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On the Multiplicative Order of an Modulo N Jonathan Chappelon On the multiplicative order of an modulo n Jonathan Chappelon To cite this version: Jonathan Chappelon. On the multiplicative order of an modulo n. Journal of Integer Sequences, University of Waterloo, 2010, 13, pp.Article 10.2.1. hal-00371233 HAL Id: hal-00371233 https://hal.archives-ouvertes.fr/hal-00371233 Submitted on 22 Mar 2016 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. On the multiplicative order of an modulo n Jonathan Chappelona,b,c aUniv Lille Nord de France, F-59000 Lille, France bULCO, LMPA J. Liouville, B.P. 699, F-62228 Calais, France cCNRS, FR 2956, France January 20, 2010 Abstract Let n be a positive integer and αn be the arithmetic function which assigns the multiplicative order of an modulo n to every integer a coprime to n and vanishes n elsewhere. Similarly, let βn assign the projective multiplicative order of a mod- ulo n to every integer a coprime to n and vanishes elsewhere. In this paper, we present a study of these two arithmetic functions. In particular, we prove that for positive integers n1 and n2 with the same square-free part, there exists an exact relationship between the functions αn1 and αn2 and between the functions βn1 and βn2 . This allows us to reduce the determination of αn and βn to the case where n is square-free. These arithmetic functions recently appeared in the context of an old problem of Molluzzo, and more precisely in the study of which arithmetic progressions yield a balanced Steinhaus triangle in Z/nZ for n odd. 2000 Mathematics Subject Classifications: 11A05, 11A07, 11A25. Keywords: multiplicative order, projective multiplicative order, balanced Stein- haus triangles, Steinhaus triangles, Molluzzo’s Problem. 1 Introduction We start by introducing some notation relating to the order of certain elements modulo n. For every positive integer n and every prime number p, we denote by vp(n) the p- adic valuation of n, i.e., the greatest exponent e > 0 for which pe divides n. The prime factorization of n may then be written as n = pvp(n), P Yp∈ where P denotes the set of all prime numbers. We denote by rad(n) the radical of n, i.e., the largest square-free divisor of n, namely rad(n)= p. p∈P Yp|n 1 For every positive integer n and every integer a coprime to n, we denote by On(a) the multiplicative order of a modulo n, i.e., the smallest positive integer e such that ae ≡ 1 (mod n), namely ∗ e On(a) = min {e ∈ N | a ≡ 1 (mod n)} and we denote by Rn(a) the multiplicative remainder of a modulo n, i.e., the multiple of n defined by On(a) Rn(a)= a − 1. The multiplicative order of a modulo n also corresponds with the order of the element πn(a), where πn : Z −։ Z/nZ is the canonical surjective morphism, in the multiplicative ∗ group (Z/nZ) , the group of units of Z/nZ. Note that On(a) divides ϕ(n), ϕ being the Euler’s totient function. For every positive integer n, we define and denote by αn the arithmetic function αn : Z −→ N O (an) , for all gcd(a, n) = 1; a 7−→ n 0, otherwise, where gcd(a, n) denotes the greatest common divisor of a and n, with the convention that gcd(0, n) = n. Observe that, for every a coprime to n, the integer αn(a) divides ϕ(n)/ gcd(ϕ(n), n). This follows from the previous remark on On(a) and the equality n αn(a)= On(a )= On(a)/ gcd(On(a), n). For every positive integer n and every integer a coprime to n, we denote by POn(a) the projective multiplicative order of a modulo n, i.e., the smallest positive integer e such that ae ≡ ±1 (mod n), namely ∗ e POn(a) = min {e ∈ N | a ≡ ±1 (mod n)} . The projective multiplicative order of a modulo n also corresponds with the order of the ∗ element πn(a) in the multiplicative quotient group (Z/nZ) /{−1, 1}. For every positive integer n, we define and denote by βn the arithmetic function βn : Z −→ N PO (an) , for all gcd(a, n) = 1; a 7−→ n 0, otherwise. Observe that we have the alternative αn = βn or αn =2βn. In this paper, we study in detail these two arithmetic functions. In particular, we prove that, for every positive integers n1 and n2 such that rad(n1)|n2 