How to Di®erentiate an Integer Modulo n Caleb Emmons,¤ Mike Krebs,y and Anthony Shaheenz 1 Introduction To generate an undergraduate research project, take a familiar mathematical object and attempt to make sense of it in a completely di®erent setting. Here, we will make sense of the concept of di®erentiation in the setting of the integers mod n. The articles [1], [6], and [7], de¯ne a \number derivative" on Z (also called an \arithmetic derivative" or \quasiderivation") to be a map that satis¯es the product rule (that is, Á(xy) = Á(x)y + xÁ(y) for all x; y 2 Z). Inspired by these papers, we de¯ne a number derivative on Zn to be a map from Zn to itself which satis¯es the product rule, and proceed to classify all number derivatives on Zn. At the end of this article, we list some undergraduate research projects one can pursue using these maps as a starting point. A motivated student will quickly be able to dig in and obtain some interesting results, without needing any sophisticated mathematics. Indeed, the author of [7] was a high school student when she wrote that paper! ¤Department of Mathematics and Computer Science, Paci¯c University, 2043 College Way, Forest Grove, Oregon 97116, emmons@paci¯cu.edu yDepartment of Mathematics, California State University - Los Angeles, 5151 State University Drive, Los Angeles, California 90032, [email protected] zDepartment of Mathematics, California State University - Los Angeles, 5151 State University Drive, Los Angeles, California 90032, [email protected] 1 2 Basic properties of number derivatives on Zn Let Zn = f 0 ; 1 ;:::; n ¡ 1 g be the set of integers modulo n. Note that we write elements of Zn with bars on top of them, so as to distinguish them from ordinary integers. De¯nition 1 We say that Á : Zn ! Zn is a number derivative on Zn if Á( x y ) = x Á( y ) + y Á( x ) for all x ; y 2 Zn. We shall sometimes refer to De¯nition 1 as the product rule. Lemma 1 If Á is a number derivative on Zn, then Á( 0 ) = 0 ;Á( 1 ) = 0 ; and Á( x m) = m x m¡1Á( x ) for all x 2 Zn and m ¸ 1. Proof. Note that Á( 0 ) = Á( 0 ¢ 0 ) = 2 Á( 0 ): Subtract Á( 0 ) from both sides to get Á( 0 ) = 0 . A similar argument shows that Á( 1 ) = 0 . The last result follows by induction using the product rule. ¤ If a 2 Zn, then by a Zn we mean the set of all multiples of a . That is, a Zn = f a ¢ k j k 2 Zng: For example, suppose that à is a number derivative on Z4. Then, 0 = Ã( 3 ¢ 3 ) = 3 Ã( 3 ) + 3 Ã( 3 ) = 2 Ã( 3 ) Ã( 2 ) = Ã( 2 ¢ 3 ) = 2 Ã( 3 ) + 3 Ã( 2 ) The ¯rst equation tells us that Ã( 3 ) 2 2 Z4 = f 0 ; 2 g. This combined with the second equation gives us that 2 Ã( 2 ) = 0 . Hence, Ã( 2 ) 2 f 0 ; 2 g. Therefore, à is one of the maps in Table 1, each of which, one easily veri¯es, is a number derivative on Z4. 2 Table 1: Number derivatives on Z4. 0 1 2 3 Ã1 0 0 0 0 Ã2 0 0 2 0 Ã3 0 0 0 2 Ã4 0 0 2 2 3 Number Derivatives Modulo a Power of 2 We begin by classifying all number derivatives on Z2e , where e is a positive integer. Lemma 1 shows that the only number derivative on Z21 is the zero map. Our ¯rst example dealt with the case e = 2. So from now on, we assume that e ¸ 3. To classify number derivatives on Z2e , we need some kind of multiplicative basis. After all, the product rule is all about multiplication. Fortunately we have the following result: Lemma 2 Let e be a positive integer. (i) Let x 2 Z2e . Then there exist integers i; k; m with 0 · i · e such that x = 2i ¢ 5k ¢ (¡1)m . (ii) Let i; j; k; `; m; n 2 Z such that 0 · i; j · e. Then 2i ¢ 5k ¢ (¡1)m = 2j ¢ 5` ¢ (¡1)n i® i = j and 5k ¢ (¡1)m ´ 5` ¢ (¡1)n (mod 2e¡i). (iii) If e ¸ 2, then 5k ¢ (¡1)m = 5` ¢ (¡1)n i® m ´ n (mod 2) and k ´ ` (mod 2e¡2). (iv) If e ¸ 2, then ( 5 )2e¡2 = 1 . 