On the Multiplicative Order of a n Modulo n

Transcription

1 Journal of Integer Sequences, Vol ), Article On the Multilicative Order of a n Modulo n Jonathan Chaelo Université Lille Nord de France F Lille France jonathan.chaelon@lma.univ-littoral.fr Abstract Let n be a ositive integer and α n be the arithmetic function which assigns the multilicative order of a n modulo n to every integer a corime to n and vanishes elsewhere. Similarly, let β n assign the rojective multilicative order of a n modulo n to every integer a corime to n and vanishes elsewhere. In this aer, we resent a study of these two arithmetic functions. In articular, we rove that for ositive integers and with the same square-free art, there exists a relationshi between the functions α n1 and α n2 and between the functions β n1 and β n2. This allows us to reduce the determination of α n and β n to the case where n is square-free. These arithmetic functions recently aeared in the context of an old roblem of Molluzzo, and more recisely in the study of which arithmetic rogressions yield a balanced Steinhaus triangle in Z/nZ for n odd. 1 Introduction We start by introducing some notation relating to the order of certain elements modulo n. For every ositive integer n and every rime number, we denote by v n) the -adic valuation of n, i.e., the greatest exonent e 0 for which e divides n. The rime factorization of n may then be written as n = P vn), where P denotes the set of all rime numbers. We denote by radn) the radical of n, i.e., the largest square-free divisor of n, namely radn) = P n 1 ULCO, LMPA J. Liouville, B.P. 699, F Calais, France and CNRS, FR 2956, France.. 1

2 For every ositive integer n and every integer a corime to n, we denote by O n a) the multilicative order of a modulo n, i.e., the smallest ositive integer e such that a e 1 mod n), namely O n a) = min {e N a e 1 mod n)} and we denote by R n a) the multilicative remainder of a modulo n, i.e., the multile of n defined by R n a) = a Ona) 1. The multilicative order of a modulo n also corresonds with the order of the element π n a), where π n : Z Z/nZ is the canonical surjective morhism, in the multilicative grou Z/nZ), the grou of units of Z/nZ. Note that O n a) divides ϕn), ϕ being the Euler s totient function. For every ositive integer n, we define and denote by α n the arithmetic function α n : Z a N { On a n ), for all gcda,n) = 1; 0, otherwise, where gcda,n) denotes the greatest common divisor of a and n, with the convention that gcd0,n) = n. Observe that, for every a corime to n, the integer α n a) divides ϕn)/ gcdϕn),n). This follows from the revious remark on O n a) and the equality α n a) = O n a n ) = O n a)/ gcdo n a),n). For every ositive integer n and every integer a corime to n, we denote by PO n a) the rojective multilicative order of a modulo n, i.e., the smallest ositive integer e such that a e ±1 mod n), namely PO n a) = min {e N a e ±1 mod n)}. The rojective multilicative order of a modulo n also corresonds with the order of the element π n a) in the multilicative quotient grou Z/nZ) /{ 1, 1}. For every ositive integer n, we define and denote by β n the arithmetic function β n : Z a N { POn a n ), for all gcda,n) = 1; 0, otherwise. Observe that we have the alternative α n = β n or α n = 2β n. In this aer, we study in detail these two arithmetic functions. In articular, we rove that, for every ositive integers and such that { radn1 ) and, if v 2 ) 1; 2 rad ) and, if v 2 ) 2, the integer α n1 a) resectively β n1 a)) divides α n2 a) res. β n2 a)), for every integer a. More recisely, we determine the exact relationshi between the functions α n1 and α n2 and between β n1 and β n2. We rove that we have α n1 a) = α n2 a) ) for all gcda, ) = 1 gcd α n2 a), gcd,r n2 a)) 2

