The Greatest Common Divisor of k Positive Integers

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1 International Mathematical Forum, Vol. 3, 208, no. 5, HIKARI Ltd, The Greatest Common Divisor of Positive Integers Rafael Jaimczu División Matemática, Universidad Nacional de Luján Buenos Aires, Argentina Copyright c 208 Rafael Jaimczu. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original wor is properly cited. Abstract We study using the inclusion-eclusion principle and very elementary methods the distribution of positive integers not eceeding with the same greatest common divisor. Mathematics Subject Classification: A99, B99 Keywords: -tuples of positive integers, sets of positive integers, greatest common divisor Introduction and Preliminary Notes Let 2 an arbitrary but fied positive integer. Let us consider a -tuple of positive integers a,..., a where a i i =,...,. The number of these -tuples such that gcda,..., a = g will be denoted N g, and the number of these -tuples such that gcda,..., a > will be denoted N 0,. On the other hand, let us consider a set of distinct positive integers {a,..., a } where a i i =,...,. The number of these sets such that gcda,..., a = g will be denoted D g, and the number of these sets such that gcda,..., a > will be denoted D 0,. Let us consider a positive integer n such that its prime factorization is n = q r qs rs. The number of -tuples of positive integers a,..., a not eceeding such that gcda,..., a = and gcda i, n = i =,..., will be denoted P,n. The number of -tuples of positive integers a,..., a

2 26 Rafael Jaimczu not eceeding such that the a i are pairwise relatively prime and such that gcda i, n = i =,..., will be denoted P n. Note that if n = then P is the number of -tuples of positive integers a,..., a not eceeding such that the a i are pairwise relatively prime. We shall need the following well-nown theorem. Theorem. Inclusion-eclusion principlelet S be a set of N distinct elements, and let S,..., S r be arbitrary subsets of S containing N,..., N r elements, respectively. For i < j <... < l r, let S ij...l be the intersection of S i, S j,..., S l and let N ij...l be the number of elements of S ij...l. Then the number K of elements of S not in any of S,..., S r is K = N i r N i + i<j r N ij i<j< r Proof. See, for eample, [3] page 84 or [2] page Main Results N ij r N 2...r Nymann [4] proved the following theorem with a better error term using a Moebius inversion formula. In this note, we prove the theorem using the inclusion-eclusion principle, the proof is very elementary and short. In this proof p n denotes the n-th prime, ζs denotes the Riemann zeta function and. denotes the integer-part function. Theorem 2. Let 2 an arbitrary but fied integer. asymptotic formula holds. The following N, = ζ + o Proof. Let A ph, be the number of -tuples of positive integers not eceeding such that the least prime factor in their greatest common divisor is p h. The inclusion-eclusion principle gives Note that A ph, = + p h j h p h p j i<j h p h p i p j = h + o 2 p h i= p i A ph, p h p h 3

3 The greatest common divisor of positive integers 27 The following equation can be proved without difficulty using mathematical induction n h n = p i p i Therefore we have h= p h i= h h= p h i= p i i= = ζ Let ɛ > 0. We shall choose n such that the following inequality holds We have see 2 and 4 N 0, = = where see 3 2 p h ζ h=n+ p h A ph, = 4 ɛ 5 h h n p h i= h h=n+ p h i= + o + F p i p i + o + F 6 0 F h=n+ p h ɛ 7 Equations 6, 7 and 5 give N 0, ζ ɛ + ɛ + ɛ = 3ɛ ɛ 8 Consequently, since ɛ > 0 can be arbitrarily small, equation 8 gives N 0, = + o 9 ζ Now N, + N 0, = = + o 0 Equations 9 and 0 give. The theorem is proved. Remar 2.2 Since ζ, if is large the number of -tuples not eceeding such that the gcd of the a i is greater than is negligible compared with the number -tuples not eceeding such that gcd of the a i is.

4 28 Rafael Jaimczu Corollary 2.3 The following asymptotic formula holds. N g, = g ζ + o Proof. We have N g, = N, g. The corollary is proved. Theorem 2.4 The following asymptotic formula holds. D, =!ζ + o Proof. The proof is the same as the proof of Theorem 2.. In this case, in equation 2 inclusion-eclusion principle, we substitute p h by...! p h p h p h etc. Note also that D 0, + D, = =! + o The theorem is proved. Corollary 2.5 The following asymptotic formula holds. D g, =!g ζ + o Theorem 2.6 The following asymptotic formula holds. P,n = si= ζ si= + o qi Proof. Let N be the number of numbers not eceeding and relatively prime to n. The inclusion-eclusion principle gives N = + = qi + o 2 q j i s i<j s i=

