On the least positive solution to a proportionally modular Diophantine inequality
aa r X i v : . [ m a t h . N T ] F e b ON THE LEAST POSITIVE SOLUTION TO A PROPORTIONALLYMODULAR DIOPHANTINE INEQUALITY
A. MOSCARIELLO
Abstract.
Given three positive integers a, b, c , a proportionally modular Diophantine in-equality is an expression of the form ax mod b ≤ cx . Our aim is to give a recursive formulafor the least solution to such an inequality. We then use the formula to derive an algorithm.Finally, we apply our results to a question of Rosales and Garc´ıa-S´anchez. Introduction A proportionally modular Diophantine inequality is an expression of the form( ax mod b ) ≤ cx where the positive integers a, b, c are called respectively the factor , modulus and proportion .It is well-known that the set of the non-negative integer solutions of this inequality is a numerical semigroup (cf. [5], [6]), i.e. a submonoid S of ( N , +) with finite complement in it.Denoting by S ( a, b, c ) the set of solutions, the structure of this set (called a proportionallymodular semigroup ) has been widely studied, but is not completely understood yet. Inparticular, it is an open problem (cf. [5]) to find explicit formulas for several classicalinvariants of numerical semigroups.In this paper we study the multiplicity of a proportionally modular semigroup S , that isthe smallest positive integer that belongs to S . Although some partial results are known(cf. [6], [8], [9]) as of today the main problem of finding a formula for this invariant stillremains unsolved. Using elementary number theory we will work on a recursive formula forthe smallest positive solution of the more general inequality( ax mod b ) ≤ cx a, b ∈ Z + , c ∈ Q + . Our work is structured as follows: in the first section prove our main theorem, whichprovides a recursive formula for the computation of the multiplicity of S . In Section 2 wedescribe the algorithm that can be derived from our main theorem. In the final section weexplain how our result can be applied to a question of Rosales and Garc´ıa-S´anchez ([5, OpenProblem 5.20]). Mathematics Subject Classification.
Key words and phrases.
Diophantine inequality; proportionally modular numerical semigroups;multiplicity. Main Result
Given two integers m and n with n > remainder operator [ m ] n as follows[ m ] n = min { i ∈ N | i ≡ m (mod n ) } Note that if m and n are positive integers such that m < n then m = [ m ] n . The followingproperties follow from the definition of floor and ceiling function, and we will use themextensively: Proposition 1.1.
Let a, b ∈ Z + . Then:(1) (cid:22) ba (cid:23) a + [ b ] a = b (2) (cid:24) ba (cid:25) a − [ − b ] a = b Let a, b ∈ Z + and let c ∈ Q + . Consider the inequality ( ax mod b ) = [ ax ] b ≤ cx , anddefine L ( a, b, c ) = min { x ∈ Z + | [ ax ] b ≤ cx } = min { S ( a, b, c ) \ { }} First of all, we note that if a ≥ b then [ ax ] b = [[ a ] b x ] b and it follows that L ( a, b, c ) = L ([ a ] b , b, c ), so the condition a < b that we will impose in the next results is not restrictive.Moreover, we have the following property: Lemma 1.2.
Let a, b ∈ Z + , c ∈ Q + . Let d ∈ Z + be such that d | a and d | b . Then L ( a, b, c ) = L (cid:0) ad , bd , cd (cid:1) .Proof. Denote a = da ′ and b = db ′ . In our notation we have [ a ] b = d [ a ′ ] b ′ , therefore[ ax ] b ≤ cx ⇐⇒ [ a ′ ] b ′ ≤ cd x and the thesis follows. (cid:3) The following proposition gives us the value of L ( a, b, c ) for two special cases: Proposition 1.3.
Let a, b ∈ Z + be such that a < b , and let c ∈ Q + be a positive rationalnumber. Then:(1) If c ≥ a then L ( a, b, c ) = 1 .(2) If c < a and a | b then L ( a, b, c ) = ba .Proof. The first one is obvious. As for the second one, if x < ba then ax < b and [ ax ] b = ax >cx , so the inequality is false for x < ba , while for x = ba we have ax = b and [ ax ] b = 0 ≤ cx .We conclude that L ( a, b, c ) = ba . (cid:3) With these premises we can reduce our problem to the case c < a < b , a b . Proposition 1.4.
