Recent content by Terry Coates

120, 60,30 cos 120 = -0.5, cos 60 = 0.5, cos 30 = 0.866

Thanks. Rather a silly question. It came about because I have been comparing what applies with a power higher than 2, to what applies with a modular power 2 then for instance, you get irrational sides to a triangle. also you get an oval curve instead of an ellipse. Nearest to an ellipse with...

n = 3,x = 1, y = 10 z = (10^3 +9^3)^(1/3 = (1000 + 1729)^1/3 Cos (Angle xy) = (x^2 +y^2-z^2)/(2 x.y) n = 2,x= 3, y=4 z = (3^2 +4^2)^0.5 = 5 Cos (Angle xy) = (3^2 +4^2 -5^2)/(2.3.4) = cos (0) = 1

See edited version

It might be argued that anything can happen at infinity, but here is a specific case where it is shown to be true.

Is what follows a possible simple proof of FLT? If you calculate the irrational values of Y = (1- X^p)(1/p). with p > 2 it is found that in a triangle with unit base and sides X and Y the angle between X and Y is a minimum with X = Y and is between 60 and 90 degrees, approaching 60 as p...

Error in above chart . should be a=e^(2/3.pi^2) b = e^2.pi()

In my original contribution I wondered how k would vary with p I have now found a disconnected curve that answers this, and it results in zero N for all p greater than 2. I'll try to insert this curve. k = y and p = x

N is the total number of pairs with a maximum sum A, not only those adding up to A. I will see if using Euler's totient function can get my count programme to run much faster and then go to say A = 10^7. It might be worth looking at the linear case as a flattened triangle.

I am only counting the primitive pairs, i.e co-prime pairs. So up to A = 8 there are for a = 3,4,5,6,7,8 there are 1,1,2,1,3,2 pairs respectively making a total N = 10

I wonder what K would be in the linear case and much higher value of A? It took me about 4 days to count N with A = 10^ 6 so I'm not going any further. Apparently Euler's totient function is related to my work. An interesting calculation of 1/2.pi *3/pi =1.5/pi^2 = 0.151982 is very close to my...

Thanks. I made an error in N = 318 at A = 1000, it should be 158 (actually from 975 to 1008) Not intended to be a proof, just a way of describing the theorem. My counts of N are based on b > c, so would be doubled if they are reversed.

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...

In the factorisation of A^3 +/- B^3 = (A+/B)(A^2-/+A.B + B^2) the larger factor is the solution for finding the third side of a triangle with sides A and B with the angle between them of 60/120 degrees.(Cosine rule) For any set of three different numbers X>Y>Z there would be three...

Note if you omit (B^n + C^n) MOD A = 0 and put n = 2 there are all the Pythaqorean triples that satisfy all the other five statements. Since no replies so far to show that there are other ways with n odd (and greater than one), it looks as if there are none, in which case here is a simple proof...