5 The greatest common divisor of positive integers 29 Let us consider a positive integer C relatively prime to n such that its prime factorization is c r c rt t. For sae of simplicity we put y = c c t. Then the number of numbers not eceeding relatively prime to n and multiple of C is inclusion eclusion principle y y + y = qi + o 3 q j c c t i s i<j s In this proof p n denotes the n-th prime. Let p h is the h-th prime, where p h i =,..., s. For sae of simplicity we put S = s i=. Let Aph be the number of -tuples a,..., a not eceeding such that gcda i, n = i =,... and such that the least prime factor in their greatest common divisor is p h. The inclusion-eclusion principle and equation 3 give A ph = = i= S + o p h p h i= S + o + j h,p j p i=,...,s h p j + o 4 j h,p j p i=,...,s j Let N 0 be the number of -tuples a,..., a not eceeding such that gcda i, n = i =,... and such that gcda,..., a >. Equation 4 and an identical proof as in Theorem 2. give N 0 = p j p h p i=,...,s h j h,p j i=,...,s + o 5 Now, we have the equality p h p i=,...,s h j h,p j i=,...,s p j = p j i=,...,s i= p j 6 since both series have the same terms. The term r r t, where the different primes r i i =,..., t satisfy the inequality r > > r t is obtained in the series of the left hand when p h = r. Note that see equation2 N 0 + P,n = + o 7 i= Equations 5, 6 and 7 give P,n = + o p j p i=,...,s j i=

6 220 Rafael Jaimczu = si= ζ si= + o qi That is, equation. The theorem is proved. Remar 2.7 Note that if n is fied then we have lim si= ζ si= q i = 0 8 Tóth [5] proved, using mathematical induction, the following theorem with a better error term Theorem 2.8 The following asymptotic formulae hold. P n = A + o + i= P = A + o where A = p p + p Now, we give a simple proof of the following theorem. Theorem 2.9 The following limit holds. The following inequality holds. lim A = 0 9 A Proof. The number of even numbers not eceeding is 2 = +o and the 2 number of odd numbers not eceeding is 2 = +o. Therefore the 2 number of -tuples a,..., a not eceeding with all a i odd is N = + 2 o and the number of -tuples with only one even a i is N 2 = +o. 2 Now, P = A + o N + N 2 = + + o. From this 2 inequality we obtain inequality 20. Limit 9 is an immediate consequence of inequality 20. The theorem is proved.

7 The greatest common divisor of positive integers 22 Remar 2.0 Since ζ and A 0 we see that if is large then the number of -tuples a,..., a not eceeding such that the a i are pairwise relatively prime is negligible compared with the number of -tuples a,..., a not eceeding such that gcda,..., a =. In the oposite side, let us consider the number of -tuples a,..., a not eceeding such that if i j i =,..., j =,..., then gcda i, a j >. Let B be the number of these -tuples not eceeding. We have the following simple theorem. Theorem 2. The following inequality hold If is even B 2 /2 ζ2 If is odd B 2 /2 ζ2 Therefore if B = B + o, for a positive constant B, then B 0. Proof. If = 2 is well-nown that B 2 = ζ2 2 + o 2. Therefore if is even we have B B 2 /2 = /2 ζ2 + o, since there are /2 consecutive pairs a i, a i+ i =, 3,..., in the -tuple a,..., a. If is odd in the same way we obtain B B 2 /2 = /2 + o. ζ2 The theorem is proved. To finish, we give another proof of Theorem 2. using mathematical induction. Before, we need the following lemma. Lemma 2.2 Let 2 an arbitrary but fied positive integer. The following asymptotic formula holds. n> n = O 2 Let an arbitrary but fied positive integer. The following asymptotic formula holds. p = + o 22 ζ + n p n

8 222 Rafael Jaimczu Proof. Equation 2 is proved in [, Chapter 3, page 55]. Equation 22 can be proved as [, Chapter 3, Theorem 3.7, page 62] since we have n p = n µd n p n n d d n =... = + ζ o + From here, using partial summation, we obtain 22. The lemma is proved. We also need the following definition. Let N + n be the number of + - tuples a,... a + such that gcda,..., a + = and a = n. Theorem 2.3 If n = we have If n 2 we have Nn + = p n N + = + o 23 p + o n 24 Finally, we have N, = ζ + o 25 Proof. The theorem is true for = 2. Suppose the theorem is true for, then we shall prove that the theorem is also true for +. Therefore, we have N, = ζ + f 26 where f M and lim f = 0. Equation 26 gives N g, = N, = g g ζ + f g g 27 Note that see 2 g>,g,n= g g> g M 2 28 On the other hand, let us consider the function F n = f 29 g g g,g,n=

9 The greatest common divisor of positive integers 223 We have F n f g g g,g,n= f g g g = g f g g + f <g g g ɛζ + M ɛ ɛ ɛ where ɛ > 0 and consequently ɛ > 0 can be arbitrarily small. Hence lim F n = 0. Using equations 26, 27, 28 and 29 we find that Nn + = N g, = p + o g,g,n= p n n n 30 That is, equation 24. Using equationes 30, 23 and 22 we obtain N,+ = The theorem is proved. n N + n = ζ o + Acnowledgements. The author is very grateful to Universidad Nacional de Luján. References [] T. M. Apostol, Introduction to Analytic Number Theory, Springer, New Yor, NY, [2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oford University Press, 960. [3] W. J. LeVeque, Topics in Number Theory, Vol., Addison-Wesley, 958. [4] J. E. Nymann, On the Probability that positive integers are relatively prime, Journal of Number Theory, 4 972, no. 5, [5] L. Tóth, The Probability that positive integers are pairwise relatively prime, Fibonacci Quarterly, , no., 3-8. Received: March, 208; March 28, 208