Let a, b ∈ Z + and c ∈ Q + be such that c < a < b and a b . Then thereexists µ ∈ Z + such that L ( a, b, c ) = (cid:24) µba (cid:25) . Proof. If x < (cid:6) ba (cid:7) then ax < b and [ ax ] b = ax > cx , so L ( a, b, c ) ≥ (cid:6) ba (cid:7) . From this bound itfollows that there exists µ ∈ Z + such that (cid:24) µba (cid:25) ≤ L ( a, b, c ) < (cid:24) ( µ + 1) ba (cid:25) Suppose now that L ( a, b, c ) = (cid:6) µba (cid:7) , that is equivalent to saying that there exists r ∈ N , r = 0 such that L ( a, b, c ) = (cid:24) µba (cid:25) + r with r < (cid:24) ( µ + 1) ba (cid:25) − (cid:24) µba (cid:25) Therefore aL ( a, b, c ) = a (cid:24) µba (cid:25) + ar < a (cid:24) ( µ + 1) ba (cid:25) = ⇒ [ aL ( a, b, c )] b ≥ a and since b > [ aL ( a, b, c )] b − a ≥ aL ( a, b, c )] b − a = [ aL ( a, b, c ) − a ] b . After theposition x = L ( a, b, c ) − ax ] b = [ a ( L ( a, b, c ) − b = [ aL ( a, b, c ) − a ] b = [ aL ( a, b, c )] b − a and cx = cL ( a, b, c ) − c . Hence we have[ ax ] b = [ aL ( a, b, c )] b − a < [ aL ( a, b, c )] b − c ≤ cL ( a, b, c ) − c = cx leading to x = L ( a, b, c ) − ∈ S ( a, b, c ), which is a contradiction. (cid:3) Note that by definition it’s clear that L ( a, b, c ) ≤ b , hence 1 ≤ µ ≤ a . Define R µ as theonly positive integer such that ( R µ − a [ b ] a < µ ≤ R µ a [ b ] a . Lemma 1.5.
Let a, b ∈ Z + and c ∈ Q + be such that c < a < b and a b . Let µ ∈ Z + besuch that L ( a, b, c ) = (cid:6) µba (cid:7) . Then we have:(1) L ( a, b, c ) = (cid:6) µba (cid:7) = µ (cid:4) ba (cid:5) + R µ (2) [ aL ( a, b, c )] b = R µ a − µ [ b ] a Proof. (1) By using Proposition 1.1 we have that b = (cid:4) ba (cid:5) a + [ b ] a , and then L ( a, b, c ) = (cid:24) µba (cid:25) = & µ (cid:0)(cid:4) ba (cid:5) a + [ b ] a (cid:1) a ' = (cid:24) µ (cid:22) ba (cid:23) + µ [ b ] a a (cid:25) Since µ (cid:4) ba (cid:5) ∈ Z + we can deduce easily from the definition of R µ that R µ = l µ [ b ] a a m .Then it follows that: L ( a, b, c ) = (cid:24) µ (cid:22) ba (cid:23) + µ [ b ] a a (cid:25) = µ (cid:22) ba (cid:23) + R µ . (2) By the first part of this lemma and the identity b − [ b ] a = (cid:4) ba (cid:5) a of Proposition 1.1we have[ aL ( a, b, c )] b = (cid:20) µ (cid:22) ba (cid:23) a + R µ a (cid:21) b = [ µb − µ [ b ] a + R µ a ] b = [ R µ a − µ [ b ] a ] b . A. MOSCARIELLO
But by definition of R µ we have 0 ≤ R µ − µ [ b ] a ≤ [ b ] a < b and consequently[ R µ a − µ [ b ] a ] b = R µ a − µ [ b ] a , that is our thesis. (cid:3) In order to find a recursion we’ll prove that R µ itself is the smallest solution of anotherproportionally modular Diophantine inequality with smaller values of factor, modulus andproportion, and then we’ll compute µ from R µ : Theorem 1.6.
Let a, b ∈ Z + , c ∈ Q + be such that c < a < b and a b . Let µ ∈ Z + be suchthat L ( a, b, c ) = (cid:6) µba (cid:7) . Then R µ = L [ a ] [ b ] a , [ b ] a , cbc (cid:4) ba (cid:5) + [ b ] a ! µ = & R µ ( a − c ) c (cid:4) ba (cid:5) + [ b ] a ' Proof.
