proof of Erdös-Anning Theorem

Let A,B and C be three non-collinear points. For an additional point P consider the triangleMathworldPlanetmath ABP. By using the triangle inequality for the sides PB and PA we find -|AB||PB|-|PA||AB|. Likewise, for triangle BCP we get -|BC||PB|-|PC||BC|. Geometrically, this means the point P lies on two hyperbolaPlanetmathPlanetmath with A and B or B and C respectively as foci. Since all the lengths involved here are by assumptionPlanetmathPlanetmath integer there are only 2|AB|+1 possibilities for |PB|-|PA| and 2|BC|+1 possibilities for |PB|-|PC|. These hyperbola are distinct since they don’t have the same major axis. So for each pair of hyperbola we can have at most 4 points of intersectionMathworldPlanetmath and there can be no more than 4(2|AB|+1)(2|BC|+1) points satisfying the conditions.

Title proof of Erdös-Anning Theorem
Canonical name ProofOfErdosAnningTheorem
Date of creation 2013-03-22 13:19:11
Last modified on 2013-03-22 13:19:11
Owner lieven (1075)
Last modified by lieven (1075)
Numerical id 4
Author lieven (1075)
Entry type Proof
Classification msc 51-00