3 in Theorem 6 of Sectio and that we have β n1 a) = β n2 a) ) for all gcda, ) = 1 gcd β n2 a), gcd,r n2 a)) in Theorem 13 of Section 3. Thus, for every integer a corime to n, the determination of α n a) is reduced to the comutation of α radn) a) and R radn) a) if v 2 n) 1 and of α 2 radn) a) and R 2radn) a) if v 2 n) 2. These theorems on the functions α n and β n are derived from Theorem 7 of Sectio, which states that O n1 a) = O n2 a) gcd, R n2 a)), for all integers a corime to and. This result generalizes the following theorem of Nathanson which, in the above notation, states that for every odd rime number and for every ositive integer k, we have the equality for all integers a not divisible by. O ka) = O a) k gcd k, R a)) Theorem 3.6 of [3]. Let be an odd rime, and let a ±1 be an integer not divisible by. Let d be the order of a modulo. Let k 0 be the largest integer such that a d 1 mod k 0 ). Then the order of a modulo k is d for k = 1,...,k 0 and d k k 0 for k k 0. For every finite sequence S = a 1,...,a m ) of length m 1 in Z/nZ, we denote by S the Steinhaus triangle of S, that is the finite multiset of cardinality ) m+1 2 in Z/nZ defined by { i ) } i S = a j+k 0 i m 1, 1 j m i. k k=0 A finite sequence S in Z/nZ is said to be balanced if each element of Z/nZ occurs in its Steinhaus triangle S with the same multilicity. For instance, the sequence 2, 2, 3, 3) of length 4 is balanced in Z/5Z. Indeed, as deicted in Figure 1, its Steinhaus triangle is comosed by each element of Z/5Z occuring twice 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 S to be balanced is that the integer n divides the binomial coefficient ) m+1 2. I976, John C. Molluzzo [2] osed the roblem to determine whether this necessary condition on m is also sufficient to guarantee the existence of a balanced sequence. In [1], it was roved that, 3

4 for each odd number n, there exists a balanced sequence of length m for every m 0 or 1 mod α ) n) and for every m 0 or 1 mod β ) n). This was achieved by analyzing the Steinhaus triangles generated by arithmetic rogressions. In articular, since β 3 k2) = 1 for all k 1, the above result imlies a comlete and ositive solution of Molluzzo s roblem in Z/nZ for all n = 3 k. 2 The arithmetic function α n The table deicted in Figure 2 gives us the first values of α n a) for every ositive integer n, 1 n 20, and for every integer a, 20 a 20. n\a Figure 2: The first values of α n a) The ositive integer α n a) seems to be difficult to determine. Indeed, there is no general formula known to comute the multilicative order of an integer modulo n but, however, we get the following helful roositions. Lemma 1. Let and be two ositive integers such that rad ) = rad ). Then, an integer a is corime to if, and only if, it is also corime to. Proof. This follows from the definition of the greatest common divisor of two integers and from the definition of the radical of an integer. 4

5 Proositio. Let and be two ositive integers such that rad ) and. Then, for every integer a, the integer α n1 a) divides α n2 a). Proof. If a is not corime to and, then, by definition of the functions α n1 and α n2 and by Lemma 1, we have α n1 a) = α n2 a) = 0. Suose that a is corime to and. If v ) = 1 for all rime factors of, then =. Otherwise, let be a rime factor of such that v ) 2. We shall show that α n1 a) divides α n1 /a). By definition of α n1 /a), there exists an integer u such that a α /a) n1 = 1 + u. Therefore, by the binomial theorem, we have ) ) a α /a) = a α /a) n1 = 1 + u n1 = 1 + u + k=2 ) u k k ) k n1. Since v ) 2, it follows that /) k is divisible by for every integer k 2 and so a α /a) 1 mod ). Hence α n1 a) divides α n1 /a). This comletes the roof. An exact relationshi between α n1 a) and α n2 a), for every integer a corime to and, is determined at the end of this section. We first settle the easy rime ower case. Proosition 3. Let be a rime number and let a be an integer. Then we have for every ositive integer k. α ka) = O a) Proof. Let k be a ositive integer. If a is not corime to, then we have α ka) = α a) = 0. Suose now that a is corime to. By Proositio, the integer α ka) divides α a). It remains to rove that α a) divides α ka). The congruence imlies that a α k a) k 1 mod k ) a α k a) k 1 mod ), and hence, by Fermat s Little Theorem, it follows that a α k a) a α k a) k 1 mod ). Therefore α a) divides α ka). Finally, we have α ka) = α a) = O a ) = O a). This comletes the roof. 5