Using Lemma 1.5 we have that cL ( a, b, c ) = cµ (cid:4) ba (cid:5) + R µ c and [ aL ( a, b, c )] b = R µ a − µ [ b ] a . Then, from cL ( a, b, c ) ≥ [ aL ( a, b, c )] b it follows cL ( a, b, c ) ≥ [ aL ( a, b, c )] b = ⇒ cµ (cid:22) ba (cid:23) + R µ c ≥ R µ a − µ [ b ] a = ⇒ µ (cid:18) c (cid:22) ba (cid:23) + [ b ] a (cid:19) ≥ R µ ( a − c ) = ⇒ µ ≥ R µ ( a − c ) c (cid:4) ba (cid:5) + [ b ] a . By definition of R µ we get µ ≤ R µ a [ b ] a and by merging these parts we obtain(1) R µ ( a − c ) c (cid:4) ba (cid:5) + [ b ] a ≤ µ ≤ R µ a [ b ] a . Since L ( a, b, c ) is the smallest integer for which the inequality is verified, then it followsthat R µ must be the smallest integer such that the interval defined by the two sides ofinequuality (1) contains an integer . Furthermore, by definition µ is the smallest integer insuch an interval, so we obtain(2) R µ = min ( z ∈ Z + | " z ( a − c ) c (cid:4) ba (cid:5) + [ b ] a , za [ b ] a ∩ N = ∅ ) . and(3) µ = min (" R µ ( a − c ) c (cid:4) ba (cid:5) + [ b ] a , R µ a [ b ] a ∩ N ) = & R µ ( a − c ) c (cid:4) ba (cid:5) + [ b ] a ' . The second part of our thesis is proved in (3). For the first one, we can easily see that " z ( a − c ) c (cid:4) ba (cid:5) + [ b ] a , za [ b ] a ∩ N = ∅ ⇐⇒ (cid:22) za [ b ] a (cid:23) ≥ z ( a − c ) c (cid:4) ba (cid:5) + [ b ] a . By Proposition 1.1 we get the two identities j za [ b ] a k = za − [ za ] [ b ] a [ b ] a and (cid:4) ba (cid:5) = b − [ b ] a a . Therefore (cid:22) za [ b ] a (cid:23) ≥ z a − cc (cid:4) ba (cid:5) + [ b ] a ⇐⇒ za − [ za ] [ b ] a [ b ] a ≥ z a − cc (cid:4) ba (cid:5) + [ b ] a ⇐⇒ ⇐⇒ z a [ b ] a − a − cc (cid:4) ba (cid:5) + [ b ] a ! = z ac (cid:4) ba (cid:5) + c [ b ] a [ b ] a (cid:0) c (cid:4) ba (cid:5) + [ b ] a (cid:1) ! ≥ [ za ] [ b ] a [ b ] a ⇐⇒ z cbc (cid:4) ba (cid:5) + [ b ] a ! ≥ [ za ] [ b ] a . Now we put this condition in Eq. (2): R µ = min ( z ∈ Z + | z cbc (cid:4) ba (cid:5) + [ b ] a ! ≥ [ za ] [ b ] a ) = L [ a ] [ b ] a , [ b ] a , cbc (cid:4) ba (cid:5) + [ b ] a ! and the thesis is thus proved. (cid:3) Combining Proposition 1.4 and Theorem 1.6 we obtain a recursive formula for L ( a, b, c ): Corollary 1.7.
Let a, b ∈ Z + , c ∈ Q + be such that c < a < b and a b . Then L ( a, b, c ) = && L ( a − c ) c (cid:4) ba (cid:5) + [ b ] a ' ba ' where L = L [ a ] [ b ] a , [ b ] a , cbc (cid:4) ba (cid:5) + [ b ] a ! The Algorithm
The main result of the previous section gives rise to the following algorithm to compute L ( a, b, c ) for any given triple ( a, b, c ) with a, b ∈ Z + and c ∈ Q + . Algorithm 2.1.Input :
The values a, b ∈ Z + , c ∈ Q (condition: a < b ). Output:
The value L ( a, b, c ) = min { x ∈ Z + | [ ax ] b ≤ cx } . Instructions: (1) If c ≥ a then return L ( a, b, c ) = 1, else go to step 2.(2) If a | b then return L ( a, b, c ) = ba , else go to step 3.(3) Make the positions a = [ a ] [ b ] a , b = [ b ] a and c = cbc (cid:4) ba (cid:5) + [ b ] a , then go to step 1.(4) Call L the output of step 4.(5) Compute L ( a, b, c ) = && L ( a − c ) c (cid:4) ba (cid:5) + [ b ] a ' ba ' . (6) Return L ( a, b, c ).We now show that this algorithm terminates. In fact: Proposition 2.2.
Algorithm 2.1 stops after a finite number of steps.Proof.
