- #1
Terry Coates
- 39
- 2
Since it is known than the number N of primitive Pythagorean triples up to a given hypotenuse length A is given on average by N = Int(A/(2.pi)) and according to my calculations with primitive triples and A = B + C, I get on average N = Int(0.152 A^2) (for A = 10^3 I get N = 152,095 compare with N = 319 with pythagoras, and for 10^6 I get N = 151,981,776,195), would it be logical to conclude that for powers p then N =Int( K.A^(3-p))? Not sure how K would vary with p, probably not much variation seeing that 0.152 in not much different from 1/(2.pi). Anyway K must be less than one, which leads to N = 0 for all powers , higher than 2, as per Fermat.