ON RELATIVELY PRIME SUBSETS, COMBINATORIAL IDENTITIES, AND DIOPHANTINE EQUATIONS MOHAMED EL BACHRAOUI
Abstract. Let n be a positive integer and let A be a nonempty finite set of positive integers. We say that A is relatively prime if gcd(A) = 1 and that A is relatively prime to n if gcd(A, n) = 1. In this work we count the number of nonempty subsets of A which are relatively prime and the number of nonempty subsets of A which are relatively prime to n. Related formulas are also obtained for the number of such subsets having some fixed cardinality. This extends previous work for the case where A is an interval of successive integers. As an application we give some identities involving M¨ obius and Mertens functions which provide solutions to some kind of Diophantine equations.
1. Introduction Throughout let n and α be positive integers and let A be a nonempty finite set of positive integers. Let #A = |A| denote the cardinality of A. We suppose in Pn this paper that α ≤ |A|. Let µ be the M¨obius function, let M (n) = d=1 µ(d) be the Mertens function, and let bxc be the floor of x. If m and n are positive integers such that m ≤ n, then we let [m, n] = {m, m + 1, . . . , n}. The set A is called relatively prime if gcd(A) = 1 and it is called relatively prime to n if gcd(A ∪ {n}) = gcd(A, n) = 1. Definition 1. Let f (A) = #{X ⊆ A : X 6= ∅ and gcd(X) = 1}, fα (A) = #{X ⊆ A : #X = α and gcd(X) = 1}, Φ(A, n) = #{X ⊆ A : X 6= ∅ and gcd(X, n) = 1}, Φα (A, n) = #{X ⊆ A : #X = α and gcd(X, n) = 1}. Nathanson in [5] introduced among others the functions f (n), fα (n), Φ(n), and Φα (n) (in our terminology f ([1, n]), fα ([1, n]), Φ([1, n], n), and Φα ([1, n], n) respectively) and found exact formulas along with asymptotic estimates for each of these functions. Formulas for these functions along with asymptotic estimates are found in El Bachraoui [3] and Nathanson and Orosz [6] for A = [m, n] and in El Bachraoui [4] for A = [1, m]. Ayad and Kihel in [1, 2] considered extensions to sets in arithmetic progression and obtained identities for these functions for A = [l, m] as consequences. Formulas connecting the functions Φk (n) and fk (n) are found in Tang [7] and formulas for other related functions along with asymptotic estimates are given in T´ oth [8]. An analysis of the functions f , fα , Φ, and Φα obtained for different cases of the set A lead us to more general formulas for any nonempty finite 1991 Mathematics Subject Classification. 11A25, 11B05, 11B75, 11D41. Key words and phrases. Phi functions, Relatively prime sets, Mertens function, M¨ obius function, Combinatorial identities, Diophantine equations. 1
2
MOHAMED EL BACHRAOUI
set of positive integers. For the purpose of this work we give these functions for A = [l, m]. Theorem 1. We have (a) f ([l, m]) = (b) fα ([l, m]) =
m X d=1 m X d=1
(c)
Φ([l, m], n) =
X
m
µ(d)(2b d c−b
l−1 d c
− 1),
m b c − b l−1 d c , µ(d) d α m
µ(d)2b d c−b
l−1 d c
,
d|n
(d) φα ([l, m], n) =
X d|n
m b d c − b l−1 c d µ(d) . α
By way of example, using our formula for f (A) we will get that if gcd(m, n) = 1, then the following expression n X
n
µ(d)2b d c−b
n−1 m−1 m d c+b d c−b d c
d=1
Pn boils down to the much more simple expression d=1 µ(d) = M (n), see Theorem 5 below. In terms of Diophantine equations, this means that the integer pair (2, 1) is a solution to n X
n
µ(d) xb d c−b
n−1 m−1 m d c+b d c−b d c
n
− y b d c−b
n−1 m−1 m d c+b d c−b d c
=1,
d=1
if gcd(m, n) = 1, see Corollary 4(a). Related to this, an open question is whether or not other real or integer solutions exist for the previous equation.
2. Phi functions for integer sets Theorem 2. We have (a)
Φ(A, n) =
X
µ(d)2
P
a−1 a a∈A (b d c−b d c)
.
d|n
(b)
Φα (A, n) =
X
P µ(d)
a a∈A (b d c
− b a−1 d c)
α
d|n
.
Proof. (a) We use induction on |A|. If A = {a} = [a, a], then by Theorem 1 (c) Φ(A, n) =
X d|n
a
µ(d)2b d c−b
a−1 d c
.
RELATIVELY PRIME SUBSETS...
