ } , then I n ( z ) ∈ F and I n ( z ) has a √ p part.Proof. It is clear that I n ( z ) ∈ F (this is also true for positive even integervalues of n ). Write z ∼ π α π β r Y j =1 π α j j , where, for all distinct j, k ∈ { , , . . . , r } , π j is prime, N ( π j ) = p , α j is a positive integer, and π j π k . Fix some j ∈ { , , . . . , r } . If π j is associated to an inert integer prime, then I n ( π α j j ) ∈ Q ,so I n ( π α j j ) does not have a √ p part. If N ( π j ) = p for some integer prime p ,then I n ( π α j j ) = a + b √ p for some a, b ∈ Q . Again, we conclude that I n ( π α j j )does not have a √ p part because p = p . Writing x = r Y j =1 π α j j , Lemma 3.2 andthe multiplicativity of I n guarantee that I n ( x ) does not have a √ p part. Wenow consider two cases.First, consider the case in which p ramifies in O Q ( √ d ) (meaning π ∼ π ). Then z ∼ π α + β x . Using part ( c ) of Theorem 2.2, we have I n ( π α + β ) = δ − n ( π α + β ) = α + β X m =0 | π m | n = α + β X m =0 √ p mn = t + t √ p , where t and t are positive rational num-bers. Thus, I n ( π α + β ) has a √ p part, so Lemma 3.3 guarantees that I n ( z ) hasa √ p part.Next, consider the case in which p splits in O Q ( √ d ) (meaning π π ). Then12e have I n ( π α π β ) = δ − n ( π α ) δ − n ( π β ) = α X m =0 | π m | n ! β X m =0 | π m | n ! = α X m =0 √ p mn ! β X m =0 √ p mn ! = ( u + u √ p )( u + u √ p ), where u , u , u , u are positive rational numbers. Then ( u + u √ p )( u + u √ p ) = u u + pu u +( u u + u u ) √ p . As u u + u u > I n ( π α π β ) has a √ p part. Once again,Lemma 3.3 guarantees that I n ( z ) has a √ p part. Lemma 3.5.
Let p be an integer prime, and let m , m , β , β be nonnegativeintegers satisfying ( p m + p m )( p β + β +1 +1) = ( p β + p β )( p m + m +1 +1) . Theneither m = β and m = β or m = β and m = β .Proof. Without loss of generality, we may write m = min( m , m , β , β ). Wemay also assume that β ≤ β so that it suffices to show that m = β and m = β . Dividing each side of the given equation by p m , we have(1 + p m − m )( p β + β +1 + 1) = ( p β − m + p β − m )( p m + m +1 + 1) . (1)Suppose m = β . Then (1) becomes (1 + p m − m )( p m + β +1 + 1) = (1 + p β − m )( p m + m +1 + 1). Now, define a function f : R → R by f ( x ) = p m + x +1 + 11 + p x − m . We may differentiate to get f ′ ( x ) = ( p m + x )( p m +1 − p x + p m ) log p > , so f is one-to-one. As f ( m ) = f ( β ), we have m = β . Therefore, we onlyneed to show that m = β .Suppose p = 2. Then, if m < β , we may read (1) modulo p to reach acontradiction. Thus, if p = 2, we are done. Now, suppose p = 2 and m < β
13o that (1) becomes(1 + 2 m − m )(2 β + β +1 + 1) = (2 β − m + 2 β − m )(2 m + m +1 + 1) . (2)The right-hand side of (2) is even, which implies that we must have m = m so that 1 + 2 m − m = 2. Dividing each side of (2) by 2 yields 2 β + β +1 + 1 =(2 β − m − +2 β − m − )(2 m +1 +1). As the left-hand side of this last equation isodd, we must have β = m +1. Therefore, 2 β + β +1 +1 = (1+2 β − β )(2 β − +1) = 2 β + β − + 2 β − β + 2 β − + 1. If we subtract 2 β + β − + 1 from eachside of this last equation, we get 3 · β + β − = 2 β − β + 2 β − . However,3 · β + β − > β + β − + 2 β + β − > β − β + 2 β − , so we have reached ourfinal contradiction. This completes the proof.We now possess the tools necessary to complete the proof of part ( e ) ofTheorem 2.2. We do so in the following two theorems. Theorem 3.1.
