Solving a ± b = 2c in elements of finite sets

摘要

We show that if A and B are finite sets of real numbers, then the number of triples (a,b,c) ∈ A × B × (A ∪ B) with a + b = 2c is at most (0.15 + o(l))(|A| + |B|)~2 as + |B| → ∞. As a corollary, if A is antisymmetric (that is, A∩(-A) = ?), then there are at most (0.3+o(1))|A|~2 triples (a, b, c) with a, b, c ∈ A and a - b = 2c. In the general case where A is not necessarily antisymmetric, we show that the number of triples (a, b, c) with a,b, c ∈ A and a - b = 2c is at most (0.5 + o(1))|A|~2. These estimates are sharp.