ON CERTAIN SUMS INVOLVING THE LEGENDRE SYMBOL. Borislav Karaivanov Sigma Space Inc., Lanham, Maryland

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1 #A14 INTEGERS 16 (2016) ON CERTAIN SUMS INVOLVING THE LEGENDRE SYMBOL Borisav Karaivanov Sigma Sace Inc., Lanham, Maryand Tzvetain S. Vassiev Deartment of Comuter Science and Mathematics, Niissing University, North Bay, Ontario, Canada Received: 8/10/15, Acceted: 2/13/16, Pubished: 2/29/16 Abstract We rove severa identities on arametric sums invoving the Legendre symbo. 1. Introduction Our work is motivated by severa recent robems ubished in the American Mathematica Monthy that deat with sums invoving the Legendre symbo, most notaby Probem from October 2013 [1]. Here we consider more genera exressions. We begin by recaing the we-known definition from any introductory number theory textbook. Definition 1. For an odd rime number and an integer a, the Legendre symbo a is defined as foows: 8 a >< 0, if a 0 (mod ) 1, if a is a quadratic residue moduo and a 6 0 (mod ) >: 1, if a is a quadratic non-residue moduo. Beow we ist severa imortant roerties of the Legendre symbo that are used throughout the text. In a of them denotes a rime. Proerty 1 (Periodicity). If a b (mod ), then a b.

2 INTEGERS: 16 (2016) 2 Proerty 2 (Mutiicity). ab a b. Whie it is cear from the definition that 1 is aways a quadratic residue moduo any rime, that is not the case with 2 and 1 (usuay 1 is used instead). Proerty 3. ( 1 ( 1) 1 2 ( 2 ( 1) Additionay, when 6 3 ( 3 ( 1) b +1 6 c 1, if 1 (mod 4) 1, if 3 (mod 4), 1, if 1 or 7 (mod 8) 1, if 3 or 5 (mod 8). 1, if 1 or 11 (mod 12) 1, if 5 or 7 (mod 12). When 6 5 ( 5 ( 1) b +2 5 c 1, if 1 or 4 (mod 5) 1, if 2 or 3 (mod 5). Proerty 4. For 1 ae ae ( 1), 8 < :, if 1 (mod 4), if 3 (mod 4). Proof. By Proerty 1 and Proerty 2, we have Now, the resut foows from Proerty In his origina work [3], Legendre gave the exicit formua a a 1 2 (mod ). Another fundamenta resut, due to Gauss [2], is the aw of quadratic recirocity. Proerty 5 (Law of Quadratic Recirocity). For any two odd rimes and q, q ( 1) 1 q q

3 INTEGERS: 16 (2016) 3 2. Preiminaries Lemma 1. For any a and b such that - a, 1 a + b 0. (1) Proof. First, notice that the numbers (a + b) for 0,..., 1 form a comete set of residues moduo. Indeed, if are such that a 1 + b a 2 + b (mod ), then a( 1 2 ) 0 (mod ) which contradicts the fact that - a. Hence, for the given sum we find 1 a + b The ast sum vanishes, since exacty haf of the numbers 1,..., 1 are quadratic residues moduo whie the other haf are quadratic non-residues. Indeed, the congruence 2 ( ) 2 (mod ) imies that the squares of the numbers in the set {1,..., ( 1)/2} are the same, u to ordering and moduo, as the squares of the numbers in the set {( + 1)/2,..., 1}. On the other hand, x 2 6 y 2 (mod ) for any x and y in the first set. Thus, there are exacty ( 1)/2 quadratic residues moduo, and exacty ( 1)/2 quadratic non-residues moduo among the numbers 1,..., 1 as caimed. Lemma 2. Let be a rime congruent to 7 moduo 8. Then (2) 1 Proof. Let S denote the sum to be shown to vanish. Using Lemma 1, we find S (4 + 1) 0<<4 1(mod 4). (3) In the ast sum, we change summation index from to 4, and use Proerty 3

4 INTEGERS: 16 (2016) 4 to obtain 4S 1 0<<4 3(mod 4) (4 ) + 0<<4 3(mod 4) 0<<4 3(mod 4), (4) where for the ast equaity we used Lemma 1 again. After adding (3) and (4) together, we find 8S 0<<4 odd because, by Proerty 3, 0<<4 0<<4 even ( + 2) 0<<2 0<< <<2 0<< , 0<< and, by Lemma 1, the ast sum vanishes. 3. Main Resuts Definition 2. For any rime and integers a and b, et 1 a + b S (a, b). (5) Caim 1. For any odd rime 6 3, S (a, b) is divisibe by. 1 Proof. Let a 6 0 be a number co-rime with the odd rime 6 3. Adding a zero term for 0, we have 1 a + b S (a, b).