3
Assume that A = {a1 , a2 , . . . , ak } and that the identity holds for {a2 , . . . , ak }. Then Φ({a1 , . . . , ak }, n) = Φ({a2 , . . . , ak }, n) + Φ({a2 , . . . , ak }, gcd(a1 , n)) X X Pk Pk ai ai −1 ai −1 ai = µ(d)2 i=2 (b d c−b d c) + µ(d)2 i=2 (b d c−b d c) d|n
d|(a1 ,n)
X
=2
µ(d)2
Pk
i=2 (b
ai d
c−b
ai −1 d c)
+
X
µ(d)2
ai i=2 (b d
Pk
c−b
ai −1 d c)
d|n d-a1
d|(a1 ,n)
=
X
a1 d
µ(d)2b
c−b
a1 −1 c d
2
ai i=2 (b d
Pk
c−b
ai −1 d c)
d|(a1 ,n)
+
X
ai i=1 (b d
Pk
µ(d)2
c−b
ai −1 d c)
d|n d-a1
X
=
µ(d)2
ai i=1 (b d
Pk
c−b
ai −1 d c)
+
X
ai i=1 (b d
Pk
µ(d)2
c−b
ai −1 d c)
d|n d-a1
d|(a1 ,n)
=
X
µ(d)2
ai i=1 (b d
Pk
c−b
ai −1 d c)
.
d|n
(b) Similar.
Corollary 1. Let l1 , l2 , . . . , lk and m1 , m2 , . . . , mk be nonnegative integers such that li < mi for i = 1, 2, . . . , k and mi ≤ li+1 for i = 1, 2, . . . , k − 1. Then X Pk li mi (a) Φ([l1 + 1, m1 ] ∪ [l2 + 1, m2 ] ∪ . . . ∪ [lk + 1, mk ], n) = µ(d)2 i=1 (b d c−b d c) . d|n
(b)
Φα ([l1 +1, m1 ]∪[l2 +1, m2 ]∪. . .∪[lk +1, mk ], n) =
X
Pk µ(d)
mi i=1 (b d c
d|n
− b ldi c)
α
Proof. Apply Theorem 2 to the set A = {l1 + 1, l1 + 2, . . . , m1 , l2 + 1, l2 + 2, . . . , m2 , . . . , lk + 1, lk + 2, . . . , mk }. Corollary 2. If n ∈ A, then Φ(A, n) ≡ 0 mod 2. Proof. Note first that X j a k a∈A
d
−
a−1 d
counts the number of multiples of d in the set A. So, if n ∈ A, then evidently P a a−1 (b a∈A d c − b d c) > 0 for all divisor d of n and thus the required congruence follows by Theorem 2(a). 3. Relatively prime subsets of integer sets Theorem 3. We have (a) f (A) =
sup XA d=1
P a−1 a µ(d) 2 a∈A (b d c−b d c) − 1 .
.
4
MOHAMED EL BACHRAOUI
(b) fα (A) =
sup XA
P a a−1 a∈A (b d c − b d c) µ(d) . α
d=1
Proof. (a) We use induction on |A|. If A = {a} = [a, a], then by Theorem 1 (a) a a X a−1 f (A) = µ(d) 2b d c−b d c − 1 . d=1
Assume now that A = {a1 , a2 , . . . , ak } and that the identity is true for {a2 , . . . , ak }. Without loss of generality we may assume that a1 < sup A. Then with the help of Theorem 2(a) we have f ({a1 , . . . , ak }) = f ({a2 , . . . , ak }) + Φ({a2 , . . . , ak }, a1 )) =
sup XA
Pk ai −1 ai µ(d) 2 i=2 (b d c−b d c) − 1
d=1
+
X
µ(d)2
ai i=2 (b d
Pk
c−b
ai −1 d c)
d|a1
=
X
Pk ai ai −1 µ(d) 2 i=2 (b d c−b d c) − 1
d|a1
+
sup XA
Pk ai ai −1 µ(d) 2 i=2 (b d c−b d c) − 1
d=1 d-a1
+
X
=2
X
µ(d)2
ai i=2 (b d
Pk
c−b
ai −1 d c)
d|a1
µ(d)2
ai i=2 (b d
Pk
c−b
−
X
µ(d)
d|a1
d|a1
+
ai −1 d c)
sup XA
Pk ai ai −1 µ(d) 2 i=2 (b d c−b d c) − 1
d=1 d-a1
=
X
Pk ai −1 ai µ(d) 2 i=1 (b d c−b d c) − 1
d|a1
+
sup XA
Pk ai ai −1 µ(d) 2 i=1 (b d c−b d c) − 1
d=1 d-a1
=
sup XA
Pk ai ai −1 µ(d) 2 i=1 (b d c−b d c) − 1 .
d=1
(b) Similar.