Consider the three sequences of integers a i , b i and c i defined recursively as b i = (cid:26) b = bb i = [ b i − ] a i − if i > a i = (cid:26) a = aa i = [ a i − ] b i if i > A. MOSCARIELLO c i = c = cc i = c i − b i − c i − j b i − a i − k + [ b i − ] a i − if i > a i +1 < a i if a i ≥ c i ≥ i ≥ a i ≤
1, hence c i ≥ a i thus meeting one condition for termination. (cid:3) Applications
The given algorithm has an application in the context of numerical semigroups. Given twocoprime integers a and a we consider the numerical semigroup generated by them, that is S = h a , a i = { λ a + λ a | λ , λ ∈ N } . We define the quotient of a numerical semigroup S by a positive integer d as follows: Sd := { x ∈ N | xd ∈ S } . The quotient Sd is a numerical semigroup, but it does not have necessarily the same structureas S : little is known about the existence of a relation between the invariants of S and Sd .In particular, given three positive integers a , a , d , it’s an open problem (cf. [5, OpenProblem 5.20]) to find a formula for the smallest multiple of d that belongs to h a , a i andfor the largest multiple of d that does not belong to h a , a i , that actually are invariantsof the quotient h a ,a i d . The class of quotients of numerical semigroups is tightly related tothe Diophantine inequalities we have studied, as it was shown that a numerical semigroupis proportionally modular if and only if is the quotient of an embedding dimension twonumerical semigroup. In particular we know that h a , a i is proportionally modular, as thenext result shows: Lemma 3.1 ([9, Lemma 18]) . Let a , a be relatively prime positive integers and let u be apositive integer such that ua ≡ a ) . Then h a , a i = { x ∈ N | [ ua x ] ( a a ) ≤ x } . By Lemma 1.2 we immediately obtain h a , a i = (cid:26) x ∈ N | [ ux ] a ≤ xa (cid:27) . Consider now the quotient h a , a i d = { x ∈ N | xd ∈ h a , a i} = (cid:26) x ∈ N | [ uxd ] a ≤ xda (cid:27) whose multiplicity is m (cid:18) h a , a i d (cid:19) = min (cid:26) x ∈ N | [ uxd ] a ≤ xda (cid:27) = L (cid:18) [ ud ] a , a , da (cid:19) and therefore it can be obtained by applying Algorithm 2.1. The second application regards the set S ( a, b, c ) itself. Since this set is a numerical semi-group, it has finite complement in N ; the greatest integer not belonging to S ( a, b, c ) is calledthe Frobenius number of S ( a, b, c ), that we will denote here with F ( a, b, c ). In [10] the au-thors give a relation between F ( a, b,
1) and the multiplicity of a particular proportionallymodular numerical semigroup. For this purpose we fix the following notation.Given p, q ∈ Q + such that p < q denote by [ p, q ] and h [ p, q ] i the sets[ p, q ] = { x ∈ Q | p ≤ x ≤ q } and h [ p, q ] i = { λ a + λ a + . . . + λ n a n | λ , . . . , λ n ∈ N , a , . . . , a n ∈ [ p, q ] n ∈ N \ { }} It is known that for any p, q ∈ Q + such that p < q the set S ([ p, q ]) = h [ p, q ] i ∩ N is aproportionally modular numerical semigroup, as the next proposition shows: Proposition 3.2 ([10, Proposition 1]) . Let a , b , a , b ∈ Z + be such that b a < b a . Then S ([ b a , b a ]) = S ( a b , b b , a b − a b ) . A direct consequence of Proposition 3.2 is that m ( S ([ b a , b a ])) = L ( a b , b b , a b − a b ).Furthermore, note that Lemma 1.2 allows us to divide each term by b , hence obtaining(4) m (cid:18) S (cid:18)(cid:20) b a , b a (cid:21)(cid:19)(cid:19) = L (cid:18) a , b , a b − a b b (cid:19) Theorem 3.3 ([10, Theorem 18]) . Let a, b ∈ Z + be such that ≤ a < b and S = S ([ b +12 ab , b − b ( a − ]) . Then F ( a, b,
1) = b − m ( S ) . By Theorem 3.3 and Eq. (4) we have F ( a, b,
1) = b − m ( S ) = b − L (cid:18) b, b + 1 , b − ab + 2 b b + 1 (cid:19) and thus we can apply Algorithm 2.1. References
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A. MOSCARIELLO (A. Moscariello)
Dipartimento di Matematica e Informatica, Universit`a di Catania, VialeAndrea Doria 6, 95125 Catania,Italy
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