Let us work in a ring O Q ( √ d ) with d ∈ K , and let n be anodd positive integer. Let π be a prime such that π ∼ π , and let k be a positiveinteger. Then π k is n -powerfully solitary in O Q ( √ d ) .Proof. We suppose, for the sake of finding a contradiction, that there exists x ∈ O Q ( √ d ) \{ } such that | x | 6 = | π k | and I n ( x ) = I n ( π k ). Suppose that π isa prime such that π | x and N ( π ) = p is an integer prime. Then, by Lemma3.4, I n ( x ) has a √ p part. This implies that I n ( π k ) has a √ p part. However,if N ( π ) = p , where p is an integer prime, then I n ( π k ) = k X m =0 √ p mn = t + t √ p for some t , t ∈ Q . Hence, we find that p = p , which means that π ∼ π . Onthe other hand, if π is associated to an inert integer prime q , then I n ( π k ) ∈ Q .Therefore, if a prime that is not associated to π divides x , that prime must beassociated to an inert integer prime. We now consider two cases.14ase 1: In this case, π ∼ q , where q is an inert integer prime. This impliesthat all primes dividing x must be associated to inert integer primes, so δ n ( x )and | x | are integers. From I n ( x ) = I n ( π k ) and | π k | n = q kn , we have δ n ( x ) q kn = δ n ( π k ) | x | n . We know that δ n ( π k ) = k X j =0 | π j | n = 1 + k X j =1 q jn , so q ∤ δ n ( π k ) in Z .Therefore, q kn divides | x | n in Z , so q k divides | x | in Z . We conclude that q k | x ,so π k | x . However, part ( d ) of Theorem 2.2 tells us that this is a contradiction.Case 2: In this case, N ( π ) = p is an integer prime. Because all of theprime divisors of x that are not associated to π must be associated to inertinteger primes, we may write x ∼ π α t Y j =1 q β j j , where α ∈ N and, for each j ∈{ , , . . . , t } , q j is an inert integer prime and β j is a positive integer. Note that α ≥ I n ( π k ) has a √ p part, which implies that I n ( x ) has a √ p part.Also, α < k because, otherwise, π k | x , from which part ( d ) of Theorem 2.2 yieldsa contradiction. We have I n ( π k ) = P kl =0 √ p ln √ p kn = √ p ( k +1) n − √ p kn ( √ p n − , and I n ( π α ) = P αl =0 √ p ln √ p αn = √ p ( α +1) n − √ p αn ( √ p n − . Now, I n ( π k ) I n ( π α ) = I n t Y j =1 q β j j ∈ Q because each integer prime q j is inert. Thisimplies that ( p ( α +1) n − I n ( π k ) I n ( π α ) ∈ Q . We have( p ( α +1) n − I n ( π k ) I n ( π α ) = ( p ( α +1) n − √ p ( k +1) n − ( √ p ( α +1) n − √ p ( k − α ) n = ( √ p ( α +1) n − √ p ( α +1) n + 1) √ p ( k +1) n − √ p ( α +1) n − √ p ( k − α ) n
15 ( √ p ( k +1) n − √ p ( α +1) n + 1) √ p ( k − α ) n ∈ Q . If k is odd, then √ p ( k +1) n − α must also beodd. Similarly, if α is odd, then k must be odd. Therefore, k and α have thesame parities, which implies that √ p ( k − α ) n is rational. This implies ( √ p ( k +1) n − √ p ( α +1) n + 1) ∈ Q . We clearly have a contradiction if k and α are both even,so they must both be odd. As k is odd, we have I n ( π k ) = δ − n ( π k ) = k X l =0 √ p ln = k − X m =0 √ p mn + k − X m =0 √ p mn (cid:18) √ p n (cid:19) = k − X m =0 p mn (cid:18) √ p n (cid:19) = h p n − (cid:18) √ p n (cid:19) , where h = k − X m =0 p mn ( p n −
1) = p k +12 n − p k − n . Similarly, if we write h = p α +12 n − p α − n , then we have I n ( π α ) = h p n − (cid:18) √ p n (cid:19) . Now, I n t Y j =1 q β j j = I n ( π k ) I n ( π α ) = h h = p k +12 n − p k − α n (cid:16) p α +12 n − (cid:17) , so δ n t Y j =1 q β j j h p k − α n i h p α +12 n − i = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t Y j =1 q β j j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n h p k +12 n − i . Notice that each bracketed expression in this last equation is an integer, andnotice that p divides the left-hand side in Z . However, p does not divide theright-hand side in Z , so we have a contradiction.16e now only have to prove part ( e ) of Theorem 2.2 for the case in which n isodd and π π . We do so as a corollary of the following more general theorem. Theorem 3.2.
