Recently I have found something, as bellow: Theorem “Pythagoras – 3D”(name is given by me)
The sum of the volumes of the three cubes with edges equal to the sides of a triangle, where it’s side lengths are in ratio 3 : 4 : 5 is equal to the volume of a cube with edge equal to the semi perimeter of the triangle. a³ + b³ + c³ = p³, where p = ½ (a + b + c) Furthermore: All these triangles are straight angle triangles e. g. a² + b² = c², as 3² +4² = 5² and the Pythagorean equation can be written as: 3²k + 4²k = 5²k where k is any positive number. If √k is an integer √k = 1, 2, 3 … ∞, e.g. whole number, these all are straight angle triangles e.g. Pythagorean triangles measuring their sides with the integers of the first primitive Pythagorean triple (3, 4, 5) and its generated derivates. Similarly the above equation for cubes can be written as 3³k + 4³k + 5³k = 6³k where k is any positive number. If ³√k is an integer ³√k = 1, 2, 3 … ∞, e.g. a whole number, these cubes are all set of four cubes measuring their edges with integers (whole numbers), that are representing the first primitive Pythagorean triple (3, 4, 5) and its generated non primitive derivates, where the sum of the volumes of the three small cubes is equal to the volume of the largest one and there are no more such cubes, to represent other primitive or non primitive Pythagorean triples or any other combination of four natural numbers!
If k = 1, then 3³ + 4³ + 5³ = 6³!!!
Reverse Theorem
Any cube with edge length m can be divided in three cubes with edge lengths 3/6m; 4/6m and 5/6m e. g.
m³ = (3/6m) ³ + (4/6m) ³ + (5/6m) ³
If m is an integer, divisible by 6, the three small cubes are measuring their edges also with integers (whole numbers) that can be divided by 3, 4 and 5 accordingly. If m = ³√k, then the cubes equation can be written in the same form, as above: 6³k = 3³k + 4³k + 5³k.
I have no information, as to whether some similar findings have been already published by anybody, although I have tried in Wikipedia, as well. Please comment and suggest. Best regards. Michael Ivanov 12 Oct. 2011 Michael Ivanov (talk) 13:07, 12 October 2011 (UTC)