and n2|n1, if v2(n1) 6 1; 2 rad(n )|n and n |n , if v (n ) > 2, 1 2 2 1 2 1 the integer αn1 (a) (respectively βn1 (a)) divides αn2 (a) (resp. βn2 (a)), for every integer a. More precisely, we determine the exact relationship between the functions αn1 and αn2 and between βn1 and βn2 . We prove that we have αn2 (a) αn (a)= for all gcd(a, n1)=1 1 gcd(n ,R (a)) gcd α (a), 1 n2 n2 n2 2 in Theorem 2.5 of Section 2 and that we have βn2 (a) βn (a)= for all gcd(a, n1)=1 1 gcd(n ,R (a)) gcd β (a), 1 n2 n2 n2 in Theorem 3.3 of Section 3. Thus, for every integer a coprime to n, the determination of αn(a) is reduced to the computation of αrad(n)(a) and Rrad(n)(a) if v2(n) 6 1 and of α2 rad(n)(a) and R2 rad(n)(a) if v2(n) > 2. These theorems on the functions αn and βn are derived from Theorem 2.6 of Section 2, which states that n1 On1 (a)= On2 (a) · , gcd(n1, Rn2 (a)) for all integers a coprime to n1 and n2. This result generalizes the following theorem of Nathanson which, in the above notation, states that for every odd prime number p and for every positive integer k, we have the equality pk k Op (a)= Op(a) · k gcd(p , Rp(a)) for all integers a not divisible by p. Theorem 3.6 of [3]. Let p be an odd prime, and let a =6 ±1 be an integer not divisible d by p. Let d be the order of a modulo p. Let k0 be the largest integer such that a ≡ 1 k0 k k−k0 (mod p ). Then the order of a modulo p is d for k =1,...,k0 and dp for k > k0. For every finite sequence S = (a1,...,am) of length m > 1 in Z/nZ, we denote by m+1 Z Z ∆S the Steinhaus triangle of S, that is the finite multiset of cardinality 2 in /n defined by i i ∆S = aj+k 0 6 i 6 m − 1 , 1 6 j 6 m − i . ( k ) k=0 ZXZ Z Z A finite sequence S in /n is said to be balanced if each element of /n occurs in its Steinhaus triangle ∆S with the same multiplicity. For instance, the sequence (2, 2, 3, 3) of length 4 is balanced in Z/5Z. Indeed, as depicted in Figure 1, its Steinhaus triangle is composed by each element of Z/5Z occuring twice. 2233 4 0 1 4 1 0 Figure 1: The Steinhaus triangle of a balanced sequence in Z/5Z Note that, for a sequence S of length m > 1 in Z/nZ, a necessary condition on m for m+1 S to be balanced is that the integer n divides the binomial coefficient 2 . In 1976, John C. Molluzzo [2] posed the problem to determine whether this necessary condition on m is also sufficient to guarantee the existence of a balanced sequence. In [1], it was proved that, for each odd number n, there exists a balanced sequence of length m for every m ≡ 0 or −1 (mod αn(2) · n) and for every m ≡ 0 or −1 (mod βn(2) · n). This was achieved by analyzing the Steinhaus triangles generated by arithmetic progressions. In particular, since β3k (2) = 1 for all k > 1, the above result implies a complete and positive solution of Molluzzo’s Problem in Z/nZ for all n =3k. 3 2 The arithmetic function αn The table depicted in Figure 2 gives us the first values of αn(a) for every positive integer n, 1 6 n 6 20, and for every integer a, −20 6 a 6 20. n\a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 3 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 4 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 5 1 4 4 2 0 1 4 4 2 0 1 4 4 2 0 1 4 4 2 0 6 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 7 1 3 6 3 6 2 0 1 3 6 3 6 2 0 1 3 6 3 6 2 8 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 9 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 10 1 0 2 0 0 0 2 0 1 0 1 0 2 0 0 0 2 0 1 0 11 1 10 5 5 5 10 10 10 5 2 0 1 10 5 5 5 10 10 10 5 12 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 13 1 12 3 6 4 12 12 4 3 6 12 2 0 1 12 3 6 4 12 12 14 1 0 3 0 3 0 0 0 3 0 3 0 1 0 1 0 3 0 3 0 15 1 4 0 2 0 0 4 4 0 0 2 0 4 2 0 1 4 0 2 0 16 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 17 1 8 16 4 16 16 16 8 8 16 16 16 4 16 8 2 0 1 8 16 18 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 19 1 18 18 9 9 9 3 6 9 18 3 6 18 18 18 9 9 2 0 1 20 1 0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 0 Figure 2: The first values of αn(a) The positive integer αn(a) seems to be difficult to determine.
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