Corollary 3. Let l1 , l2 , . . . , lk and m1 , m2 , . . . , mk be nonnegative integers such that li < mi for i = 1, 2, . . . , k and mi ≤ li+1 for i = 1, 2, . . . , k − 1. Then (a) f ([l1 +1, m1 ]∪[l2 +1, m2 ]∪. . .∪[lk +1, mk ]) =
sup XA d=1
Pk mi li µ(d) 2 i=1 (b d c−b d c) − 1 .
RELATIVELY PRIME SUBSETS...
(b) fα ([l1 +1, m1 ]∪[l2 +1, m2 ]∪. . .∪[lk +1, mk ], n) =
sup XA d=1
5
Pk mi li i=1 (b d c − b d c) µ(d) . α
Proof. Apply Theorem 3 to the set A = {l1 + 1, l1 + 2, . . . , m1 , l2 + 1, l2 + 2, . . . , m2 , . . . , lk + 1, lk + 2, . . . , mk }. Alternatively, we have the following formulas for f (A) and fα (A). Theorem 4. Let A = {a1 , a2 , . . . , ak }, let τ be a permutation of {1, 2, . . . , k}, and let Aτ (j) = {aτ (1) , aτ (2) , . . . , aτ (j) } for j = 1, 2, . . . , k. Then (a) f (A) =
X
µ(d) +
k X X
i=1 (b
aτ (i) d
c−b
aτ (i) −1 d
c)
.
j=2 d|aτ (j)
d|aτ (1)
(b) fα (A) =
Pj−1
µ(d)2
k X X j=1 d|aτ (j)
Pj−1 aτ (i) aτ (i) −1 c) i=1 (b d c − b d . µ(d) α−1
Proof. For simplicity we assume that τ is the identity permutation. As to part (a) we have with the help of Theorem 2 f ({a1 , . . . , ak }) = f ({a1 , . . . , ak−1 }) + Φ({a1 , . . . , ak−1 }, ak ) = f ({a1 }) + Φ({a1 }, a2 ) + . . . + Φ({a1 , . . . , ak−1 }, ak ) X X a1 a1 −1 µ(d)2b d c−b d c µ(d) + = d|a2
d|a1
+ ... +
X
µ(d)2
Pk−1 i=1
(b
ai d
c−b
ai −1 d c)
d|ak
=
X
µ(d) +
d|a1
k X X
µ(d)2
ai i=1 (b d
Pj−1
c−b
ai −1 d c)
,
j=2 d|aj
where the third formula follows from Theorem 2. Part (b) follows similarly.
4. Combinatorial identities and Diophantine equations We now give some identities involving Mertens function which provide solutions to a type of Diophantine equations. Theorem 5. Let m and n be positive integers such that 1 < m < n. Then ( n X M (n), if gcd(m, n) > 1; c−b n−1 c+b m c−b m−1 c bn d d d d = µ(d)2 1 + M (n), if gcd(m, n) = 1. d=1 Proof. If gcd(m, n) > 1, then clearly have f ({m, n}) = 0. If 1 < m ≤ n and gcd(m, n) = 1, then clearly f ({m, n}) = 1. On the other hand, by Theorem 3 (a) applied to the set {m, n} we have n X n−1 m−1 n m f ({m, n}) = µ(d) 2b d c−b d c+b d c−b d c − 1 d=1
=
n X d=1
n
µ(d)2b d c−b
n−1 m−1 m d c+b d c−b d c
− M (n).
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MOHAMED EL BACHRAOUI
Combining the identities for f ({m, n}) for the case gcd(m, n) > 1 gives M (n) =
n X
n
µ(d)2b d c−b
n−1 m−1 m d c+b d c−b d c
d=1
and for the case 1 < m ≤ n and gcd(m, n) = 1 gives 1 + M (n) =
n X
n
µ(d)2b d c−b
n−1 m−1 m d c+b d c−b d c
.
d=1
This completes the proof.
In terms of Diophantine equations Theorem 5 translates into the following. Corollary 4. Let 1 < m < n be positive integers. Then (a) If gcd(m, n) = 1, then (2, 1) is a solution to the equation n X
n
µ(d) xb d c−b
n−1 m−1 m d c+b d c−b d c
n
− y b d c−b
n−1 m−1 m d c+b d c−b d c
= 1.
d=1
(b) If gcd(m, n) > 1, then (1, 2) and (2, 1) are solutions to the equation n X
n
µ(d) xb d c−b
n−1 m−1 m d c+b d c−b d c
n
− y b d c−b
n−1 m−1 m d c+b d c−b d c
= 0.
d=1
Proof. Immediate from Theorem 5.