5 INTEGERS: 16 (2016) 5 By Proerty 1, there is no oss of generaity in assuming that a and b are reduced moduo. Mutiying both sides of the ast identity by a, we obtain 1 a + b 1 a + b 1 a + b as (a, b) a a (a + b), where the ast ste foows from Lemma 1. As discussed in the roof of Lemma 1, the numbers a + b, 0, 1,..., 1, form a comete set of residues moduo. After roer reordering, we denote these numbers by q +, 0,..., 1. Then, Thus, 1 q + 1 as (a, b) (q + ) q + 1 as (a, b) (mod ). In the case 4n + 1, using Proerty 4, we obtain 1 ( 1)/2 ae + ( ) 2 ( 1)/ (mod ). In the case 4n + 3, using Proerty 4 again, we can air the terms in the sum in the foowing fashion 1 ( )1 ( )1 ae + (2 ) 2 ( )1 ( ) ( 1) 2 2 As discussed in the roof of Lemma 1, the indices for which 1 2, 2 2,..., (( 1)/2) 2 (mod ), u to ordering. Therefore, 2 ( )1 ( 1)/2 2 2 (mod ) ( 1)( + 1) 12 ( )1 (mod ). 1 are exacty (mod ) 0 (mod ). The ast ste is ustified by the fact that is co-rime with 12 when 6 3. Caim 2. If is rime and - a, then S (a, b) 1 S (a, a b). (6)

6 INTEGERS: 16 (2016) 6 Proof. Changing summation index in S (a, b) from to 1, we obtain 2 a( 1 ) + b S (a, b) ( 1 ) 2 a (a b) 2 a (a b) ( 1) 1 a (a b) a + b ( 1) ( 1) 1 1 a + (a b) a + b + ( 1) 1 1 S (a, a b). Caim 3. For any integers a, b, and c, S (ca, cb) c S (a, b). (7) Proof. The caim foows directy from Definition 2 and Proerty 2. The ast caim indicates that it su ces to study the vaues of S (a, b) ony for co-rime airs (a, b). To reduce even further the set of uncharted airs, we introduce the foowing resut. Caim 4. For any integers a and b, the foowing formua hods b S (a, a + b) S (a, b) +. (8) Proof. By changing the summation index from to 1, we obtain 1 a + a + b a + b S (a, a + b) ( 1) 1 2 a + b a + b a + b a + b b a + b b S (a, b) +, where for the ast equaity we used Lemma 1. + a + b

7 INTEGERS: 16 (2016) 7 Coroary 1. For any integers a, b, and m S (a, ma + b) S (a, b) + Proof. Ay Caim 4 inductivey m times. 1 the foowing formua is vaid m 1 a + b. (9) Thus, for any fixed a, we ony need to know the vaues of S (a, b) for the set b 2 {0,..., a 1}. By Caim 3, this set can be further reduced to those b s that are co-rime with a, assuming the sums S (a, b) for smaer vaues of a are aready known. It is evident from the above formua, as it is from Definition 2, that the sum S (a, ma + b) is eriodic in m with eriod of. An interesting secia case of Caim 4 is a b. On one hand, we have S (a, a) S (a, 0). On the other hand, 1 from Caim 2 we have S (a, a) S (a, 0). Thus, S (a, a) S (1, 1) 0 1 whenever 1, which according to Proerty 3 is recisey when 4n + 1. In the case when 4n + 3, S (a, a) S (a, 0) is not necessariy zero as suggested by the roof of Caim 1. These considerations are su cient to comute S (1, m) for any m. We ony need to know S (1, 0) or some S (1, ) for that matter. Aso note that formua (9) can be aied backwards, i.e. S (a, a + b) S (a, b) a + b, (10) and therefore m S (a, ma + b) S (a, b) 1 a + b. (11) When a > 1, due to the eriodicity, we need more initia vaues in order to comute the vaues of S (a, b) for a b. In Caims 5 and 6 beow, we rovide the vaues of S (a, b) for some other co-rime airs (a, b). Caim 5. For any rime congruent to 1 moduo 4, Proof. From Caim 2, we have S (2, 1) S (2, 1) 0. (12) S (2, 1), because Caim 6. For any rime congruent to 7 moduo 8, 1 1. S (2, 1) 0 (13)