3 For a proof see [5, pg. 105]. Note, however, that (iv) is a special case of (iii), with k = 2e¡2 and m = ` = n = 0. Lemma 2 tells us that every element of Z2e can be written as a power of 2, times a power of 5, times a power of ¡1. Moreover, if two elements of Z2e are written this way, Lemma 2 tells us precisely how to determine whether or not they are equal. Using Lemma 1 and the product rule, we ¯nd that if Á is any number i k m derivative on Z2e and x = 2 ¢ 5 ¢ (¡1) , then Á( x ) = i ¢ 2i¡1 ¢ 5k ¢ (¡1)m ¢ Á( 2 )+ k ¢ 2i ¢ 5k¡1 ¢ (¡1)m ¢ Á( 5 )+ (1) m ¢ 2i ¢ 5k ¢ (¡1)m¡1 ¢ Á( ¡1 ): Consequently, we can determine what Á does to any element of Z2e , as long as we know Á( 2 ), Á( 5 ), and Á( ¡1 ). Hence we are compelled to deter- mine what restrictions there are on the values of Á( 2 ), Á( 5 ), and Á( ¡1 ). Lemma 3 Let e ¸ 3. If Á is a number derivative on Z2e , then Á( 2 ) 2 2 Z2e , e¡1 Á( 5 ) 2 4 Z2e , and Á( ¡1 ) 2 2 Z2e . Proof. First note that from Lemma 1, we have 2 ¢ Á( ¡1 ) = ¡Á( ¡1 2) = ¡Á( 1 ) = 0 : Therefore Á( ¡1 ) is a multiple of 2e¡1 . Similarly, from Lemmas 1 and 2, we have 0 = Á( 1 ) = Á( 5 2e¡2 ) = 2e¡2 ¢ 52e¡2¡1 ¢ Á( 5 ): Since 52e¡2¡1 is odd, we must have that Á( 5 ) is a multiple of 4 . Finally, we show that Á( 2 ) 2 2 Z2e . If e is odd, then 0 = Á( 0 ) = Á( 2 e) = e ¢ 2e¡1 ¢ Á( 2 ) implies that Á( 2 ) 2 2 Z2e . Now suppose that e is even. We have that Á( 2e¡1 ) = Á( 2e + 2e¡1 ) = Á( 2e¡1 ¢ 3 ) (2) = 3 Á( 2e¡1 ) + 2e¡1 Á( 3 ): 4 Now, 3 is odd, and so it follows from Lemma 2 that 3 = 5k ¢ (¡1)m for some k; m. Therefore Á( 3 ) = k ¢ 5k¡1 ¢ (¡1)m ¢ Á( 5 ) + m ¢ 5k ¢ (¡1)m¡1 ¢ Á( ¡1 ): e¡1 In particular, using the facts that Á( 5 ) 2 4 Z2e and Á( ¡1 ) 2 2 Z2e , we have that Á( 3 ) is even. Consequently, the term 2e¡1 Á( 3 ) in (2) disappears and we get Á( 2e¡1 ) = 3 Á( 2e¡1 ). Therefore 2 Á( 2e¡1 ) = 0 , which in turn implies that e ¡ 1 ¢ 2e¡1 ¢ Á( 2 ) = 0 . We know that e ¡ 1 is odd (since e was even by assumption), and so Á( 2 ) 2 2 Z2e . ¤ It would be tempting at this point simply to de¯ne a number derivative Á on Z2e by setting Á( 2 ), Á( 5 ), and Á( ¡1 ) equal to arbitrary multiples of e¡1 2 , 4 , and 2 , respectively, and extending Á to all of Z2e by (1). Indeed, this is precisely how we shall eventually de¯ne all number derivatives on Z2e . However, we must exercise some caution, as an element of Z2e may have many di®erent representations in the form 2i ¢ 5k ¢ (¡1)m . For example, if 2 0 0 2 2 0 e = 5 (so we are in Z32), we have 2 ¢ 5 ¢ (¡1) = 2 ¢ 5 ¢ (¡1) . In order for (1) to provide a well-de¯ned function, we must be certain that when di®erent values of i; k; m give us the same element of Z2e , plugging them into (1) yields the same result. The following lemma shows that this is exactly what happens. e¡1 i k m Lemma 4 Let a 2 2 Z2e , b 2 4 Z2e , and and c 2 2 Z2e . If 2 ¢ 5 ¢ (¡1) = 2j ¢ 5` ¢ (¡1)n and 0 · i; j · e, then i ¢ 2i¡1 ¢ 5k ¢ (¡1)m ¢ a + k ¢ 2i ¢ 5k¡1 ¢ (¡1)m ¢ b + m ¢ 2i ¢ 5k ¢ (¡1)m¡1 ¢ c = j ¢ 2j¡1 ¢ 5` ¢ (¡1)n ¢ a + ` ¢ 2j ¢ 5`¡1 ¢ (¡1)n ¢ b + n ¢ 2j ¢ 5` ¢ (¡1)n¡1 ¢ c We leave the proof to the reader; it follows rapidly from Lemma 2. Lemma 4 says that (1) yields a well-de¯ned function Á from Z2e to Z2e , provided that Á( 2 ) is a multiple of 2 , Á( 5 ) is a multiple of 4 , and Á( ¡1 ) is a multiple of 2e¡1 . We may therefore make the following de¯nition with a clear conscience. e¡1 De¯nition 2 Let e ¸ 3, a 2 2 Z2e , b 2 4 Z2e , and c 2 2 Z2e . De¯ne Á a ; b ; c : Z2e ! Z2e by i k m i¡1 k m Á a ; b ; c ( 2 ¢ 5 ¢ (¡1) ) = i ¢ 2 ¢ 5 ¢ (¡1) ¢ a + k ¢ 2i ¢ 5k¡1 ¢ (¡1)m ¢ b + m ¢ 2i ¢ 5k ¢ (¡1)m¡1 ¢ c : 5 We now state a result which completely characterizes number derivatives modulo powers of 2. e¡1 Theorem 1 Let e ¸ 3, a 2 2 Z2e , b 2 4 Z2e , and c 2 2 Z2e .