Theorem 6. Let l, m, and n be integers such that 1 < l < m < n. Then n X
n
µ(d)2b d c−b
n−1 m−1 l−1 m l d c+b d c−b d c+b d c−b d c
=
d=1
4 + M (n), if gcd(l, m) = gcd(l, n) = gcd(m, n) = 1; 3 + M (n), if exactly two pairs from {l, m, n} are co-prime; 2 + M (n), if exactly one pair from {l, m, n} is co-prime; 1 + M (n), if no pair from {l, m, n} is co-prime and gcd(l, m, n) = 1; M (n), otherwise. Proof. Suppose that 1 < l < m < n and gcd(l, m) = gcd(l, n) = gcd(m, n) = 1. Then the relatively prime subsets of {l, m, n} are {l, m}, {l, n}, {m, n}, and {l, m, n}, implying that f ({l, m, n}) = 4. Combining this with the formula for f ({l, m, n}) obtained by using Theorem 3 (a) we get n X
n
µ(d)(2b d c−b
n−1 m−1 l−1 m l d c+b d c−b d c+b d c−b d c
− 1) = 4,
d=1
which is equivalent to the first case of the desired identity. As to the second case, if exactly two pairs are co-prime, then f ({l, m, n}) = 3 and the result follows from Theorem 3 (a). The remaining three cases follow similarly and the proof is completed. In terms of Diophantine equations Theorem 6 means the following.
RELATIVELY PRIME SUBSETS...
7
Corollary 5. Let l, m, n be integers such that 1 < l < m < n. Then (a) If gcd(l, m) = gcd(l, n) = gcd(m, n) = 1, then (2, 1) is a solution to n X
n
µ(d) xb d c−b
n−1 m−1 m d c+b d c−b d c
n
− y b d c−b
n−1 m−1 m d c+b d c−b d c
= 4.
d=1
(b) If exactly two pairs from {l, m, n} are co-prime, then (2, 1) is a solution to n X
n
µ(d) xb d c−b
n−1 m−1 m d c+b d c−b d c
n
− y b d c−b
n−1 m−1 m d c+b d c−b d c
= 3.
d=1
(c) If exactly one pair from {l, m, n} is co-prime, then (2, 1) is a solution to n X
n
µ(d) xb d c−b
n−1 m−1 m d c+b d c−b d c
n
− y b d c−b
n−1 m−1 m d c+b d c−b d c
= 2.
d=1
(d) If no pair from {l, m, n} is co-prime and gcd(l, m, n) = 1, then (2, 1) is a solution to n X n−1 m−1 n−1 m−1 n m n m µ(d) xb d c−b d c+b d c−b d c − y b d c−b d c+b d c−b d c = 1. d=1
(e) Otherwise, the integer pairs (1, 2) and (2, 1) are solutions to n X
n
µ(d) xb d c−b
n−1 m−1 m d c+b d c−b d c
n
− y b d c−b
n−1 m−1 m d c+b d c−b d c
= 0.
d=1
Proof. Straightforward from Theorem 6.
We close the paper by some open questions which are suggested by our results. Open Questions. Question 1. Do the Diophantine equations in Corollary 4 and Corollary 5 have any other real solutions? Question 2. Do the Diophantine equations in Corollary 4 and Corollary 5 have any other integer solutions? Question 3. It is clear that any real pair (x, x) is a solution to the equation in part (b) of Corollary 4 and to the equation in part (e) of Corollary 5. These solutions might be called trivial. Is the number of non-trivial integer solutions to the equations in Corollary 4 and Corollary 5 finite? Acknowledgment. The author is grateful to the referee for valuable comments and interesting suggestions. References [1] M. Ayad and O. Kihel, On the Number of Subsets Relatively Prime to an Integer, J. Integer Sequences 11, (2008), Article 08.5.5. [2] M. Ayad and O. Kihel, On Relatively Prime Sets, Integers 9 (2009), 343-352. [3] M. El Bachraoui, The number of relatively prime subsets and phi functions for sets {m, m + 1, . . . , n}, Integers 7 (2007), A43. [4] M. El Bachraoui, On the Number of Subsets of [1, m] Relatively Prime to n and Asymptotic Estimates, Integers 8 (2008), A41. [5] M. B. Nathanson, Affine invariants, relatively prime sets, and a phi function for subsets of {1, 2, . . . , n}, Integers 7 (2007), A1.
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MOHAMED EL BACHRAOUI
[6] M. B. Nathanson and B. Orosz, Asymptotic estimates for phi functions for subsets of {m + 1, m + 2, . . . , n}, Integers 7 (2007), A54. [7] M. Tang, Relatively prime sets and a phi function for subsets of {1, 2, . . . , n}, J. Integer Sequences 13 (2010), Article 10.7.6. [8] L. T´ oth, On the number of certain relatively prime subsets of {1, 2, . . . , n}, Integers 10 (2010), 407-421. Dept. Math. Sci, United Arab Emirates University, PO Box 17551, Al-Ain, UAE E-mail address:
[email protected] 