8 INTEGERS: 16 (2016) 8 and S (4, 1) 0 S (4, 3). (14) Proof. We estabish (13) as foows: because 1 2k k + 1 2S (2, 1) 2 k k0 k <<2 odd 0<< << << ( + ) 0<<2 even + 0<< (2k + 1) 2 0<< 0, 1 2k + 1 k0 2 In (14), S (4, 1) 0 by Lemma 2. The second equaity foows from the first by Caim 2. The foowing rocedure summarizes our findings: Theorem 1. Let be an odd rime, and et a be an integer, reativey rime to. For any integer b, reativey rime to a, the vaue of S (a, b) can be comuted in the foowing way: 1. Let k be the smaest (ositive) integer such that ka b (mod ). Then, by Proerty 1, 2. Using Caim 3, reduce S (a, ka) by S (a, b) S (a, ka). (15) S (a, ka) a S (1, k). (16) 3. Further reduce S (1, k) according to k 1 S (1, k) S (1, 0) + which is a articuar case of (9), the Coroary of Caim 4. (17)

9 INTEGERS: 16 (2016) 9 Or, summarizing 1 3 in a singe formua, S (a, b) a Beow, we iustrate the use of Theorem 1. Exame. Comute S 53 (7, 13). 0 k (1, 0) + 1 A. (18) Foowing the stes outined above, we sove the inear congruence 7k 13 (mod 53). The smaest ositive integer soution is k 17 and 7 7 S 53 (7, 13) S 53 (1, 17) S 53 (1, 17), for Thus, we need to find S 53 (1, 17). Since 53 is of the form 4n + 1, the sum S 53 (1, 0) vanishes. Therefore 16 S 53 (1, 17) Among the integers from 1 to 16, quadratic residues moduo 53 are 1, 4, 6, 7, 9, 10, 11, 13, 15, 16. Hence, the ast sum above has one zero term, 10 ositive terms and 6 negative terms. Thus, S 53 (1, 17) In Tabe 1 we show the reativey rime airs (a, b) with 1 ae a ae 16 and b ae 25 for which the vaue of S (a, b) is either 0, ±, or ± 2 for every rime of the secified form. 8n + 1 8n + 3 8n + 5 8n + 7 S (1, 2) 2 0 S (1, 0) 0 0 S (1, 2) S (1, 3) 2 0 S (2, 1) S (2, 1) S (2, 3) S (4, 1) 0 S (6, 1) 0 0 Tabe 1: Reativey rime airs (a, b) whose vaue of S (a, b) 0, ±, ±2. Note that for 8n + 3 the vaues of S (a, b) do not foow any sime attern.

10 INTEGERS: 16 (2016) 10 Acknowedgements. The resented roof of Lemma 2 is a sighty modified version of a roof by Dr. Henryk Iwaniec, contributed by ersona communication [4]. Our many thanks go to him. We thank Dr. Ognian Trifonov for his exert suggestions and enthusiastic encouragement without which this aer woud not have materiaized. He aso suggested Integers as a ubication venue. Dr. Tzvetain S. Vassiev acknowedges the financia suort of his research rovided by the Natura Sciences and Engineering Research Counci of Canada (NSERC) through Discovery Grant. References [1] Gerad A. Edgar, Doug Hensey and Dougas B. West, editors. Probems and Soutions. Amer. Math. Monthy, 120(8) (2013), 754. [2] Johann Car Friedrich Gauß. Disquisitiones Arithmeticæ 4, arts , Leizig, [3] Adrien-Marie Legendre. Essai sur a Theorie des Nombres,. 186, Paris, [4] Henryk Iwaniec. Persona Communication