6 Remark 4. If = 2, then, for every ositive integer k, we obtain α 2 ka) = O 2 a) = { 0, for a even; 1, for a odd. Proosition 5. Let and be two corime numbers and let a be an integer. Then α n1 a) divides lcmα n1 a),α n2 a)), the least common multile of α n1 a) and α n2 a). Proof. If gcda, ) 1, then gcda, ) 1 or gcda, ) 1 and so α n1 a) = lcmα n1 a),α n2 a)) = 0. Suose now that gcda, ) = 1 and hence that the integers a, and are corime airwise. Let i {1, 2}. The congruences imly that a αn i a) n i 1 mod n i ) a lcmα n1 a),α n2 a)) 1 mod n i ). Therefore α n1 a) divides lcmα n1 a),α n2 a)) by the Chinese remainder theorem. Let and be two ositive integers such that { radn1 ) and, if v 2 ) 1; 2 rad ) and, if v 2 ) 2. By definition, we know that α n1 a) = α n2 a) = 0 for every integer a not corime to and. We end this section by determining the exact relationshi between α n1 a) and α n2 a) for every integer a corime to and. Theorem 6. Let and be two ositive integers such that { radn1 ) and, if v 2 ) 1; 2 rad ) and, if v 2 ) 2. Then, for every integer a corime to and, we have α n1 a) = α n2 a) ) gcd α n2 a), gcd,r n2 a)) This result is a corollary of the following theorem. Theorem 7. Let and be two ositive integers such that { radn1 ) and, if v 2 ) 1; 2 rad ) and, if v 2 ) 2. Then, for every integer a corime to and, we have O n1 a) = O n2 a) 6 gcd, R n2 a)).

7 The roof of this theorem is based on the following lemma. Lemma 8. Let n be a ositive integer and let a be an integer corime to n. Let m be an integer such that radm) radn. Then, there exists an integer u m, corime to n if m is odd, or corime to n/2 if m is even, such that a Ona) m = 1 + u m R n a) m. Proof. We distinguish different cases based uon the arity of m. First, we rove the odd case by induction on m. If m = 1, then, by definition of the integer R n a), we have a Ona) = 1 + R n a). Therefore the assertion is true for m = 1. Now, let be a rime factor of m and suose that the assertion is true for the odd number m/, i.e., there exists an integer u m/, corime to n, such that Then, we obtain a Ona) m = a Ona) m ) a Ona) m = 1 + u m/ R n a) m 1 = 1 + u m/ R n a) m + k=2 k) 1 = 1 + um/ R n a) m. ) ) u m/ R n a) m k m = 1 + u m/ + u m/) k R n a) k 1 k=2 1 ) ) 1 +u m/ ) Rna) m R n a) m = 1 + u m R n a) m. Since n divides R n a) which divides u m u m/ = 1 k=2 k) u m/) k R n a) k 1 ) k + u m/ R n a) m ) ) k 1 + ) k 1 m 1 + u m/) Rna) ) 1 m, it follows that gcdu m,n) = gcdu m/,n) = 1. This comletes the roof for the odd case. Suose now that n and m are even. We roceed by induction on v 2 m). If v 2 m) = 1, then m/2 is odd and by the first art of this roof, a m 2 Ona) = 1 + u m/2 m 2 R na) where u m/2 is corime to n and hence to n/2. Now assume that v 2 m) > 1 and that a m 2 Ona) = 1 + u m/2 m 2 R na) 7

8 with u m/2 corime to n/2. Then, we obtain a Ona) m = a Ona) m 2 ) 2 ) 2 = 1 + u m/2 R n a) m 2 = 1 + u m/2 R n a) m + u m/2 R n a) m 2 = 1 + u m/2 + u m/2 ) 2 Rna) 2 = 1 + u m R n a) m. ) 2 m ) R n a)m 2 Since n/2 divides R n a)/2 which divides u m u m/2, it follows that gcdu m,n/2) = gcdu m/2,n/2) = 1. This comletes the roof. We are now ready to rove Theorem 7. Proof of Theorem 7. The roof is by induction on the integer /. If =, then we have gcd, R n2 a)) = gcd, R n1 a)) = = 1, since R n1 a) is divisible by, and thus the statement is true. Let be a rime factor of and such that divides / and suose that First, the congruence imlies that O n1 /a) = O n2 a) / gcd /, R n2 a)). a O a) 1 mod ) a O a) 1 mod ) and so O n1 /a) divides O n1 a). We consider two cases. First Case: v ) v R n2 a)). Since divides /, it follows that O n2 a) divides O n1 /a). Let r = O /a) O n2 a). Hence R n1 /a) = a O n /a) 1 1 = a O a) r 1 = a O a) 1 ) ) r 1 a ko a) = R n2 a) r 1 ) a ko a) k=0 and so R n1 /a) is divisible by R n2 a). This leads to v ) v R n2 a)) v Rn1 /a) ). k=0 8