Let us work in a ring O Q ( √ d ) with d ∈ K , and let n be an oddpositive integer. Let π be a prime such that π π , and let k , k be nonnegativeintegers. Then π k π k is n -powerfully solitary in O Q ( √ d ) unless, possibly, if k and k are both odd. In the case that k and k are both odd, any friend of π k π k , say x , must satisfy x ∼ π α π α t Y j =1 q γ j j , where α , α are odd positiveintegers and, for each j ∈ { , , . . . , t } , q j is an inert integer prime and γ j is apositive integer.Proof. First note that Fact 1.2 tells us that N ( π ) = N ( π ) = p , where p is aninteger prime.We suppose, for the sake of finding a contradiction, that there exists x ∈O Q ( √ d ) \{ } such that | x | 6 = | π k π k | and I n ( x ) = I n ( π k π k ). Suppose that π is a prime such that π | x and N ( π ) = p is an integer prime. Then, byLemma 3.4, I n ( x ) has a √ p part. This implies that I n ( π k π k ) has a √ p part.However, as N ( π ) = N ( π ) = p , we must have I n ( π k π k ) = I n ( π k ) I n ( π k ) = k X m =0 √ p mn ! k X m =0 √ p mn ! = t + t √ p for some t , t ∈ Q , so we find that p = p . Therefore, if a prime that is not associated to π or π divides x , thatprime must be associated to an inert integer prime. Hence, we may write x ∼ π α π α t Y j =1 q γ j j , where α , α ∈ N and, for each j ∈ { , , . . . , t } , q j isan inert integer prime and γ j is a positive integer.We have I n ( π k ) = P k l =0 √ p ln √ p k n = √ p ( k +1) n − √ p k n ( √ p n − ,I n ( π k ) = P k l =0 √ p ln √ p k n = √ p ( k +1) n − √ p k n ( √ p n − , n ( π α ) = P α l =0 √ p ln √ p α n = √ p ( α +1) n − √ p α n ( √ p n − , and I n ( π α ) = P α l =0 √ p ln √ p α n = √ p ( α +1) n − √ p α n ( √ p n − . Now, I n ( π k ) I n ( π k ) I n ( π α ) I n ( π α ) = I n ( π k π k ) I n ( π α π α ) = I n t Y j =1 q γ j j ∈ Q because each inte-ger prime q j is inert. This implies that( p ( α +1) n − p ( α +1) n − I n ( π k ) I n ( π k ) I n ( π α ) I n ( π α ) ∈ Q . We have( p ( α +1) n − p ( α +1) n − I n ( π k ) I n ( π k ) I n ( π α ) I n ( π α )= ( p ( α +1) n − p ( α +1) n −
1) ( √ p ( k +1) n − √ p ( k +1) n − √ p ( α +1) n − √ p ( α +1) n − √ p ( k + k − α − α ) n = ( √ p ( k +1) n − √ p ( k +1) n − √ p ( α +1) n + 1)( √ p ( α +1) n + 1) √ p ( k + k − α − α ) n ∈ Q . We now consider several cases. In what follows, we will write m = ( k + 1) n −
12 , m = ( k + 1) n −
12 , β = ( α + 1) n −
12 , and β = ( α + 1) n −
12 . This will simplify notation because, for example, if k iseven, then √ p ( k +1) n = p m √ p and m is a nonnegative integer.Case 1: α α ≡ k ≡ k ≡ √ p ( k +1) n − √ p ( k +1) n − √ p ( α +1) n + 1) ∈ Q , so √ p ( α +1) n + 1 √ p ( k + k − α − α ) n ∈ Q . However,this is impossible because ( α + 1) n is odd. By the same argument, we mayshow that it is impossible to have exactly one of k , k , α , α be even.Case 2: α α ≡ k ≡ k ≡ √ p ( α +1) n − ∈ Q , and18 p ( k + k − α − α ) n = µ √ p for some µ ∈ Q . This implies that( √ p ( k +1) n − √ p ( k +1) n − √ p ( α +1) n + 1)= ( p m √ p − p m √ p − p β √ p + 1) = λ √ p for some λ ∈ Q . We may expand to get( p m √ p − p m √ p − p β √ p + 1)= (( p m + m +1 + 1) − ( p m + p m ) √ p )( p β √ p + 1)= ( p m + m +1 + 1 − p β +1 ( p m + p m )) + ( p β ( p m + m +1 + 1) − ( p m + p m )) √ p. As m , m , β ∈ N , we find that p m + m +1 + 1 − p β +1 ( p m + p m ) and p β ( p m + m +1 + 1) − ( p m + p m ) are integers. Therefore, from the equation( p m + m +1 +1 − p β +1 ( p m + p m ))+( p β ( p m + m +1 +1) − ( p m + p m )) √ p = λ √ p ,we have p m + m +1 + 1 − p β +1 ( p m + p m ) = 0. Reading this last equationmodulo p , we have a contradiction. The same argument eliminates the case α α ≡ k ≡ k ≡ k k ≡ α ≡ α ≡ √ p ( k +1) n − ∈ Q , and √ p ( k + k − α − α ) n = µ √ p for some µ ∈ Q . This implies that( √ p ( k +1) n − √ p ( α +1) n + 1)( √ p ( α +1) n + 1)= ( p m √ p − p β √ p + 1)( p β √ p + 1) = λ √ p for some λ ∈ Q . We may expand just as we did in Case 2, and we will find p m + m +1 + 1 + p β +1 ( p m + p m ) = 0, which is clearly a contradiction. Thissame argument eliminates the case k k ≡ α ≡ α ≡ k ≡ k ≡ α ≡ α ≡ √ p ( k +1) n − √ p ( k +1) n − √ p ( k + k − α − α ) n ∈ Q , so we must have( √ p ( α +1) n + 1)( √ p ( α +1) n + 1) = ( p β √ p + 1)( p β √ p + 1) ∈ Q . However, thisis impossible because β and β are nonnegative integers.Case 5: k ≡ k ≡ α ≡ α ≡ √ p ( α +1) n + 1)( √ p ( α +1) n + 1) √ p ( k + k − α − α ) n ∈ Q , so we must have( √ p ( k +1) n − √ p ( k +1) n −
1) = ( p m √ p − p m √ p − ∈ Q . However, thisis impossible because m and m are nonnegative integers.Case 6: k ≡ k ≡ α ≡ α ≡ √ p ( k + k − α − α ) n ∈ Q ,so ( √ p ( k +1) n − √ p ( k +1) n − √ p ( α +1) n + 1)( √ p ( α +1) n + 1)= ( p m √ p − p m √ p − p β √ p + 1)( p β √ p + 1) ∈ Q . One may verify that, after expanding this last expression and noting that m , m , β , and β must be positive integers, we arrive at the requirement( p m + p m )( p β + β +1 + 1) = ( p β + p β )( p m + m +1 + 1). Lemma 3.5 then guar-antees that either m = β and m = β or m = β and m = β , whichmeans that either k = α and k = α or k = α and k = α . Then I n ( π k ) I n ( π k ) I n ( π α ) I n ( π α ) = I n t Y j =1 q γ j j = 1, which implies that t Y j =1 q γ j j is a unit. How-ever, we then find that | π k π k | = | π α π α | = | x | , which we originally assumedwas not true. Therefore, this case yields a contradiction.Case 7: k ≡ α ≡ k ≡ α ≡ √ p ( k +1) n − √ p ( α +1) n + 1) √ p ( k + k − α − α ) n ∈ Q , so we must have( √ p ( k +1) n − √ p ( α +1) n + 1) = ( p m √ p − p β √ p + 1) ∈ Q . Writing( p m √ p − p β √ p + 1) = ( p m + β +1 −
1) + ( p m − p β ) √ p and noting that m β are nonnegative integers, we find that m = β . Therefore, k = α , so I n t Y j =1 q γ j j = I n ( π k ) I n ( π k ) I n ( π α ) I n ( π α ) = I n ( π k ) I n ( π α ) . Because I n t Y j =1 q γ j j >
1, we seethat α < k . As k is odd, we have I n ( π k ) = δ − n ( π k ) = k X l =0 √ p ln = k − X r =0 √ p rn + k − X r =0 √ p rn (cid:18) √ p n (cid:19) = k − X r =0 p rn (cid:18) √ p n (cid:19) = h p n − (cid:18) √ p n (cid:19) , where h = k − X r =0 p rn ( p n −
1) = p k n − p k − n . Similarly, if we write h = p α n − p α − n , then we have I n ( π α ) = h p n − √ p n ). Now, I n t Y j =1 q γ j j = I n ( π k ) I n ( π α ) = h h = p k n − p k − α n ( p α n − , so δ n t Y j =1 q γ j j h p k − α n i h p α n − i = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t Y j =1 q γ j j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n h p k n − i . Now, each bracketed part of this last equation is an integer, and p divides theleft-hand side in Z . However, p does not divide the right-hand side in Z , so wehave a contradiction. We may use this same argument to find contradictions inthe three other cases in which k k (mod 2) and α α (mod 2).One may check that we have found contradictions for all of the possiblechoices of parities of k , k , α , and α except the case in which all four are21dd. Therefore, the proof is complete. Corollary 3.1.
