Second Proof: Every Positive Integer is a Frobenius Number of Three Generators

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1 International Mathematical Forum, Vol., 5, no. 5, - 7 HIKARI Ltd, Second Proof: Ever Positive Integer is a Frobenius Number of Three Generators Firu Kamalov Mathematics Department Canadian Universit of Dubai Dubai, UAE Ho-Hon Leung Mathematics Department Canadian Universit of Dubai Dubai, UAE Copright c 5 Firu Kamalov and Ho-Hon Leung. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in an medium, provided the original work is properl cited. Abstract Let n be an positive integer. We give a simple proof to the theorem that there alwas exist positive integers a, b and c such that the Frobenius number generated b a, b, c is equal to n. Mathematics Subject Classification: A99, A, 4A6 Kewords: Frobenius number Introduction Let n, n,..., n k be positive integers such that gcd(n, n,..., n k ) =. Let S := S(n, n,..., n k ) be the following set of integers: S(n, n,..., n k ) = {n n = x n + x n x k n k, x, x,..., x k Z }.

2 Firu Kamalov and Ho-Hon Leung It is well-known that the set Z S is finite. The Frobenius number, F (n, n,..., n k ), is defined as the largest integer in Z S. The Conductor, N(n, n,..., n k ), is the smallest integer in S such that all integers greater than it belong to S. It is clear that N(n,..., n k ) = F (n,..., n k ) +. For k =, Slvester proved in [6] that F (n, n ) = n n n n. Equivalentl, N(n, n ) = (n )(n ). For k =, Curtis showed in [] that there is no closed formula for F (n, n, n ). Nevertheless, it would be interesting to determine properties of the Frobenius number of three generators. In particular, the authors proved in [5] that F (n, n, n ) is surjective, i.e. there exist positive integers a, b, c such that F (a, b, c) = n for ever nonnegative integer n. In [], Arnold provided a geometrical method to find the Conductors of three generators. The same method can be easil extended to find the Conductors of more than three generators, see []. In this paper, we give a simple geometric proof (in the sense of Arnold) that ever positive integer greater than can be represented as the Conductor N(a, b, c) for some positive integers a, b, c. This result can be taken as an alternative proof that the Frobenius number F (a, b, c) is a surjection onto the set of nonnegative integers since N(a, b, c) is alwas one greater than F (a, b, c). Main Results. Arnold s domain D a (b, c) Let a, b, c be positive integers such that gcd(a, b, c) =. In the first quadrant {, }, for each α =,,..., a, define the valuation function l a (, ) := b + c and define I α be the set of points I α := {(, ) l a (, ) = α mod a}. Let M α I α be the set of points such that l a (, ) is minimal. The Arnold s domain D a := D a (b, c) is the union of the sets M α over all α =,,..., a. D b and D c can be defined similarl. Let L a := L a (b, c) = max {l a (, ) (, ) D a (b, c)}. In [], it is shown that the Conductor N(a, b, c) can be computed as follows: N(a, b, c) = L a a +. () It is also shown that the domains D a, D b and D c have the form of Young diagrams: The boundar of D a in the -plane consists of a base segment along the -axis, a vertical segment along the -axis and a staircase line.

3 Ever positive integer is a Frobenius number... Example. For the domain D 5 (9, ), it is shown in the following Fig. as the region bounded b the bold line. The bold number in the position (,) in the (, )-plane corresponds to L 5 (9, ) = ()(9) + ()() = 4. Hence, b (), N(5, 9, ) = = 7. () (88) (66) (44) 4 () 4 () 4 () (9) (8) (57) (76) (95) Figure : The domain D 5 (9, ).. Main proofs In Lemma., Lemma. and Lemma.4. we deal with the cases for n modulo, and 4 respectivel. Lemma. If n is an even number, then there exist a, b, c such that N(a, b, c) = n. Proof. If a = b, then N(a, a, c) = N(a, c) = (a )(c ) b Slvester s formula in [6]. Let n = k where k is an positive integer. Set a = b = and c = k +, then we get N(,, k + ) = n b (). Lemma. If n > and n mod, then there exist a, b, c such that N(a, b, c) = n. Proof. If n =, then it is done b Lemma.. If n = k + where k is an positive integer, then set a =, b = k + and c = k + 4. The domain D (k +, k + 4) is shown in Fig.. L (k +, k + 4) = c = k + 4 and hence b (), N(, k +, k + 4) = (k + 4) + = n.

4 4 Firu Kamalov and Ho-Hon Leung If n = k where k is an positive integer, then set a =, b = k + and c = k+. The domain D (k+, k+) is shown in Fig.. L (k+, k+) = c = k + and hence b (), N(, k +, k + ) = (k + ) + = n. Hence the result. Figure : The domain D (k +, k + 4). Figure : The domain D (k +, k + ). Lemma.4 If n > and n mod 4, then there exist a, b, c such that N(a, b, c) = n. Proof. The cases when n = or mod 4 are done b Lemma.. Let n = 4k + where k is an positive integer. Set a = 4 and b = k +. Let c be an integer such that b + c = mod 4 and b > c > b. The domain D 4 (b, c) is shown in Fig.4 for the case b = mod 4 and c = mod 4. (The domain for the case b = mod 4 and c = mod 4 is ver similar.) Since b > c, L 4 (b, c) = b = (k + ) = 4k + 6. B (), N(4, b, c) = (4k + 6) 4 + = 4k + = n. Hence the result. A combination of Lemma., Lemma. and Lemma.4 gives us the following corollar:

5 Ever positive integer is a Frobenius number... 5 Figure 4: The domain D 4 (b, c) in Lemma.4. Corollar.5 If n > and n mod, then there exist a, b, c such that N(a, b, c) = n. The following lemma will be used inductivel in the proof of Theorem.8. Lemma.6 Let p be a prime. If n > p(p )(p ) and n mod p, then there exist b, c such that N(p, b, c) = n. Proof. For such n, set c = n + (p ). It immediatel implies that c > (p ) and c mod p. We need to choose b such that b + c = mod p and (p )b < c < (p )b. There alwas exist b satisfing these two conditions if the following inequalit holds: c p c p > p. It is equivalent to c > p(p )(p ). It must hold as we fix c such that c > (p ). For such a pair of b, c, the domain D p (b, c) has a L-shaped diagram which covers the positions (, ), (, ), (, ), (, ),..., (p, ) in the, -plane and L p (b, c) = c. (For example, see Fig.5 for the case D p (, 7, ).) Now, b (), N(p, b, c) = c p + = n. Hence the result. The following lemma describes certain arithmetic properties of prime numbers. It will be used in the inductive argument of Theorem.8. Lemma.7 Let p i be the i th prime number. For n, let R(n) and Q(n) be the following statements: Then R(n) > Q(n) for all n 5. R(n) = + p p...p n Q(n) = p n+ (p n+ )(p n+ ).

6 6 Firu Kamalov and Ho-Hon Leung (4) () 9 () () (7) (4) (5) (684) (855)(6)(97)(68)(59)(7)(88)(5) Figure 5: In D (7, ), L (7, ) =. Proof. For n, define R (n) and Q (n) b the following statements: R (n) = p p...p n Q (n) = p n+. It is clear that R(n) > R (n) and Q (n) > Q(n) for all n. So we onl need to prove that R (n) > Q (n) for n 5. It is obvious that R (n) > Q (n) for n = 5, 6, 7, 8 since R (5) =, R (6) =, R (7) = 55, R (8) = and Q (5) = 97, Q (6) = 49, Q (7) = 6859, Q (8) = 67 respectivel. In [4], Nagura showed that < p n+ p n <., for n > 9. Hence, for n 9, the following inequalities are true: R (n + ) R (n) = p n+ > >.78 = (.) > ( p n+ ) = Q (n + ). () p n+ Q (n) Since R (9) = 987 > 489 = Q (9), we have R (n) > Q (n) for all n 9 b (). Hence the result. Finall, we will prove the main theorem of this article: Theorem.8 For an positive integer n greater than, there alwas exist a, b, c such that N(a, b, c) = n. Proof. A straightforward check shows that N(5, 8, 9) =, N(5,, 8) = 5, N(5,, 9) = 7 (shown in Example. and Fig.) and N(5,, ) = 49. Hence, b Corollar.5 and an application of Lemma.6 to the case p = 5, it remains to show that there alwas exist a, b, c such that N(a, b, c) = n where n > and n = mod 6.

7 Ever positive integer is a Frobenius number... 7 It can be shown that N(7, 7, ) = 6, N(7, 8, 5) = and N(7, 58, 7) = 8. B an application of Lemma.6 to the case p = 7, it remains to check the cases when n > and n = mod 4. It can be shown that N(, 9, ) = 4 and N(, 6, 5) = 84. B an application of.6 to the case p =, it remains to check the cases when n > and n = mod 46. B the notations of Lemma.7, these n s are greater than or equal to R(5) and hence great than Q(5) = ()()(). But b Lemma.6 when p 6 =, it remains to check the cases when n is greater than or equal to R(6) and n = mod p p p p 4 p 5 p 6. But then these n s are all greater than Q(6) b Lemma.7 and we can appl Lemma.6 again for p 7 = 7 to reduce the proof to checking the cases when n is great than or equal to R(7) and n = mod p p p p 4 p 5 p 6 p 7. As these are all greater than Q(7) b Lemma.7 and the same argument based on using Lemma.6 for the next prime numbers just as before can be applied inductivel. The proof is complete for all positive integers n greater than. References [] Aicardi, F., On the geometr of the Frobenius problem, Funct. Analsis and Other Math, (9), no. -4, [] Arnold, V. I., Geometr of continued fractions associated with Frobenius numbers, Funct. Analsis and Other Math, (9), no. -4, [] Curtis, F., On formulas for the Frobenius number of a numerical semigroup, Math. Scand, 67 (99), 9-9. [4] Nagura, J., On the interval containing at least one prime number, Proc. Japan Acad, 8 (95), no. 4, [5] Rosales, J. C., García-Sánche P. A. and Garćia-García J. I., Ever positive integer is the Frobenius number of a numerical semigroup with three generators, Math. Scand., 94 (4), 5-. [6] Slvester, J. J., Mathematical questions with their solutions, Educational Times 4 (884),. Received: March, 5; Published: April 4, 5