A new proof of a theorem of Mansour and Sun

We give a new proof of a theorem of Mansour and Sun by using number theory and Rothe’s identity. It is well-known that the number of ways of choosing k points, no two consecutive, from a collection of n points arranged on a cycle is n n−k ( n−k k ) (see [11, Lemma 2.3.4]). A generalization of this result was obtained by Kaplansky [5], who proved that the number of k-subsets {x1, . . . , xk} of Zn such that |xi − xj | / ∈ {1, 2, . . . , p} (1 ≤ i < j ≤ k) is n n−pk ( n−pk k ) , where n ≥ pk + 1. Some other generalizations and related problems were studied by several authors (see [2,6,7,9]). Very recently, Mansour and Sun [8] extended Kaplansky’s result as follows. Theorem 1. Let m, p, k ≥ 1 and n ≥ mpk + 1. Then the number of k-subsets {x1, . . . , xk} of Zn such that |xi − xj | / ∈ {m, 2m, . . . , pm} for all 1 ≤ i < j ≤ k, denoted by fm,n, is given by n n−pk ( n−pk k ) . Their proof needs to establish a recurrence relation and compute the residue of a Laurent series. Mansour and Sun [8] also asked for a combinatorial proof of Theorem 1. In this note, we shall give a new but not purely combinatorial proof of Theorem 1. Let p and k be fixed throughout. Let (a, b) denote the greatest common divisor of the integers a and b. We first establish the following three lemmas. Lemma 2. Let (a,m) = 1 and let d be a positive integer. Then at least one of a, a + m, a + 2m, . . . , a + (d− 1)m is relatively prime to d. Proof. If (a, d) = 1, we are done. Now assume that (a, d) = p1 1 . . . p rs s and d = p l1 1 · · · pt t , where 1 ≤ s ≤ t and p1, . . . , pt are distinct primes and r1, . . . , rs, l1, . . . , lt ≥ 1. We claim that a + ps+1 · · · ptm is relatively prime to d. Indeed, since (a,m) = 1, we have (p1 · · · ps,m) = 1 and therefore (p1 · · · ps, a + ps+1 · · · ptm) = (p1 · · · ps, ps+1 · · · ptm) = 1, (ps+1 · · · pt, a + ps+1 · · · ptm) = (ps+1 · · · pt, a) = 1. This completes the proof. 2 Lemma 3. Let (m,n) = d. Then there exist integers a, b such that (a, n) = 1 and am + bn = d. Proof. Since (m,n) = d, we may write m = m1d and n = n1d, where (m1, n1) = 1. Then there exist integers a and b such that am1 + bn1 = 1. It is clear that (a, n1) = 1. Noticing that (a + n1t)m1 + (b−m1t)n1 = 1, by Lemma 2, we may assume that (a, d) = 1 and so (a, n) = 1. 2