Let d ∈ K , and let k, n ∈ N with n odd. If π is a prime in O Q ( √ d ) such that π π , then π k is n -powerfully solitary in O Q ( √ d ) .Proof. Setting k = k and k = 0 in Theorem 3.1, we find that π k is n -powerfullysolitary in O Q ( √ d ) because k is even. Corollary 3.2.
Let d ∈ K , and let p be an integer prime. Let k be a positiveinteger that is either even or equal to , and let n be an odd positive integer. If z ∼ p k , then z is n -powerfully solitary in O Q ( √ d ) .Proof. If p is inert or ramified in O Q ( √ d ) , then z ∼ π α for some prime π andsome positive integer α . Therefore, z is n -powerfully solitary in O Q ( √ d ) by part( e ) of Theorem 2.2. If p splits in O Q ( √ d ) and k = 1, then z ∼ ππ . Therefore,by Theorem 3.2, any friend of z , say x , must satisfy x ∼ π α π α t Y j =1 q γ j j , where α , α are odd positive integers and, for each j ∈ { , , . . . , t } , q j is an inertinteger prime and γ j is a positive integer. However, this implies that ππ | x , so z | x . This is a contradiction. Finally, if p splits in O Q ( √ d ) and k is even, then z ∼ π k π k . As k is even, the result follows from Theorem 3.2.Note that Corollary 3.1 delivers the final blow in the proof of part ( e ) ofTheorem 2.2. After the introduction of our generalization of the abundancy index, we quicklybecome inundated with new questions. We pose a few such problems, acknowl-edging that their difficulties could easily span a large gamut.22o begin, we note that we have focused exclusively on rings O Q ( √ d ) with d ∈ K . One could generalize the definitions presented here to the other quadraticinteger rings. While complications could surely arise in rings without uniquefactorization, generalizing the abundancy index to unique factorization domains O Q ( √ d ) with d > O Q ( √ d ) with d ∈ K ,we may ask some interesting questions. For example, for a given n , what aresome examples of n -powerful friends in these rings? We might also ask whichnumbers (or which types of numbers), are n -powerfully solitary for a given n .For example, the number 21 is solitary in Z , so it is not difficult to show that 21 isalso solitary in O Q ( √− . Furthermore, for a given element of some ring O Q ( √ d ) ,we might ask to find the values of n for which this element is n -powerfullysolitary. Conjecture 4.1.
Let d ∈ K . If p is an integer prime and k is a positive integer,then p k is n -powerfully solitary in O Q ( √ d ) for all positive integers n . As astronger form of this conjecture, we wonder if π α π α is necessarily n -powerfullysolitary in O Q ( √ d ) whenever π is a prime in O Q ( √ d ) and α , α , n ∈ N . Notethat part ( e ) of Theorem 2.2 and Theorem 3.2 resolve this issue for many cases. The author would like to thank Professor Pete Johnson for inviting him to the2014 REU Program in Algebra and Discrete Mathematics at Auburn Universityand for making that program an extremely relaxed environment that provedexceptionally conducive to research. The author would also like to thank theunknown referee for his or her careful reading.23 eferences ∼ kconrad/blurbs/ugradnumthy/quadraticundergrad.pdf.[2] Klazar, Martin. A question on linear independence of square roots. 2009.Available at http://kam.mff.cuni.cz/ ∼∼