At the suggestion of a colleague and in reference to the extension of the Pythagorean Theorem to a cuboid, I’ve begun to explore the analogous application of an+bn+cn=dn.
Please reference prior post regarding the Harmonious Pythagorean Tetrahedra Conjecture: Pythagorean Theorem Generalization as Consequence of Three-Dimensions
Solutions to an+bn=cn for non-right triangles with mutual exponents are requisite common a:b ratios and ∠C’s. Established are an a:b of 1:2 and the assignment of a=1, b=2 and c=4 (Ref. Fig. 1). This would seem to necessitate a solid of non-orthogonal construction with a parallelogram base that can divided from opposite corners into isosceles triangles. Thus, a parallelepiped is realized. Overlooking the extreme exponents encountered with isosceles triangles, at least for now, the use of C1=cos-1(-(c12/2ab)+(a/2b)+(b/2a)) yields 75.52248781407008° which must also be designated to C2 and C3. Thereafter, from (an+bn)(1/n)=√a2+b2-(2ab x cos C), c1=2 and a1=4. This leaves d to equal 4 as well, and in this case, defines an inequality of an+bn+cn≠dn.
The isosceles nature of this example, made necessary by the ratio constraint to solve other than right triangles by the means shown, again demonstrates the exclusivity of Pythagoras’s (et al.) rule unencumbered by all but an angle. Greater depth to be sought for conformance with the analogy.