9 Therefore R n1 /a) is divisible by and hence we have a O /a) = 1 + R n1 /a) 1 mod ). This imlies that O n1 a) = O n1 /a). Moreover, the hyothesis v ) v R n2 a)) imlies that gcd /, R n2 a)) = gcd, R n2 a))/. Finally, we obtain O n1 a) = O n1 /a) = O n2 a) / gcd /, R n2 a)) = O a) gcd, R n2 a)). Second Case: v ) > v R n2 a)). If v 2 ) 1, then /)/ gcd /, R n2 a)) is odd. Otherwise, if v 2 ) 2, then v 2 ) 2 and every integer corime to /2 is also corime to. In both cases, v 2 ) 1 or v 2 ) 2, we know, by Lemma 8, that there exists an integer u, corime to, such that a O n /a) 1 = a O a) / gcd n /,R a)) 1 / = 1 + u R n2 a) gcd /, R n2 a)) R n2 a) = 1 + u gcd /, R n2 a)) n1. As v R n2 a)) v /), it follows that R n2 a)/ gcd /, R n2 a)) is corime to, and hence O n1 /a) is a roer divisor of O n1 a) since a O /a) 1 mod ). Moreover, by Lemma 8 again, there exists an integer u such that This leads to a O /a) = 1 + u R n1 /a) 1 mod ). O n1 a) = O n1 /a) = O n2 a) This comletes the roof of Theorem 7. gcd /, R n2 a)) = O a) gcd, R n2 a)). We may view Theorem 7 as a generalization of Theorem 3.6 of [3], where = is an odd rime number and = k for some ositive integer k. Note that the conclusion of Theorem 7 fails in general in the case where v 2 ) 2 and = rad ). For instance, for = 24 = 3 2 3, = 6 = 3 2 and a = 7, we obtain that O n1 a) = 2 while O n2 a) / gcd, R n2 a)) = 24/ gcd24, 6) = 4. We now turn to the roof of the main result of this aer. Proof of Theorem 6. From Theorem 7, we obtain α n1 a) = O n1 a ) = = O n1 a) gcdo n1 a), ) = O n2a) gcd O n2 a) O n2 a) gcdo n2 a),, R n2 a)). 9 gcd,r n2 a)) gcd,r n2 a)), )

10 Thus, α n2 a) α n1 a) = O n2 a) gcdo n2 a), ) O n2 a) gcdo n2 a),,r n2 a)) = gcd = gcdo a),, R n2 a)) gcdo n2 a), ) O n2 a) gcdo n2 a), ), gcdo n2 a), ) gcd, R n2 a)) Finally, since we have ) O n2 a) gcd gcdo n2 a), ), gcdo n2 a), ) it follows that α n2 a) α n1 a) = gcd O n2 a) gcdo n2 a), ), gcdn ) 1, R n2 a)) = gcdo a), ) gcdo n2 a), ) = 1, ). = gcd α n2 a), gcdn ) 1, R n2 a)). Thus, the determination of α n is reduced to the case where n is square-free. Corollary 9. Let n be a ositive integer such that v 2 n) 1. Then, for every integer a, corime to n, we have α n a) = α radn) a) ). gcd α radn) a), gcdn,r radn)a)) radn) Corollary 10. Let n be a ositive integer such that v 2 n) 2. Then, for every integer a, corime to n, we have α n a) = α 2radn) a) ). gcd α 2radn) a), gcdn,r 2 radn)a)) 2 radn) 3 The arithmetic function β n First, we can observe that, by definition of the functions α n and β n, we have for every integer a not corime to n and α n a) = β n a) = 0 α n a) β n a) {1, 2} for every integer a corime to n. There is no general formula known to comute α n a)/β n a) but, however, we get the following roosition. 10

