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 are sharp.
@article{bwmeta1.element.bwnjournal-article-doi-10_4064-aa163-2-3,
author = {Vsevolod F. Lev and Rom Pinchasi},
title = {Solving a $\pm$ b = 2c in elements of finite sets},
journal = {Acta Arithmetica},
volume = {166},
year = {2014},
pages = {127-140},
zbl = {1302.11081},
language = {en},
url = {http://dml.mathdoc.fr/item/bwmeta1.element.bwnjournal-article-doi-10_4064-aa163-2-3}
}
Vsevolod F. Lev; Rom Pinchasi. Solving a ± b = 2c in elements of finite sets. Acta Arithmetica, Tome 166 (2014) pp. 127-140. http://gdmltest.u-ga.fr/item/bwmeta1.element.bwnjournal-article-doi-10_4064-aa163-2-3/