Partitions of Natural Numbers with the Intersection Not Empty

Let be the set of nonnegative integers. For a given set the representation functions , are defined as the number of solutions of the equation with condition , , respectively. In this paper, we prove that if and , then cannot hold for all sufficiently large integers where . 1. Introduction Let be the set of nonnegative integers. For , let , , denote the number of solutions of respectively. Sárk？zy asked ever whether there exist two sets and of nonnegative integers with infinite symmetric difference, that is: for . As Dombi [1] has shown, the answer is negative for by the simple observation that is odd if and only if for some . For , Dombi presented a partition of the set of all positive integers into two disjoint subsets and such that for all sufficiently large . For , Chen and Wang [2] constructed sets such that for all integers , where is the set of all those positive integers such that the number of zeroes in the binary representation of is even, and is the set of all those positive integers such that the number of zeroes in the binary representation of is odd. For the other related results, the reader is referred to see [3–7]. It is natural to ask the following: for , are there subsets with for all large enough integers such that and ? Recently, the authors of this paper [8] obtained some results in this direction. In this paper, we obtain the following results. Theorem 1. If and , then cannot hold for all sufficiently large integers . Theorem 2. If and , then cannot hold for all sufficiently large integers . Remark 3. For , similarly, we can prove that if and , then cannot hold for all sufficiently large integers , where . 2. Proof of Theorem 1 Proof. Suppose that there exist integer and sets with and such that for all . Without loss of generality, we may assume that , is a positive integer; then there exists a polynomial of degree at most such that Let Then we have Moreover, Therefore, Write By (3) and (7), we have Writing both sides of (9) in power series, then we have Comparing the coefficients of on the both sides of (10), we have Then we have Since , we have . Then, by (12), has the following 30 cases: By (13), we have or Thus has the following 10 cases: By (14) and (25), we have which has the following 8 cases: By (15) and (26), we have which has the following 6 cases: By (16) and (27), we have which has the following 6 cases: By (17) and (28), we have which has the following 2 cases: By (18) and (29), we have which has the following 2 cases: By (19) and (30), we have which has the following 2 cases: By (20) and (31), we have which has the