The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

The search session has expired. Please query the service again.

Currently displaying 1 – 1 of 1

Showing per page

Order by Relevance | Title | Year of publication

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

Vsevolod F. LevRom Pinchasi — 2014

Acta Arithmetica

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...

Page 1

Download Results (CSV)