11 Proositio1. Let and be two ositive integers such that rad ) = rad ). Let a be an integer corime to and. If v 2 ) 1, then we have α n1 a) β n1 a) = α a) β n2 a). If v 2 ) 2, then we have α n1 a) = β n1 a). Proof. Let be a ositive integer such that v 2 ) 1 and a be an integer corime to. Let be an odd rime factor of such that v ) 2. We will rove that If α n1 a) = 2β n1 a), then and thus α n1 a) β n1 a) = α /a) β n1 /a). a β a) 1 mod ) a βn n1 1 a) 1 mod ). This imlies that α n1 /a) = 2β n1 /a). Conversely, if α n1 /a) = 2β n1 /a), then we have Since v ) 2, it follows that and thus a β /a) n1 1 mod ). a β /a) n1 1 mod ) ) a β /a) +1 = 1 a β /a) n1 ) 1 ) = 1 + a β /a) n1 a β /a) n1 k 0 mod n1 ). This imlies that α n1 a) = 2β n1 a). Continuing this rocess we have and since rad ) = rad ), α n1 a) β n1 a) = α rad )a) β radn1 )a) α n1 a) β n1 a) = α a) β n2 a). Now, let be a ositive integer such that v 2 ) 2, and let a be a non-zero integer. Suose that we have α n1 a) = 2β n1 a). Since k=0 a β a) 1 mod ) 11

12 it follows that ) a βn n1 4 1 a) 4 1 mod 4) in contradiction with ) a βn n1 4 1 a) 4 1 mod 4). Thus α n1 a) = β n1 a). If n is a rime ower, then β n = β radn), in analogy with Proosition 3 for α n. Proositio2. Let be a rime number and let a be an integer. Then we have for every ositive integer k. β ka) = β a) Proof. This result is trivial for every integer a not corime to. Suose now that a is corime to. For = 2, then, by Proositio1, we have β 2 ka) = α 2 ka) = 1 for every ositive integer k. For an odd rime number 3, Proositio1 and Proosition 3 lead to β ka) = α ka) α a) β a) = β a) for every ositive integer k. This comletes the roof. Let and be two ositive integers such that { radn1 ) and, if v 2 ) 1; 2 rad ) and, if v 2 ) 2. It immediately follows that β n1 a) = β n2 a) = 0 for every integer a not corime to and. Finally, we determine the relationshi between β n1 a) and β n2 a) for every integer a corime to and. Theorem 13. Let and be two ositive integers such that { radn1 ) and, if v 2 ) 1; 2 rad ) and, if v 2 ) 2. Let a be an integer corime to and. Then, we have β n1 a) = β n2 a) ). gcd β n2 a), gcd,r n2 a)) 12

13 Proof. If v 2 ) 1, then Theorem 6 and Proositio1 lead to β n2 a) β n1 a) = α a) α n1 a) = gcd α n2 a), gcdn ) 1, R n2 a)). Since v 2 ) = v 2 ) 1, it follows that gcd, R n2 a))/ is odd and hence, we have β n2 a) β n1 a) = gcd α n2 a), gcdn ) 1, R n2 a)) = gcd β n2 a), gcdn ) 1, R n2 a)). If v 2 ) 2, then β n1 a) = α n1 a) and β n2 a) = α n2 a) by Proositio1 and the result follows from Theorem 6. Thus, as for α n, the determination of β n is reduced to the case where n is square-free. Corollary 14. Let n be a ositive integer such that v 2 n) 1. Then, for every integer a, corime to n, we have β n a) = β radn) a) ). gcd β radn) a), gcdn,r radn)a)) radn) Corollary 15. Let n be a ositive integer such that v 2 n) 2. Then, for every integer a, corime to n, we have β n a) = β 2radn) a) ). gcd β 2radn) a), gcdn,r 2 radn)a)) 2radn) 4 Acknowledgments The author would like to thank Shalom Eliahou for his hel in rearing this aer. He also thanks the anonymous referee for his/her useful remarks. References [1] Jonathan Chaelon, On a roblem of Molluzzo concerning Steinhaus triangles in finite cyclic grous, Integers ), #A37. [2] John C. Molluzzo, Steinhaus grahs, Theory and Alications of Grahs, Sringer, volume 642 of Lecture Notes in Mathematics, Berlin / Heidelberg, 1978, [3] Melvyn B. Nathanson, Elementary Methods in Number Theory, Sringer, 2000,

14 2000 Mathematics Subject Classification: Primary 11A05; Secondary 11A07, 11A25. Keywords: multilicative order, rojective multilicative order, balanced Steinhaus triangles, Steinhaus triangles, Molluzzo s roblem. Received October ; revised version received January Published in Journal of Integer Sequences, January Return to Journal of Integer Sequences home age. 14