Solving a ± b = 2c in elements of finite sets
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(1))(|A|+|B|)² as |A| + |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...